Improper Integrals
Improper Integrals.
The concept of Riemann integrals as developed in previous chapter requires that the range of integration is finite and the integrand remains bounded on that domain. if either (or both) of these assumptions is not satisfied it is necessary to attach a new interpretation to the integral Definition 3.1. In case the integrand f becomes infinite in the interval a ≤ 푥 ≤ 푏, That is f has points of infinite discontinuity (singular points) in [a, b] or the limits of integration a or b (or both becomes infinite, the 푏 symbol ( ) ∫푎 푓푑푥 is called an improper or infinite or generalised integral. Thus, ∞ 푑푥 ∞ 푑푥 1 푑푥 ∞ 푑푥 ∫ , ∫ . ∫ , ∫ 1 푥2 −∞ 1+푥2 0 푥(1−푥) −1 푥(1−푥) are examples of improper integrals. The integrals which are not improper are called proper integral , thus 1 푠푖푛푥 ∫ 푑푥 is a proper integral. 0 푥 Integration of Unbounded Function with finite limits of integration. Definition 3.2. Let a function f be defined in a interval [a, b] everywhere except possible at finite number of points.
(i) Convergence at left –end. Let a be the only points of infinite
discontinuity of f so that according to assumption made in the
last section, the integral
푏 exists ∫푎+휆 푓푑푥 ∀ 휆, 0 < 휆 < 푏 − 푎. 2
푏 is defined as the 푇ℎ푒 푖푚푝푟표푝푒푟 푖푛푡푒푔푟푎푙 ∫푎 푓푑푥 푏 푙푖푚 ∫ 푓푑푥 so that, 휆→0+ 푎+휆 푏 푏 ∫ 푓푑푥 = 푙푖푚 ∫ 푓푑푥 . 푎 휆→0 푎+휆 푏 Limit exists and is finite, the improper integral is said If this ∫푎 푓푑푥 to converge at (a) if otherwise, it is called divergent. 퐍퐨퐭퐞. For any c , 푎 < 푐 < 푏 푏 푐 푏 + . ∫푎 푓푑푥 = ∫푎 푓푑푥 ∫푐 푓푑푥 Then, 푏 푐
∫푎 푓푑푥 푎푛푑 ∫푎 푓푑푥 푏 converges and diverges together and is proper. ∫푐 푓푑푥
(ii) Convergence at right-end. Let b be the only point of infinite
discontinuity the improper integral is then defined by the relation
푏 푏−휇 ∫ 푓푑푥 = lim ∫ 푓푑푥 0< 휇 < 푏 − 푎 . 푎 휇→0+ 푎 If the limit exists, the improper integral is said to be convergent at 푏. Otherwise is called divergent. 퐍퐨퐭퐞 ∶ For the same reason as above, 푏 푏 and converges and diverges together . ∫푐 푓푑푥 ∫푎 푓푑푥 ∀ 푐, 푎 < 푐 < 푏
(iii) Convergence at both the end points. If the end points a and
b are the only points of infinite discontinuity of f , then for any
point c, a < c < b,
푏 푐 푏
∫푎 푓푑푥 = ∫푎 푓푑푥 + ∫푐 푓푑푥
If both the integrals are convergent as by case (i) and (ii),then 3
푏 is convergent , otherwise it is divergent. The improper integral ∫푎 푓푑푥
is also defined as:
푏 푏−휇 ∫ 푓푑푥 = 푙푖푚 ∫ 푓푑푥. 푎 휆→0+ 푎+휆 휇→0+
The improper integral exists if the limit exists.
(iv) Convergence at Interior points. If an interior point c,
a < c < b, is the only point of infinite discontinuity of f, we get
푏 푐 푏 (1) ∫푎 푓푑푥 = ∫푎 푓푑푥 + ∫푐 푓푑푥
푏 exists of the both integral on R.H.S of the improper integral ∫푎 푓푑푥
(1) are exists.
Example 3.1. Examine the convergence of:
1 푑푥 1 푑푥 2 푑푥 (푖) ∫ (ii) ∫ (iii) ∫ . 0 푥2 0 √푖−푥 6 2푥−푥2
(i) 0 is the point of infinite discontinuity of integrand [0, 1].
Thus,
1 푑푥 1 푑푥 ∫ = 푙푖푚 ∫ , o < 휆 < 1 0 푥2 휆→0+ 휆 푥2 1 = 푙푖푚 ( − 1) = ∞ 휆→0+ 휆
Thus the proper integral is divergent.
(ii) Home Assignment. Home Assignment
4
풃 Comparison Tests for Convergence At ‘a’ of . ∫풂 풇풅풙
Theorem 3.1. A necessary and sufficient condition for the 푏 convergence of the improper integral at ‘a’ where f is positive in ∫푎 푓푑푥 [a, b]. This is , ∃ a psoitive number 푀, independent of 휆 , such that
푏 < M, 0< . ∫푎+휆 푓푑푥 휆 < 푏 − 푎
푏 Proof. We know that the improper integral ‘a’ if for 0 ∫푎 푓푑푥 converges at
푏 0 < tends to finite limit as + 휆 < 푏 − 푎, ∫푎+휆 푓푑푥 휆 → 0 .
푏 Since f is positive in [ is 푎 + 휆, 푏], the positive function of 휆 , ∫푎+휆 푓푑푥 monotonic incereasing as λ , decreases and will therefore tend to a finite limit iff
it is bounded above , This is, ∃ a positive number 푀 independent of 휆, such that
푏 < M, 0 . ∫푎+휆 푓푑푥 < 휆 < 푏 − 푎
Hence the theorem is proved.
Note. If no such number M exists, the monotonic increasing function 푏 is not bounded above and therefore tend to + + and ∫푎+휆 푓푑푥 ∞ as λ → 0 , 푏 hence the improper integral diverges to + ∫푎 푓푑푥 ∞.
Comparison Test.
Theorem 3.2: If f and g are two positive functions and ‘a’ is only singular point of f and g on [a. b], such that
f(x) ≤ 푔(푥), 푓표푟 푎푙푙 푥 ∈ [푎, 푏]
푏 푏 ( ) converges . i ∫푎 푓푑푥 converges, if ∫푎 푔푑푥 5
푏 푏 diverges, if converges . (ii) ∫푎 푔푑푥 ∫푎 푓푑푥
Proof. Since f and g are two positive functions on [a, b] and ‘a’ is only singular point of f and g. Therefore 푓 and 푔 are bound in [푎 + 휆, 푏], for all 0 < 휆 < 푏 − 푎.
Also Since, f(x) ≤ 푔(푥), 푓표푟 푎푙푙 푥 ∈ [푎, 푏], implies
푏 푏 (i) ∫푎+휆 푓푑푥 ≤ ∫푎+휆 푔푑푥
푏 (1) Suppose be convergent, so that for all ∫푎 푔푑푥 ∃ 푚 > 0 , 푠푢푐ℎ 푡ℎ푎푡 휆, 0 < 휆 < 푏 − 푎,
푏 < m . ∫푎+휆 푓푑푥
From (i) we have
푏 < m , for all ∫푎+휆 푓푑푥 휆 , 0 < 휆 < 푏 − 푎 .
푏 Hence is convergent. ∫푎 푓푑푥
푏 (2) Now suppose is divergent then the positive function ∫푎 푓푑푥 푏 is not bounded above. ∫푎+휆 푓푑푥
푏 Therefore from (i) it follows that the positive function is not ∫푎+휆 푔푑푥
bounded above.
푏 Hence is divergent. This completes the Theorem. ∫푎+휆 푔푑푥
Comparison Test (limit form).
Theorem 3.3. If f and g are two positive functions [a, b] and ‘a’ is
the only singular point of f and g in [a, b], such that
푓(푥) 푙푖푚 = l where ‘l’ is a non – zero finite number. 푥→푎+ 푔(푥) 6
푏 푏 Then, the two integrals converges and diverges ∫푎 푓푑푥 and ∫푎 푔푑푥
together at ‘a’.
Proof. Evidently, 1 > 0. Let 휀 be positive number such that 1 − 휀 > 0 .
푓(푥) Since, 푙푖푚 = 1. 푥→푎+ 푔(푥)
Therefore there exists a nbd of ] a, c [ , a < c < b , such that for all
x ∈ ]푎, 푐[
푓(푥) | − 푙| < 휀 푔(푥) or (푙 − 휀)푔(푥) < 푓(푥) < (푙 + 휀)푔(푥) .
This implies that
(푙 − 휀)푔(푥) < 푓(푥) (2)
and
푓(푥) < (푙 + 휀)푔(푥) (3) ∀ 푥 ∈]푎, 푐[
푏 If converges, then from (i) ∫푎 푓푑푥
푏 also converges at . ∫푎 푔(푥)푑푥 푎
푏 If t (ii) ∫푎 푓푑푥 diverges , hen from
푏 . ∫푎 푓푑푥 diverges at 푎
푓(푥) 푏 If in the above Theorem, 퐿푖푚 → 0 and ∫ 푔푑푥 푐표푛푣푒푟푔푒푠, then 푥→푎+ 푔(푥) 푎 7
푏 converges and if ∫푎 푓푑푥
푓(푥) 푏 푏 퐿푖푚 → ∞ and ∫ 푔푑푥 diverges, then, ∫ 푓푑푥 also diverges. 푥→푎+ 푔(푥) 푎 푎
Useful Comparison Integral.
푏 푑푥 Theorem 3.4. The improper integral ∫ Converges if and only 푎 (푥−푎)푛
if n < 1.
Proof. It is proper integral if n ≤ 0 and improper for all other values
of n, ‘푎’ being only singular point of the integrand.
Now for n ≠ 1
푏 푑푥 푏 푑푥 ∫ = 푙푖푚 ∫ , 푎 (푥−푎)푛 휆→0+ 푎+휆 (푥−푎)푛
1 = 푙푖푚 [(푏 − 푎)−푛+1-휆−푛+1] 휆→0+ −푛+1
1 [(푏 − 푎)−푛+1, 푖푓 푛 < 1 −푛+1 = { ∞ 푖푓 푛 > 1 .
Also for n = 1
푏 푑푥 퐿푖푚 ∫ = 푙푖푚 [푙표푔(푏 − 푎) − 푙표푔휆] = ∞. 휆→0+ 푎+휆 푥−푎 휆→0+
푏 푑푥 Thus, ∫ converges for n < 1 . 푎 (푥−푎)푛
푏 푑푥 Note. A similar result holds for convergence of ∫ 푎푡 푏. 푎 (푏−푥)
Example 3.2. Test the convergence of
1 푑푥 휋/2 푠푖푛푥 (i) ∫ (ii) ∫ dx 0 √1−푥3 0 푥푝
Solution. Let f(x) = 1 √1−푥3 8
= 1 √(1−푥)(1+푥+푥2)
1 1 = 1 . 1 (1+푥+푥2)2 (1−푥)2
1 Clearly, 1 is a bounded function. (1+푥+푥2)2
Let M be its upper bound, then ,
1 1 푀 1 . 1 ≤ 1 , x ∈ [1, 0]. (1+푥+푥2)2 (1−푥)2 (1−푥)2
1 푚푑푥 1 Also since ∫ is convergent as n = < 1. 0 1 2 (1−푥)2
1 Therefore, is convergent. √1−푥3
(i) For 푝 ≤ 1, it is a proper integral for 푝 > 1 , it is an improper integral 0 being the point of infinite discontinuity
푠푖푛푥 1 푠푖푛푥 Now = ( ) 푥푝 푥푝−1 푥
푠푖푛푥 푠푖푛푥 The function is bounded and ≤ 1. 푥 푥
푠푖푛푥 1 Therefore, ≤ 푥푝 푥푝−1
휋/2 푑푥 Also ∫ converges only if 푝 − 1 < 1 or 푝 < 2 . 0 푥푝−1
휋 푠푖푛푥 Therefore by comparison test ∫2 dx converges for p < 2 and 0 푥푝 diverges for p ≥ 2.
Note. If 푙푖푚 [(푥 − 푎)푛푓 (푥)] exists and is non- zero finite, then, the 푥→0+ 푏 integral ∫푎 푓푑푥 convergs iff 푛 < 1.
Example 3.3. Find the values of 푚 and 푛 for which the following integrals converges. 9
1 (i) −푚푥 푛dx . ∫0 푒 푥 1 1 (ii) ∫ (푙표푔 )푚dx . 0 푥
Solution (i) Let 푘 be positive number greater than 1,
Then, 푒−푚푥푥푛 ≤ k푥푛, ∀ x ∈ [0 ,1] and m ;
1 1 푑푥 Also ∫ 푥푛 = ∫ converges for – 푛 < 1 , 0 0 푥−푛 that is, n > -1 only . 1 Thus, −푚푥 푛 converges only for n >-1 and m. ∫0 푒 푥 푑푥 ∀ 1 (iii) Let(푥) = (푙표푔 )푚 converges at x= 0 and 푥 1 1 ∫2(푙표푔 )푚 푑푥 is proper integral if 푚 ≤ 0. 퐴푙푠표‘0’ is the only singular 0 푥 point if m > 0.
For m > 0,
1 Take 푔(푥) = , 0 < p < 1, so that 푥푝
푓(푥) 1 푙푖푚 = 푙푖푚 푥푝(푙표푔 )푚 푥→0+ 푔(푥) 푥→0+ 푥
= 0, for 0 < 푝 < 1.
1 1 Therefore, ∫2(푙표푔 )푚 푑푥 converges for all m. 0 푥
Convergence at x=1
1 1 푚 ∫1 (푙표푔 ) is proper integral for 푚 ≥ 0 and ‘1’ is singular point , if m< 2 푥 0.
1 푚 1 푙표푔 퐹표푟 푚 < 0, take 푔(푥) = , so that 푙푖푚 [ 푥] = 1. (1−푥)−푚 푥→1 1−푥
1 Since ∫1 푔푑푥 converges for – 푚 < 1 that is for 푚 > −1. 2 10
1 1 Thus, ∫ (푙표푔 )푚dx converges for > −1 . 0 푥
1 1 Hence ∫ (푙표푔 )푚dx converges for 0 > 푚 > −1. 0 푥
1 푙표푔푥 Example 3.4. Show that (1) ∫ dx is convergent. 0 √푥
2 푥 (2) ∫ √ 푑푥 is divergent. 1 푙표푔푥
Solution. (1) Since 푙표푔푥 is negative on [0, 1]. √푥
Therefore we take f(x) = -푙표푔푥 √푥
푙표푔푥−1 푙표푔 1/푥 = = , √푥 √푥
‘0’ is the only singular point.
Let
1 3 푔(푥) = , n = < 1 3 4 푥4
We have
푓(푥) 1 1 푙푖푚 = 푙푖푚 푥4 푙표푔 = 0 푥→0 푔(푥) 푥→0 푥
1 Since ( ) converges. ∫0 푔 푥 푑푥
1 1 푙표푔푥 Therefore, ∫ 푓(푥)푑푥 converges implies that ∫ 푑푥 converges. 0 0 √푥
2 √푥 (2) Let f(x) = ∫ 푑푥 . 1 푙표푔푥
Here x=1 is only singular point.
1 Take 푔(푥) = , then 푥−1
푓(푥) (푥−1) 푥 푙푖푚 = 푙푖푚 √ 푥→1 푔(푥) 푥→1 푙표푔푥 11
3 1 푥2−푥2 = 푙푖푚 푥→1 푙표푔 푥
1 1 3 1 − 푥2− 푥 2 2 2 = 푙푖푚 1 푥→1 푥
3 3 1 1 = 푙푖푚 푥2 − 푥2 푥→1 2 2
3 1 = − = 1(≠ 0). 2 2
2 2 Thus, and behave same. ∫1 푓푑푥 ∫1 푔푑푥
2 Since is divergent. ∫1 푔푑푥
2 Hence is divergent. ∫1 푓푑푥
휋 푠푖푛푚푥 Example 3.5. Show that ∫2 ( ) 푑푥 푒푥푖푠푡푠 푖푓푓 푛 < 푚 + 1 0 푥푛
푚 Solution. Let f(x) = (푠푖푛 푥) 푥푛
푠푖푛푥 1 = ( )푚. 푥 푥푛−푚
Here as 푥 → 표+, 푓(푥) → 0 if 푛 − 푚 < 0, and f (x)→∞ if 푛 − 푚 > 0 .
Thus it is proper integral if 푛 ≤ 푚 and improper if 푛 > 푚.
‘0’ being the only point of infinite discontinuity.
When 푚 > 푛 ,
1 Let 푔(푥) = , so that 푥푛−푚
푓(푥) 푠푖푛푥 푙푖푚 = 푙푖푚( )푚 = 1. 푥→0 푔(푥) 푥→0 푥
휋 휋 1 Also, ∫2 푔푑푥 =∫2 푑푥 converges, Iff n−m < 1. 0 0 푥푛−푚
That is, n < m + 1 . 12
휋 푠푖푛푚푥 Therefore ∫2 ( ) 푑푥 also converges iff 푛 < 푚 + 1 . 0 푥푛
1 Example 3.6. (Beta Function). Show that 푚−1 푛−1 ∫0 푥 (1 − 푥) 푑푥 exists iff 푚, 푛 are both positive.
Proof. It is a proper integral for 푚 ≥ 1, 푛 ≥ 1 , 0 and 1 are the only points of infinite discontinuity; 0 when m < 1 and 1.
When n < 1, we have
1 1 푚−1 푛−1 푚−1 푛−1 1 푚−1 푛−1 2 1 ∫0 푥 (1 − 푥) 푑푥 = ∫0 푥 (1 − 푥) 푑푥 + ∫ 푥 (1 − 푥) 푑푥 2
convergence at ‘0’ , when m < 1.
Let f(x) = 푥푚−1(1 − 푥)푛−1
( )푛−1 = 1−푥 . 푥1−푚
1 Take 푔(푥) = , 푥1−푚
푓(푥) Then 퐿푖푚 = 1 . 푥→0 푔(푥)
1 Since 2 converges if and only if, 1-m < 1 or m> 0. ∫0 푔푑푥 1 Thus, 2 푚−1 푛−1 converges for m > 0. ∫0 푥 (1 − 푥) 푑푥
Convergence at x=1 ,
When n< 1,
Let f(x) = 푥푚−1(1 − 푥)푛−1
( )푚−1 = 1−푥 푥1−푛
1 Take 푔(푥) = , then 푥1−푛 13
푓(푥) 퐿푖푚 = 1. 푥→1 푔(푥)
1 1 1 1 1 Also, ∫ 푔푑푥 = ∫ 1−푛dx converges if and only if 1−n < 1 or n > 0. 2 2 (1−푥)
1 푚−1 푛−1 Thus, ∫1 푥 (1 − 푥) 푑푥 converges if n > 0. 2
1 Hence 푚−1 푛−1 converges if m > 0 , n > 0 . ∫0 푥 (1 − 푥) 푑푥
Example 3.7. For what values of m and n is the integral
1 푚−1 푛−1 convergent. ∫0 푥 (1 − 푥) 푙표푔푥 푑푥
Solution . The integrand is negative in [0, 1], therefore we shall test for the convergence of
1 푚−1 푛−1 ∫0 −푥 (1 − 푥) 푙표푔푥푑푥
1 1 = ∫ 푥푚−1(1 − 푥) 푙표푔 푑푥 0 푥
Since 0 and 1 are only possible singular points of integrand. We have
1 1 ∫ 푥푚−1(1 − 푥)푛−1 푙표푔 푑푥 0 푥
1 푚−1 푛−1 1 1 푚−1 푛−1 1 = 2 + 1 . ∫0 푥 (1 − 푥) 푙표푔 푑푥 ∫ 푥 (1 − 푥) 푙표푔 푑푥 푥 2 푥
Convergence at 0.
It is proper integral for m-1 > 0 and improper for m ≤1.‘0’ being the only point of infinite discontinuity.
Then, for m ≤ 1
1 Let f(x) = 푥푚−1(1 − 푥)푛−1 푙표푔 푥 14
1 = (1 − 푥)푛−1 푙표푔 푥 푥1−푚
1 Take 푔(푥) = 푥푝
푓(푥) 1 Also , 푙푖푚 = 푙푖푚 푥푝+푚−1(1-x)푛−1 푙표푔 푥→0+ 푔(푥) 푥→0+ 푥
= 0
If p + m-1 > or m > 1- p .
1 1 Also ∫2 dx converges for 1- p > 0. 0 푥푝
Thus
1 1 ∫2 푥푚−1(1 − 푥)푛−1 푙표푔 푑푥 converges for 푚 > 1 − 푝 > 0. 0 푥
converges at x=1
For n < 0 ,
1 Let f(x) = 푥푚−1(1-x)푛−1 푙표푔 푥
1 푥푚−1 푙표푔 = 푥 (1−푥)−푛+1
1 Take 푔(푥) = . (1−푥)푞
1 Therefore ∫1 푔(푥) converges for q-1< 0. 2
푚−1 1 푓(푥) 푥 푙표푔 Also 푙푖푚 = 푙푖푚 푥 = l 푥→1_ 푔(푥) 푥→1_ (1−푥)1−푛−푞
where l is infinite if 1-n-q≤1.
That is if n ≥ -q > -1.
1 Thus, ∫1 푓푑푥 푐표푛푣푒푟푔푒푠 푖푓 푛 > −1. 2 15
Hence the given integral is convergent when m > 0, n >-1.
휋 Example 3.8. Show that 2 converges and also evaluate it. ∫0 푙표푔 푠푖푛푥푑푥
Solution. Let f(x) = log sinx , then f is negative in [0,π/2] .
Therefore we consider –f instead of f.
Clearly ‘0’ is only point of infinite discontinuity.
Let g(x) = 1 , m < 1, 푥푚
Then,
−푓(푥) 푙푖푚 = 푙푖푚 − 푥푚logsinx = 0, m < 1 푥→0+ 푔(푥) 푥→0+
휋 1 Since ∫2 dx converges for m< 1, thus 0 푥푚
휋 2 converges. ∫0 푙표푔 푠푖푛푥푑푥
휋 Let I = 2 . ∫0 푙표푔 푠푖푛푥푑푥
We know that, sin2x = 2sinxcosx.
Therefore, logsin2x = log2 + logsinx + logcosx.
This implies that
휋 휋 휋 휋 2 2 2 2 ∫0 푙표푔 푠푖푛2푥푑푥 = ∫0 푙표푔 2푑푥 + ∫0 푙표푔 푠푖푛푥푑푥 + ∫0 푙표푔 푐표푠푥푑푥
휋 휋 = log2+I+∫2 푙표푔 푐표푠푥푑푥 2 0 Put 2x = t. 휋 In the Ist integral and x = − 푦 in the last integral, therefoe we get 2 1 휋 휋 0 log2 + I + 휋 ∫0 푙표푔 푠푖푛푡푑푡 = ∫ 푙표푔푠푖푛푦(−푑푦) 2 2 2 휋 휋 = 푙표푔2 + 퐼 + ∫2 푙표푔 푠푖푛푥푑푥 . 2 0 16
휋 1 휋 휋 2 휋 ∫0 푙표푔 푠푖푛푥푑푥 + ∫ 푙표푔 푠푖푛푥푑푥 = 푙표푔2 + 2퐼 . 2 2 2
휋 1 휋 휋 This implies [퐼 + ∫2 푙표푔 푠푖푛 (푦 + ) 푑푦] = 푙표푔2 + 2퐼. 2 0 2 2
휋 1 휋 Thus, [퐼 + ∫2 푙표푔 푐표푠푥푑푥 ] = 푙표푔2 + 2퐼 2 0 2
1 휋 = [퐼 + 퐼] = 푙표푔2 + 2퐼 2 2
휋 = 푙표푔2 + 2퐼 2
휋 ℎℎℎℎℎ 퐼 = log2 + 2I 2
휋 This Implies that 퐼 = − 푙표푔2 . 2
휋 휋 휋 Hence ∫2 푙표푔 푠푖푛푥푑푥 = ∫2 푙표푔 푐표푠푥푑푥 = − 푙표푔2. 0 0 2
Exercises.
1 푑푥 1 푥푛 (1)∫ dx (2) ∫ dx 0 1−푥 0 1+푥 1 (3) 푠푖푛푥 dx ∫0 3 푥2 3 푥2+1 휋 푥 (4)∫ dx (5)∫ √ dx 1 푥2−4 0 푠푖푛푥 1 푥푛푙표푔푥 (6) ∫ dx 0 (1+푥)2 Answer (1) divergent (2) convergent for n >-1 (3) convergent (4) divergent (5) divergent (6) convergent for n > -1
17
General Test for Convergence. (Integrand May Change Sign). We now discuss a general test for convergence of an improper integral (finite limits of integration, but discontinuous integrand) which holds whether or not integrand keeps the same sign. Theorem 3.5 (Cauchy’s Tests). 푏 The improper integral converges at a iff to every ε> 0 , ∫푎 푓푑푥 there corresponds δ > 0, such that
푎+휇 |∫ 2 푓푑푥|< ε 0 푎+휇1
< 휇1<휇2 < δ .
푏 Proof. The improper integral ∫푎 푓푑푥 is said to be exists .
푏 When , 푙푖푚 ∫ 푓푑푥 exists finitely. 휇→0+ 푎+휇
푏 Let F(μ) = . ∫푎+휇 푓푑푥
So F(μ) is a function of μ .
According to Cauchy’s Criterion for finite limits F(μ) tends to a finite limit as μ→0 . If and only if to every ε > 0 , there corresponds δ>0, such that for all possible 휇1 , 휇2 < δ ;
|퐹(휇1) − 휇2)|<ε
푏 푏 That is, |∫ 푓푑푥 − ∫ 푓푑푥| < ε 푎+휇1 푎+휇2
푎+휇 or |∫ 2 푓푑푥| . 푎+휇1
Absolute Convergence.
푏 Definition 3.3 . The improper integral ∫푎 푓푑푥 is said to be absolutely 18
푏 convergent if | | is convergent . ∫푎 푓 푑푥
Theorem 3.6. Every absolutely convergent integral is convergent .
푏 푏 That is, | | exist. ∫푎 푓푑푥 exist if ∫푎 푓 푑푥
푏 Proof. Since | | exist. ∫푎 푓 푑푥
Therefore by Cauchy’s test, to every ε >0 ∃ δ >0, such that
푎+휇2 |∫ |푓|푑푥| < ε , o<휇1<휇2< δ 푎+휇1 (4)
푎+휇 푎+휇 Since |∫ 2 푓푑푥| ≤ ∫ 2|푓|푑푥 (5) 푎+휇1 푎+휇1
푎+휇 푎+휇 and |∫ 2|푓|푑푥| = |∫ 2|푓|푑푥| . 푎+휇1 푎+휇1
Therefore (4) and (5) gives
푎+휇2 |∫ 푓푑푥|< ε, ∀ ε>0, 0<휇1<휇2<δ. 푎+휇1
푏 Thus , is convergent. ∫푎 푓푑푥
Alternative Method 3.6.
Since f ≤ |푓| implies that |푓| − 푓 ≥ 0 .
Also , |푓| − 푓 ≤ 2 |푓| (6)
Thus, |푓| − 푓 is a non- negative function on [푎, 푏] and satisfying (6). 푏 Also | | is convergent. Therefore by (1) and comparison test, ∫푎 2 푓 푑푥 we get
푏 ( | |) is convergent . ∫푎 푓 − 푓 푑푥
푏 {( | |) | |} is convergent . This gives that ∫푎 푓 − 푓 + 푓 푑푥 19
푏 is convergent . Hence ∫푎 푓푑푥
1 1 푠푖푛 Example 3.9. Show that ∫ 푥 dx , p>0 converges absolutely for 0 푥푝 p<1.
1 푠푖푛 Solution. Let f(x) = 푥 , p > 0; 푥푝
‘0’ is the only point of infinite discontinuity and f does not keeps the same sign in [0, 1].
1 |푠푖푛 | 1 |푓(푥)| = 푥 < 푥푝 푥푝
1 1 Also, ∫ 푑푥 converges for 푝 < 1 . 0 푥푝
1 1 푠푖푛 Thus ∫ | 푥| 푑푥 converges if and only if p > 0. 0 푥푝
1 1 푠푖푛 Hence ∫ | 푥| 푑푥 is absolutely convergent if and only if 푝 < 1 . 0 푥푝
Infinite range of integration.
We shall now consider the convergence of improper integral of bounded integrable function with infinite range of integration (푎 or b both infinite).
Definition 3.4. (Convergence at ∞).
∞ The symbol , ≥ (7) ∫푎 푓푑푥 푥 푎
푋 is defined as limit of ∫푎 푓푑푥 when 푋 → ∞, so that
∞ 푋 ∫ 푓푑푥 = 푙푖푚 ∫ 푓푑푥 (8) 푎 푋→∞ 푎
If the limit exists and is finite then the improper integral (8) is said to be divergent. 20
푋 푎1 푋 Note. For 푎1 > 푎 , ∫ 푓푑푥 = ∫ 푓푑푥 + ∫ 푓푑푥 푎 푎 푎1
∞ ∞ which implies that the integrals ∫ 푓푑푥 푎푛푑 ∫ 푓푑푥 are either both 푎 푎1 convergent or both divergent.
Exercises.
∞ 푥푑푥 ∞ 푑푥 (i) ∫ (ii) ∫ 0 1+푥2 1 √푥 ∞ (iii) . ∫푎 푠푖푛푥푑푥
Solution. (i) For X > 0, we have
푋 푥푑푥 1 푋 2푥푑푥 ∫ = ∫ 0 1+푥2 2 0 1+푥2
1 = [푙표푔(1 + 푥2)] 푋 2 0
1 = [푙표푔(1 + 푥2)] 2
푋 푥푑푥 Clearly , 푙푖푚 ∫ = ∞ 푋→∞ 0 1+푥2
∞ 푥푑푥 Hence ∫ is divergent. 0 1+푥2
Solution (iii) We have
푋 = ( ) 푋 X > ∫푎 푠푖푛푥푑푥 −푐표푠푥 푎 푎
= cos 푎 − cosX
Clearly, 푙푖푚(푐표푠푎 − 푐표푠푋) exists finitely but not uniquely. 푋→∞
푋 Thus, 푙푖푚 ∫ 푠푖푛푥푑푥 does not exit . 푋→∞ 푎
∞ Hence diverges. ∫푎 푠푖푛푥푑푥
21
Convergence at −∞.
푏 , (9) ∫−∞ 푓푑푥 푥 ≤ 푏 is defined by equation
푏 푏 ∫ 푓푑푥, = 푙푖푚 ∫ 푓푑푥 (10) −∞ 푋→−∞ 푋
If the limit exists and is finite the integral (9) converges otherwise it diverges.
Convergence at both ends.
∞ , (11) ∫−∞ 푓푑푥 ∀ 푥
Is understood to mean
푐 ∞ + (12) ∫−∞ 푓푑푥 ∫푐 푓푑푥
where c is any real number .
If both integrals in (12) converges according to definition (I) and (II),
∞ then, the integral also converges, otherwise it diverges. ∫−∞ 푓푑푥
Exercises.
Examine for convergence the integrals
∞ ∞ 푑푥 ∞ 2푥2푑푥 (I)∫ 푠푖푛푥푑푥 (II)∫ (III) ∫ 0 −∞ 1+푥2 2 푥4−1
∞ 푑푥 ∞ 2 (IV)∫ (V)∫ 푥3푒−푥 푑푥 −∞ (푥2+1)2 0
Solution :-(I) try yourself (limit does not exist)
∞ 푑푥 푋 푑푥 Solution:-(II) ∫ = 푙푖푚 ∫ −∞ 1+푥2 푋→∞ 푌 1+푥2 푌→−∞
= 푙푖푚 (푡푎푛−1푥) 푋 푋→∞ 푌 푌→−∞ 22
= 푙푖푚 (푡푎푛−1푋 − 푡푎푛−1푌) 푋→∞ 푌→−∞
휋 휋 = + 2 2
= π .
Thus the integral converges and is equal to π .
∞ 2푥2푑푥 푋 (III) ∫ = 푙푖푚 ∫ 2푥2 푑푥 2 푥4−1 푋→∞ 2
1 푋−1 1 = 푙푖푚[푡푎푛−1푥 − 푡푎푛−12 + 푙표푔 + 푙표푔3] 푋→∞ 2 푋+1 2
휋 1 = − 푡푎푛−12 + 푙표푔3 . 2 2
Thus the integral converges.
∞ 푑푥 ∞ 푑푥 (IV) ∫ = 2 ∫ −∞ (푥2+1)2 0 (푥2+1)2
푋 푑푥 = 2 푙푖푚 2 ∫ 푋→∞ 0 (푥2+1)2
푥 = 푙푖푚[푡푎푛−1푥 + ] . 푋→∞ 1+푥2
By Putting 푥 = tan 휃
= π/2
∞ 2 (V) 3 −푥 = ½ , converges. ∫0 푥 푒 푑푥
Comparison test for convergence at ∞.
Theorem 3.7. A necessary and sufficient condition for the ∞ convergence of [ ∞), that there exists ∫푎 푓푑푥 , where 푓 is positive in 푎, a positive number 푀, independent of X, such that
푋 < M, X ≥ a. ∫푎 푓푑푥 ∀ 23
푋 푋 Proof. The integral tends to a finite ∫푎 푓푑푥 is said to be convergent if ∫푎 푓푑푥
푋 limit as X→∞. Since f is positive in [a, X] , X ≥ and ∀ 푎 ∫푎 푓푑푥 is
푋 monotonic increasing function on X i.e. ∫푎 푓푑푥 increases as X increases.
푋 Also since . ∫푎 푓푑푥 < 푀, 푓표푟 푠표푚푒 푚 > 0 푎푛푑 ∀ 푋 ≥ 푎
푋 That is, ∫푎 푓푑푥 is bounded above.
푋 Therefore, 푙푖푚 ∫ 푓푑푥 exist finitely. 푋→0 푎
∞ 푋 Conversely, suppose ∫ 푓푑푥 is convergent, then 푙푖푚 ∫ 푓푑푥 푒푥푖푠푡푠 푓푖푛푖푡푒푙푦. 푎 푋→∞ 푎
Therefore, ∃ M > O, such that ∀ X ≥ 푎
푋
∫퐴 푓푑푥 < 푀
푋 as increases as X increases. ∫퐴 푓푑푥
Hence the theorem is proved completely.
Comparison Test I.
Theorem 3.8. If 푓 and 푔 are positive and 푓(푥) ≤ 푔(푥) , for all 푥 ∈ [푎, 푏].
∞ ∞ Then, (I) converges. ∫푎 푓푑푥 converges if ∫푎 푔푑푥
∞ ∞ (II) if ∫푎 푔푑푥 푑푖푣푒푟푔푒푠 ∫푎 푓푑푥 diverges.
∞ Proof. Suppose converges. ∫푎 푔푑푥
Therefore ∃ M> 0 such that ∀ X ≥푎 ,
푋 푋 This gives ∫푎 푓푑푥 < 푀 24 ∞ Hence . ∫푎 푓푑푥 converges ∞ (II) Suppose ∫푎 푓푑푥 diverges then ∃ 푋1, such that 푋 1 > M, M>0 ∫푎 푓푑푥 ∀ 푋 This implies that 1 > M, M>0 ∫푎 푔푑푥 ∀ ∞ This gives diverges. ∫푎 푔푑푥 Note. Since f and 푔 are bounded in [푎, X]. Therefore, f(x) ≤ 푔(x). 푋 푋 This implies that ≤ ∫푎 푓푑푥 ∫푎 푔푑푥 ∀ 푋 ≥ 푎 . Comparison Test –II. Theorem 3.9. If 푓 and 푔 are positive functions in [푎 , X] and 푓(푥) 푙푖푚 = l (≠0), 푋→∞ 푔(푥) then two integrals converges or diverges together. 푓 ∞ ∞ Also if 푙푖푚 =0 and ∫ 푔 푑푥 converges, then ∫ 푓푑푥 converges and if 푋→∞ 푔 푎 푎 푓(푥) ∞ ∞ 푙푖푚 = ∞ and ∫ 푔 푑푥 diverges, then ∫ 푓푑푥 also diverges. 푋→∞ 푔(푥) 푎 푎 Proof. Evidently l > 0 choose > 0 , such that 푙 − 휀 > 0 푓(푥) Since 푙푖푚 = l 푋→∞ 푔(푥) Therefore ∀ 휀 > 0 , ∃ 푘 > 0 such that 푓(푥) | − 푙| < ε whenever |푥| > k . 푔(푥) 25 푓(푥) That is 푙 − 휀 < < 푙 + 휀 ∀ 휀 > 0, 푤푖푡ℎ 푥 > 푘 푔(푥) (푙 − 휀)푔(푥) < 푓(푥) (13) 푓(푥) < ( 푙 + 휀)푔(푥) (14) for 푥 > 푘 푎푛푑 ∀ 휀 > 0. Clearly 푙 − 휀 > 0, by choosing ε so small. Therefore by comparison test and (13) and (14) we get ∞ ∞ ( ) diverges if converges. ∫푎 푔 푥 푑푥 ∫푎 푓푑푥 Again 푓(푥) 푙푖푚 = 0 푋→∞ 푔(푥) implies that 푓(푥) < 푔(푥), ∀ 푥 > 푘 ∞ ∞ Therefore if is divergent, then is convergent and if ∫푎 푓푑푥 ∫푎 푔푑푥 ∞ ∞ is convergent then is convergent. ∫푎 푔푑푥 ∫푎 푓푑푥 푓(푥) Also if, 푙푖푚 = ∞ 푋→∞ 푔(푥) 푓(푥) This implies > M, ∀ 푥 > 푘 푔(푥) Therefore 푓(푥) > 푀푔(푥), ∀ x > k ∞ ∞ Hence if is divergent, then is divergent. ∫푎 푔푑푥 ∫푎 푓푑푥 Useful Comparison Integral. ∞ ∞ 푐 Theorem 3.10. Show that the improper integral ∫ 푓푑푥 =∫ 푑푥, a>0 푎 푎 푥푛 where c is a positive constant, converges if and only if 푛 > 1 . Proof. We have 26 푥 푐 푙표푔 , 푛 = 1 ∞ 푐 푎 ∫ 푛 푑푥 = { 1 1 1 } 푎 푥 [ − ] , 푛 ≠ 1 1−푛 푥푛−1 푎푛−1 푋 푐 ∞ , 푖푓 푛 ≤ 1 푙푖푚 ∫ 푑푥 = { 푐 } . 푋→∞ 푎 푥푛 푖푓 푛 > 1 (푛−1)푛−1 ∞ 푐 Thus, ∫ 푑푥 converges if and only if n>1. 푎 푥푛 From this useful integral and comparison test, the improper integral ∞ onverges if there exists a positive number such that ∫푎 푓푑푥 푐 푛 > 1 푀 푓(푥) ≤ for some 푀 > 0 and for some all 푥 ≥ 푎. 푥푛 ∞ Also if, 푙푖푚 푥푛푓(푥) exists and is non- zero, then integral ∫ 푓푑푥 푥→∞ 푎 converges if and only if 푛 > 1. Exercises. ∞ 푑푥 ∞ 푥2푑푥 (I)∫ (II) ∫ 1 푥√푥2+1 0 √푥5+1 ∞ 2 (III) −푥 ∫0 푒 푑푥 ∞ 푙표푔푥 ∞ (IV)∫ dx (V)∫ 푥푛 푒−푥dx 0 푥2 1 ∞ 푠푖푛2푥 (VI)∫ dx 0 푥2 푑푥 Solution :- (I) Take 푓(푥) = and 푥√푥2+1 1 푔(푥) = 푥2 푓(푥) Then 푙푖푚 = 1(≠0) 푥→∞ 푔(푥) ∞ ∞ 푑푥 Thus ∫ 푓푑푥 = ∫ converges. 1 1 푥√푥2+1 x2dx (II) Let 푓(푥 ) = √x5+1 27 1 Take 푔(푥 = , √푥 푓(푥) Then 푙푖푚 = 1 (≠0) 푥→∞ 푔(푥) ∞ 푥2푑푥 Thus ∫ diverges. 0 √푥5+1 푙표푔푥 (IV) Let 푓(푥) = 푥2 3 Take 푔(푥) = 푥2 , 3 푥2푙표푔 푙표푔푥 Then 푙푖푚 = 푙푖푚 푥2 1 푋→∞ 푥→∞ 푥2 1 푥 = 푙푖푚 1 1 − 푥→∞ 푥 2 2 1 퐿푖푚 2 ( 1) = 0 푥→∞ 푥2 ∞ Since 푑푥 is convergent. ∫1 3 푥2 ∞ 푙표푔푥 Therefore ∫ dx is convergent. 0 푥2 (V) Let f(x) = xne−xdx Take 푔(푥) = 푥2 . Then, 푙푖푚 푥2 . 푥푛푒−푥 = 푙푖푚(푛 + 2)! 푒−푥 = 0 and 푥→∞ 푥→∞ ∞ 1 ∫ 푑푥 is convergent. 1 푥2 ∞ Therefore 푛 −푥 is convergent. ∫1 푥 푒 푑푥 ∞ 2 (III) Let = −푥 . 푓(푥) ∫0 푒 푑푥 Clearly 0 is not point of infinite discontinuity, we may write 28 ∞ 2 1 2 ∞ 2 −푥 −푥 + −푥 = + . ∫0 푒 푑푥 = ∫0 푒 푑푥 ∫1 푒 푑푥 퐼1 퐼2 Clearly I1 is proper and I2 is improper integral. ∞ 2 We test for −푥 . 퐼2 = ∫1 푒 푑푥 2 We have 푒−푥 > 푥2 ∀ 푥 ∈ 푅 1 1 2 < ∀ 푥 ∈ 푅 푒−푥 푥2 2 1 This implies that 푒−푥 < ∀ 푥 ∈ 푅. 푥2 ∞ 1 Again , ∫ 푑푥 is convergent. 1 푥2 ∞ 2 Therefore, −푥 is convergent. ∫1 푒 푑푥 ∞ 2 Hence −푥 ∫0 푒 푑푥 is convergent. ∞ 푠푖푛2푥 (VI) ∫ 푑푥 is convergent because 푠푖푛2푥 ≤ 1, ∀ 푥 ∈ 푅. 0 푥2 Exercises. ∞ 푥푡푎푛−1푥 ∞ 푑푥 ∞ 1 1 (I) ∫ 푑푥 (퐼퐼) ∫ (III) ∫ ( − ). 0 1 푒2 푥푙표푔푙표푔푥 0 푥 푠푖푛ℎ푥 (1+푥4)3 1 푥푡푎푛−1푥 − Solution (I). Let 푓(푥) = (~푥 3) (1+푥4)1/3 1 푓(푥) Take 푔(푥) = , then 푙푖푚 = π/2 . 1 푔(푥) 푥3 푥→∞ ∞ Since 1 ∫0 1 푑푥 푖푠 푑푖푣푒푟푔푒푛푡. 푥3 ∞ Therefore is divergent. ∫0 푓푑푥 (II) Put 푙표푔푥 = 푡, we get ∞ 푑푥 푡 푑푡 ∫ = ∫ 푒2 푥푙표푔푙표푔푥 2 푙표푔푡 29 ∞ 푑푥 = ∫ 2 푙표푔푥 1 Let 푓(푥) = 푙표푔푥 1 Take 푔(푥) = , then 푥푚 푥푚 푥 푙푖푚 = 푙푖푚 by taking 푚 = 1 푥→∞ 푙표푔푥 푥→∞ 푙표푔푥 푥 Therefore 푙푖푚 =푙푖푚푥 = ∞ . 푥→∞ 푙표푔푥 푥→∞ ∞ 푑푥 ∞ 푑푥 Since ∫ is divergent, so that ∫ is also divergent . 2 푥 2 푙표푔푥 ∞ 푑푥 Hence ∫ is divergent. 푒2 푥푙표푔푙표푔푥 1 1 (I) 푓(푥) = ( − )/푥 푥 푠푖푛ℎ푥 Clearly 0 is not point of infinite discontinuity, because 1 푙푖푚 푓(푥) = (By L Hospital’s rule) 푥→0+ 6 1 1 1 We have 푓(푥) = ( − ) 푥 푠푖푛ℎ푥 푥 1 1 = − 푥2 푥푠푖푛ℎ푥 1 1 1 = − [ ] 푥2 푥 푒푥−푒−푥/ 2 1 1 2 = − [ ] 푥2 푥 푒푥−푒−푥 1 1 2푒−푥 = − [ ]. 푥2 푥 1−푒−2푥 1 Take 푔(푥) = , then 푥2 푓(푥) 푙푖푚 = l (≠0) 푥→∞ 푔(푥) ∞ Thus, is convergent. ∫0 푓푑푥 30 푓(푥) 1 1 2푒−푥 Note. 푙푖푚 = 푙푖푚푥2 [ − ] 푥→∞ 푔(푥) 푥→∞ 푥2 푥 1−푒−2푥 2푥푒−푥 =푙푖푚 [1 − ] . 푥→∞ 1−푒−2푥 푥 We have 푙푖푚 푥푒−푥 = 푙푖푚 푥→∞ 푥→∞ 푒푥 1 =푙푖푚 = 0. 푥→∞ 푒푥 2푥푒−푥 0 Therefore 푙푖푚 [1 − ] = 1 − = 1 (≠0). 푥→∞ 1−푒−2푥 1−0 Example 3.10. (Gamma Function). ∞ The integral 푚−1 −푥 is convergent if and only if ∫0 푥 푒 푑푥 푚 > 0 . Solution. Let 푓(푥) = 푥푚−1푒−푥 . If 푚 < 1 , the ‘0’ infinite discontinuity. So we must examine the convergence of above improper integral at both 0 and ∞. ∞ 1 ∞ 푚−1 −푥 = 푚−1 −푥 + 푚−1 −푥 ∫0 푥 푒 푑푥 ∫0 푥 푒 푑푥 ∫1 푥 푒 푑푥 Convergence at 0 for 풎 < 1: 1 Let 푔(푥) = , 푥1−푚 푓(푥) Then , 푙푖푚 = 푙푖푚 푒−푥 =1 (≠0). 푥→0+ 푔(푥) 푥→0+ 1 1 Since ∫ 푑푥 converges, if and only if m>0. 0 푥1−푚 1 Therefore 푚−1 −푥 converges if and only if ∫0 푥 푒 푑푥 푚 > 0 . Converges at ∞. 1 Let 푔(푥) = , so that 푥2 31 푓(푥) 푙푖푚 = 푙푖푚 푥푚+1 / 푒푥 푥→∞ 푔(푥) 푥→∞ (푚+1)! = 푙푖푚 = 0. 푥→∞ 푒푥 ∞ 1 Since ∫ 푑푥 is convergent. 1 푥2 ∞ Thus, 푚−1 −푥 is convergent ∫1 푥 푒 푑푥 ∀ 푚 . ∞ Hence 푚−1 −푥 is convergent if and only if and is denoted ∫1 푥 푒 푑푥 푚 > 0 by < 푚) . ∞ Thus, Γ (m) = 푚−1 −푥 , m>0. ∫0 푥 푒 푑푥 Thus Γ(0), Γ(-1), etc. are not exists . ∞ 푑푥 Example 3.11. Examine for the convergence of ∫ and 3 푥2+푥−2 1 푓(푥) 푥2 퐺(푥) = , then 푙푖푚 = 푙푖푚 푥2 푥→∞ 푔(푥) 푥→∞ 푥2+푥−2 1 = 푙푖푚 1 2 푥→∞ 1+ − 푥 푥2 = 1 . ∞ 푑푥 Thus, ∫ is convergent. 3 푥2+푥−2 Again let us decompose the integrand into partial fraction. 1 1 1 We have = − . 푥2+푥−2 3(푥−1) 3(푥+2) ∞ 1 ∞ 1 It is obvious ∫ 푑푥 푎푛푑 ∫ 푑푥 are both divergent . 3 3(푥−1) 3 3(푥+2) ∞ 푑푥 ∞ 1 ∞ −푑푥 Thus, ∫ = ∫ 푑푥 + ∫ is not correct. 3 푥2+푥−2 3 3(푥−1) 3 3(푥+2) Now we evaluate above improper integral. ∞ 푑푥 푥 푑푥 We have ∫ = 푙푖푚 ∫ 3 푥2+푥−2 푥→∞ 3 푥2+푥−2 32 푥 푑푥 푥 푑푥 =푙푖푚 [∫ − ∫ ] 푥→∞ 3 3(푥−1) 3 3(푥+2) 1 =푙푖푚 [ {푙표푔(푥 − 1) − 푙표푔 (푥 + 2} 푥] 푥→∞ 3 3 1 푥−1 = 푙푖푚 [푙표푔( )] 푥 푥→∞ 3 푥+2 3 1 푥−1 2 =푙푖푚 [푙표푔 ( )] − 푙표 푔 ( ) 푥→∞ 3 푥+2 5 1 5(푥−1) = 푙푖푚 [푙표푔 [ ]] 푥→∞ 3 2(푥+2) 5 푙표푔[5− ] 1 푥 =푙푖푚 [ 4 ] 3 2+ 푥→∞ 푥 =1/3 푙표푔 5/2 . General test for convergence at ∞ (Integrand may change sign). Theorem 3.11. (Cauchy’s Test). 푋 The integral converges at ∞ if and only if to every , ∫푎 푓푑푥 > 0 ∃ X0, such that 푋2 |∫ 푓푑푥| < ϵ ∀ 푋1, 푋2 > 푋0 . 푋1 ∞ x Proof. The improper integrand ∫ fdx exists if lim ∫ fdx exists finitely a x→∞ a ∞ Let = 퐹(푋) ∫푎 푓푑푥 , a function of X . According to Cauchy’s criterion for finite limits, 퐹(푥) tends to a finite limits as 푥 → ∞ if to a finite 휖 > 0 ∃ X0, such that ∀ 푋1, 푋2 > 푋0 |퐹(푋1) − 퐹(푋2)| < ϵ 푋 That is |∫ 2 푓푑푥| < ϵ . 푋1 33 ∞ 푠푖푛푥 Example 3.12. Show that ∫ 푑푥 is convergent . 0 푥 푠푖푛푥 Solution. Since 푙푖푚 = 1. 푥→0 푥 Therefore ‘0’ is not infinite discontinuity, we may put ∞ 푠푖푛푥 1 푠푖푛푥 ∞ 푠푖푛푥 ∫ 푑푥 = ∫ 푑푥+∫ 푑푥 . 0 푥 0 푥 1 푥 ∞ 푠푖푛 1 푠푖푛푥 We now test for the convergence of ∫ 푑푥 as ∫ 푑푥 is proper 1 푥 0 푥 integral. For any휖 > 0 , 2 Let 푥 ,푥 be two numbers both greater than , 1 2 ϵ 푥 푠푖푛푥 푐표푠푥 푥 푐표푠푥 Now ∫ 2 푑푥 = [− ] 푥2 − ∫ 2 푑푥 푥1 푥 푥 푥1 푥1 푥2 푥2 푠푖푛푥 푐표푠푥1 푐표푠푥2 푥2 푐표푠푥 so that, |∫ 푑푥| ≤ | − | + |∫ 2 푑푥 | 푥1 푥 푥1 푥2 푥1 푥 1 1 푥2 푑푥 ≤ + + ∫ 2 푥1 푥2 푥1 푥 휖 =2. = 휖 . 2 ∞ 푠푖푛푥 Therefore, by Cauchy’s test the improper integral ∫ 푑푥 is 0 푥 convergent. Absolute Convergence. ∞ Definition 3.4. The improper integral is said to be absolutely ∫푎 푓푑푥 ∞ convergent if | | is convergent. ∫푎 푓 푑푥 Theorem 3.12. ∞ ∞ Absolute convergences of implies convergence ∫푎 푓푑푥 of ∫푎 푓푑푥 ∞ ∞ i.e., | | exist. ∫푎 푓푑푥 exists if ∫푎 푓 푑푥 34 ∞ Proof. Suppose | | exists, then by Cauchy’s Test, ∫푎 푓 푑푥 ∀ 휀 > 0, ∃ 푋0 , 푠푢푐ℎ 푡ℎ푎푡 푥2 | | | | < ϵ , , ,> . ∫푥1 푓 푑푥 푥1 푥2 푥0 푥2 푥2 We have |∫ 푓푑푥| ≤ ∫ |푓|푑푥 < ϵ 푥1, 푥2,> 푥0 . 푥1 푥1 ∞ Thus by Cauchy’s test . ∫푎 푓푑푥 converges ∞ 푠푖푛푥 Example 3.13. Show that ∫ 푑푥 converges absolutely if 푝 > 1 1 푥푝 푠푖푛푥 |푠푖푛푥| 1 Solution. We have | | = ≤ , ∀ 푥 ≥ 1 , 푥푝 푥푝 푥푝 ∞ 1 and ∫ 푑푥 converges for 푝 > 1 . 1 푥푝 ∞ 푠푖푛푥 Thus, ∫ | | 푑푥 converges for 푝 > 1 . 1 푥푝 ∞ 푠푖푛푥 Therefore, ∫ 푑푥 converges absolutely for 푝 > 1. 1 푥푝 Integrand as a product of functions (convergent at ‘∞’). A test for absolutely convergence. Theorem 3.13. If a function 휑 is bounded in [푎, ∞] and integrable in [푎, 푥], ∀ 푥 ≥ 푎. ∞ ∞ Also if is absolutely convergent at ∞, then is also ∫푎 푓푑푥 ∫푎 푓휑푑푥 absolutely convergent at ∞. Proof. Since f is bounded in [푎, ∞), therefore ∃ 푘 > 0, such that |휑(푥)| ≤ 푘, ∀ 푥 ∈ [푎, ∞) (15) ∞ Again since | |is positive in [ ) and | | is convergent. f 푎, ∞ , ∫푎 푓 푑푥 35 Therefore we can find m, such that 푥 | | < , ∫푎 푓 푑푥 푚 ∀ 푥 ≥ 푎 (16) Using (1) we have |푓휑| = |푓||휑| ≤ k|푓|, ∀ 푥 ∈ [푎, ∞). 푥 푥 Therefore, | | ≤ k | | ∫푎 푓휑 푑푥 ∫푎 푓 푑푥 < 푘푚 ∀ 푥 ≥ 푎 . 푥 Thus, | | ≤ ∫푎 푓휑 푑푥 푘푚 ∀ 푥 ≥ 푎 . 푥 Therefore, | | is convergent. ∫푎 푓휑 푑푥 푥 Hence is absolutely convergent. ∫푎 푓휑푑푥 Test for convergence. Theorem (Abel’s Test) 3.14. If 휑 is bounded and monotonic in [푎, ∞) ∞ ∞ and is convergent at ∞, then, is convergent at ∞. ∫푎 푓푑푥 ∫푎 푓휑푑푥 Proof. Since φ is monotonic in [푎, ∞), 푡ℎ푒푛 is integrable in [푎, 푥], ∀ 푋 ≥ 푎 . Also since 푓 is integrable in [푎, 푥] , we have by 2nd mean value theorem 푋2 푦 푋2 ∫ 푓휑 푑푥 = φ(푋1)∫ 푓푑푥 +φ(푋2)∫ 푓푑푥 (17) 푋1 푋1 푦 푓표푟 푎 < 푋1 < 푌 ≤ 푋2. Let 휖 > 0 be arbitrary. Since φ is bounded in [푎, ∞) , a positive number k exists, such that 36 |휑(푥)| ≤ 푘, ∀ 푋 ≥ 푎 . In particular, |휑(푥1)| ≤ k, |휑(푥2)|≤ k, (18) ∞ Again since is convergent, therefore their exists , such that ∫푎 푓푑푥 푋0 푥2 휖 |∫ 푓푑푥| < , ∀ 푋1, 푋2>푋0 (19) 푥1 2푘 Since, 푋1≤ Y≤ 푋2 . 푦 휖 Therefore, |∫ 푓푑푥| < 푎nd 푥1 2푘 푋 휖 |∫ 2 푓푑푥| < (20) 푦 2푘 Thus from (17), (18),(19), and (20), we deduce that ∃ X0, such that for all , X1, X2>X0 and 휖 > 0 푋 푦 푋 휖 휖 |∫ 2 푓휑푑푥| ≤ |휑(푥1)| |∫ 푓푑푥| + |휑(푥2)| |∫ 2 푓푑푥| < k + 푘 = 휖 . 푋1 푋1 푦 2푘 2푘 ∞ Hence is convergent. ∫푎 푓휑푑푥 Theorem Drichlet’s Test 2.15. If 휑 is bounded and monotonic in 푥 and tends to 0 as and is bound for [푎, ∞) 푥 → ∞ ∫푎 푓푑푥 푋 ≥ 푎 , ∞ then convergent at ∞. ∫푎 푓휑푑푥 Proof. Since 휑 is bounded and integrable in [푎, 푥] . Also since f is integrable in [푎, 푥], therefore by second mean value theorem: 푥2 푦 푥2 ∫ 푓휑푑푥= φ(푥1)∫ 푓푑푥 + φ(푥2)∫ 푓푑푥 (21) 푥1 푥1 푦 for 푎 < 푋1 ≤ 푌 ≤ 푋2. 푥 Again, since is bound when therefore such that ∫푎 푓푑푥 푋 ≥ 푎, ∃ 푘, 푥 | | ≤ k, . ∫푎 푓푑푥 ∀ 푋 ≥ 푎 37 Therefore, 푦 푦 푥 | ∫ 푓푑푥| = |∫ 푓푑푥 − ∫ 1 푓푑푥| 푥1 푎 푎 푦 푥 ≤| | +| 1 | ∫푎 푓푑푥 ∫푎 푓푑푥 ≤ 2k, for 푋1, ≥ 푎 . Similarly, 푥 | 2 | ≤ 2k, for . ∫푦 푓푑푥 X2 ≥ 푎 Let 휖 > 0 be arbitrary. Since 휑 → 0 as → ∞ , there exists a positive X0, such that 휖 휖 |휑(푋 )| < , |휑(푋 )| < where 푋 ≥ 푋 ≥ 푋 . 1 4푘 2 4푘 2 1 0 Let the numbers X1, X2 in (21) be ≥X0, so that from (17), (18), (19) & (20), we get 푋 휖 휖 |∫ 2 푓휑푑푥| < 2푘 + 2푘 푋1 4푘 4푘 = 휖 ∀ 푋2≥ 푋1≥ 푋0 . ∞ Hence by Cauchy’s test is convergent at ∞. ∫a fφdx ∞ 푠푖푛푥 Example 3.14.The improper integral ∫ 푑푥 is divergent for p>0. 1 푥푝 1 Solution. Take φ(x) = , 푝 > 0 and 푥푝 푓(푥) = 푠푖푛푥 . Then 휑(푥) is monotonic decreasing and tends to 0 as 푥 → ∞. 푥 푥 Also, | | = | | ∫1 푓푑푥 ∫1 푠푖푛푥 푑푥 =|푐표푠1 − 푐표푠푥| 38 ≤ |푐표푠1|+|푐표푠푥| ≤ 1+1 = 2, ∀ 푋 ≥ 1 . 푥 Thus, | | ≤ 2 . ∫1 푠푖푛푥푑푥 ∀ 푋 ≥ 2 푥 Therefore is bounded. ∫1 푠푖푛푥 푑푥 ∞ 1 ∞ 푠푖푛푥 Hence by Drichlet’s test ∫ 푠푖푛푥 = ∫ 푑푥 is convergent 푝 > 0. 1 푥푝 1 푥푝 ∞ 푠푖푛푥 Also, we know that ∫ 푑푥 is absolutely convergent if and only if 푝 > 1 1 푥푝 ∞ sinx Thus, ∫ dx is conditional convergent for 0 < 푝≤1 . 1 xp Conditionally Convergent. ∞ ∞ An improper integral is conditionally convergent at ∞ if is ∫푎 푓푑푥 ∫푎 푓푑푥 ∞ convergent at ∞, but | | is not convergent .That is the improper ∫푎 푓 푑푥 integral is said to be conditionally convergent if it is convergent but not absolutely. ∞ 푠푖푛푥 Example 3.15. Show that ∫ 푑푥 is convergent, but not absolutely. 0 푥 ∞ 푠푖푛푥 1 푠푖푛푥 ∞ 푠푖푛푥 Solution. We have ∫ 푑푥 = ∫ 푑푥 + ∫ 푑푥 0 푥 0 푥 1 푥 1 푠푖푛푥 Now , ∫ 푑푥 is proper integral. 0 푥 ∞ 푠푖푛푥 To examine the convergence of ∫ 푑푥 at ∞, we see that 1 푥 푋 | | | | ≤ | | | | , so that ∫1 푠푖푛푥푑푥 = 푐표푠1 − 푐표푠푋 푐표푠1 + 푐표푠푋 < 2 푋 | | ∫1 푠푖푛푥푑푥 is bounded above for all X ≥ 1. Also, 1/푥 is a monotonic decreasing function tending to 0 푎푠 푥 → ∞. 39 ∞ 푠푖푛푥 Therefore by Dirchlet’s test ∫ 푑푥 is convergent. 0 푥 ∞ 푠푖푛푥 Hence ∫ 푑푥 is convergent. 0 푥 ∞ 푠푖푛푥 To show that ∫ 푑푥 is not absolutely convergent, we proceed as 0 푥 follows: 푛휋 푠푖푛푥 푟휋 |푠푖푛푥| ∫ | | 푑푥 = ∑푛 ∫ 푑푥 0 푥 푟=1 (푟−1)휋 푥 Now, ∀ 푥 ∈ [(푟 − 1)휋, 푟휋] 푟휋 |푠푖푛푥| 푟휋 |푠푖푛푥| ∫ 푑푥 ≥ ∫ 푑푥 (푟−1)휋 푥 (푟−1)휋 푟휋 Putting, 푥 = (푟 − 1)휋 + 푦 푟휋 |푠푖푛푥| 휋 |푠푖푛(푟−1)휋+푦|푑푦 ∫ 푑푥 = ∫ (푟−1)휋 푟휋 0 푟휋 1 휋 2 = ∫ 푠푖푛푦푑푦 = . 휋푟 0 푟휋 푛휋 |푠푖푛푥| 푟휋 |푠푖푛푥| 2 Hence ∫ 푑푥 = ∑푛 ∫ 푑푥 ≥ ∑푛 . 0 푥 푖=푟 (푟−1)휋 푥 푖=1 푟휋 2 But ∑푛 is a divergent series. 푟=1 푟휋 푛휋 |푠푖푛푥| 푛 2 Therefore, 푙푖푚 ∫ 푑푥 ≥ 푙푖푚 ∑푖=1 . 푛→ 0 푥 푛→∞ 푟휋 푛휋 |푠푖푛푥| This implies that 푙푖푚 ∫ 푑푥 is infinite. 푛→∞ 0 푥 Now, let 푡 be a real number, there exists positive integer 푛 , such that 푛휋 ≤ 푡 < (푛 + 1)휋 . 푡 |푠푖푛푥| 푛휋 |푠푖푛푥| We have, ∫ 푑푥 ≥∫ 푑푥. 0 푥 0 푥 Let 푡 → ∞, so that 푛 → ∞, thus we see that 푡 |푠푖푛푥| ∫ 푑푥 → ∞ . 0 푥 40 ∞ |푠푖푛푥| This implies ∫ 푑푥 does not converge. 0 푥 ∞ 푠푖푛푥 This example show that ∫ 푑푥 , 0 < 푝 ≤ 1, is convergent but not 표 푥푝 absolutely . ∞ 푐표푠푥 Example 3.16. Show that ∫ 푑푥 is conditionally convergent. 2 푙표푔푥 1 Solution. Let φ(x) = , 푓(푥) = 푐표푠푥 . 푙표푔푥 푋 | | = | |≤| | + | | ≤ 2, so that ∫2 푐표푠푥푑푥 푠푖푛푋 − 푠푖푛2 푠푖푛푋 푠푖푛2 푋 is bounded for all X ∫2 푐표푠푥 푑푥 ≥ 2 1 Also, φ(x) = is monotonic decreasing function tending to 0 as 푥 → ∞ . 푙표푔푥 ∞ 푐표푠푥 Hence by Dirichlet’s test ∫ 푑푥 is convergent. 2 푙표푔푥 For absolute convergence consider 3휋 5휋 ∞ 푐표푠푥 |푐표푠푥| |푐표푠푥| 2 2 I= ∫ | | 푑푥 = ∫ 푑푥 + ∫3휋 푑푥----- 2 푙표푔푥 2 푙표푔푥 푙표푔푥 2 (2푛+1)휋 |푐표푠푥| +∫ 2 푑푥+---- (2푛−1)휋 푙표푔푥 2 Therefore, 3휋 (2푛+1)휋 2 |푐표푠푥| |푐표푠푥| 2 |푐표푠푥| I= ∫휋 푑푥+ ∫ 2 푑푥+-----∫ 푑푥+ ….. 푙표푔푥 2 푙표푔푥 (2푛−1)휋 푙표푔푥 2 2 2 |푐표푠푥| − ∫휋 푑푥 2 푙표푔푥 (2푛+1)휋 푛 2 |푐표푠푥| 2 |푐표푠푥| = ∑ ∫ 푑푥 - ∫휋 푑푥 푟=1 (2푛−1)휋 푙표푔푥 푙표푔푥 2 2 Now, (2푛+1)휋 (2푟+1)휋 |푐표푠푥| 1 ∫ 2 푑푥 ≥ |∫ 2 푐표푠푥푑푥| (2푛−1)휋 푙표푔푥 푙표푔(2푟+1)휋/2 (2푟−1)휋 2 2 41 1 휋 휋 = (2푟+1)휋 |푠푖푛 [(2푟 + 1) ] − 푠푖푛 [(2푟 − 1) ]| 푙표푔[ ] 2 2 2 |2(−1)푟| = (2푟+1)휋 푙표푔[ ] 2 2 = (2푟+1)휋 . 푙표푔 2 ∞ 2 2 |푐표푠푥| Therefore, I ≥ ∑푟=1 (2푟+1)휋 − ∫휋 푑푥 . 푙표푔 푙표푔푥 2 2 ∞ 1 2 |푐표푠푥| But ∑푟=2 is divergent and ∫휋 푑푥 is proper integral. 푙표푔푥 2 푙표푔푥 ∞ |푐표푠푥| ∞ 푐표푠푥 Hence I=∫ 푑푥 is divergent and so ∫ 푑푥 is conditionally 0 푙표푔푥 2 푙표푔푥 convergent. ∞ 푠푖푛푥 푛 Example 3.17. Using ∫ 푑푥 = , show that 0 푥 2 ∞ 푠푖푛2푥 휋 ∫ 푑푥 = . 0 푥2 2 Solution. To compute it let us integrate by parts, therefore ∞ 푠푖푛2푥 −푠푖푛2푥 ∞ 푠푖푛2푥 ∫ 푑푥 = [ ] ∞ + ∫ 푑푥 0 푥2 푋 0 0 푋 ∞ 푠푖푛2푥 ∞ 푠푖푛푡 휋 Hence ∫ 푑푥 = ∫ 푑푡 = . 0 푥2 0 푡 2 Example 3.18.The function 푓 is defined on [0, ∞ [by f(x) = (-1)n−1 , ∞ show that the integral ( ) does not 푓표푟 푛 − 1 ≤ 푥 < 푛, 푛 ∈ 푁 , ∫0 푓 푥 푑푥 converge. Solution. Consider 2푛 1 2 ( ) = 0 ( ) + ∫0 푓 푥 푑푥 ∫0 (−1) 푑푥 + ∫1 −1 푑푥 3 2푛 2 2푛−1 + ∫2 (−1) 푑푥 + ⋯ ∫2푛−1(−1) 푑푥 42 =1 − 1 + 1 − 1 + 1 − 1 … … … . . +1 − 1. and 2푛+1 1 2 2푛+1 ( ) = ( ) +… …. 2푛 ∫0 푓 푥 푑푥 ∫0 푑푥 + ∫1 −1 푑푥 … ∫2푛 (−1) 푑푥 =1 − 1 + 1 … … … . . −1 + 1 =1 2푛 푙푖푚 ∫ 푓(푥)푑푥 = 0 푛→∞ 0 2푛+1 and 푙푖푚 ∫ 푓(푥)푑푥 = 1. 푛→∞ 0 Hence the integral does not exist and therefore it is not convergent. Example 3.19. Test the convergence of ∞ 푥푑푥 ∞ 푑푥 (I) ∫ (퐼퐼) ∫ . 0 1+푥4푐표푠2푥 0 1+푥4푐표푠2푥 Solution. The integral is positive for positive value of x but the tests obtained for the convergence of positive integrands so far, are not applicable. In order to show the integral convergent we proceed as follows: 푛휋 푥푑푥 Consider ∫ . 0 1+푥4푐표푠2푥 푛휋 푥푑푥 푟휋 푥푑푥 Therefore ∫ = ∑푛 ∫ . 0 1+푥4푐표푠2푥 푟=1 (푟−1)휋 1+푥4푐표푠2푥 Now, ∀ 푥 ∈ [(푟 − 1)휋, 푟휋]. We have ( ) 푥 ≥ 푟−1 휋 1+푥4푐표푠2푥 1+푟4푐표푠2푥 푟휋 푥푑푥 푟휋 (푟−1)휋푑푥 Therefore ∫ ≥ ∫ (푟−1)휋 1+푥4푐표푠2푥 (푟−1)휋 1+푥4푐표푠2푥 Putting 푥 = (푟 − 1)휋 + 푦 , we see that 43 푟휋 (푟−1)휋푑푥 휋 (푟−1)휋푑푦 ∫ = ∫ (푟−1)휋 1+푥4푐표푠2푥 0 1+푟4휋4푐표푠2{(푟−1)휋+푦)} 휋 (푟−1)휋푑푦 = ∫ 0 1+푟4휋4푐표푠2푦 휋 푑푦 = 2(r-1)π∫2 0 1+푟4휋4푐표푠2푦 휋 푠푒푐2푦푑푦 = 2(r-1)π∫2 0 1++푡푎푛2푦+푟4휋4 2(푟−1)휋 푡푎푛푦 휋 (푟−1)휋2 = 푡푎푛−1( ) |2 = . √1+푟4휋4 √1+푟4휋4 0 √1+푟4휋4 푟휋 푥푑푥 (푟−1)휋2 Therefore, ∑푛 ∫ ≥ ∑푛 . 푟=1 (푟−1)휋 1+푥4푐표푠2푥 푟=1 √1+푟4휋4 2 푛휋 푥푑푥 푛 (푟−1)휋 Hence 푙푖푚 ∫ ≥ 푙푖푚 ∑푟=1 . 푛→∞ 0 1+푥4푐표푠2푥 푛→∞ √1+푟4휋4 (푟−1)휋2 1 But ∑푛 is a divergent series (~ ∑푛 ) . 푟=1 √1+푟4휋4 푟=1 푟 ∞ 푥푑푥 Therefore ∫ is divergent. 0 1+푥4푐표푠2푥 ∞ 푑푥 (II) ∫ try yourself 0 1+푥4푐표푠2푥 Inequalities. Definition 3.5. If 푎1, 푎2, 푎3, … … 푎푛 are n real numbers, then their Arithmetic mean is defined as A = 푎1+푎2+⋯푎푛 푛 푎 = ∑푛 푖 . 푖=1 푛 If the above numbers are positive then, their Geometric mean is defined by 44 1 G = (푎1, 푎2, 푎3, … … 푎푛)푛 and the recipocial of the arithmetic mean of the recipocials of a1, a2, a3, … … an is defined to be the harmonic mean 푛 H= 1 1 1 where H is the Harmonic Mean. + +⋯+ 푎1 푎2 푎푛 Arithmetic Mean - Geometric mean Inequality. Theorem 3.16. Let 푎1, 푎2, 푎3, … … 푎푛 be n positive numbers. If A denotes their Arithmetic mean and G denotes their Geometric mean, then 퐴 ≥ 퐺 . Equality sign holds if and only if 푎1= 푎2= 푎3=…..=푎푛 . Proof. For 푛 = 1, there is nothing to prove as 퐴 = 푎1 = 퐺 . For 푛 = 2, we have to show that 푎 +푎 1 2 ≥ (푎 푎 )1/2 . 2 1 2 2 We know that (√푎1- √푎2) ≥ 0 푎1 + 푎2 − 2√푎1 √푎2 ≥ 0 1 푎1+푎2 Therefore , ≥ (푎 푎 )2 . 2 1 2 Thus result holds for n=2 푎푛푑 푡ℎ푒 푒푞푢푎푙푖푡푦 ℎ표푙푑푠 푖푓 푎푛푑 표푛푙푦 푖푓 푎1 = 푎2 . Now we used induction on n. For 푛 = 4 = 2 2 , we have a a a a 1 2 3 4 푎1+푎2+푎3+푎4 푎1+푎2+푎3+푎4 2 2 2 = = 2 4 2 1 1 (푎1푎2)2+(푎3푎4)2 (By previous case). 2 45 1 1 1 2 ≥ [((푎1푎2)2(푎3푎4)2 ] 1 푎1+푎2+푎3+푎4 Thus, ( ) ≥ (푎 푎 푎 푎 )4 . 4 1 2 3 4 Thus result holds for 푛 = 4 i.e. , 2 2 . Suppose result holds for 푛 = 푚 that is, for 2푚 . Let n be a positive integer not of the form 2푚. We choose k suitable, such that 2 m n . Thus 2 m n is a positive integer. a +a +a . . . a Let K 1 2 3 n (22) n and a a . . . a m K (23) n 1 n 2 2 푚 Consider the product ( a . a .a . . a . a .a .. . a m ) which has 2 form. 1 n 3 n n 1 n 2 2 Since the inequality is supposed to be true for all positive integral powers of 2, we have 1 m a a ... a a . . . a m 2 1 2 n n 1 2 a 1 . a 2 . . . a n . . . a m 2 2 m ( a a ... a ) ( a . . . a m ) m 1 2 n n 1 2 2 a 1 . a 2 . . . a n . . . a m } 2 2 m m 2 nK 2 m n K = With equality iff 푎1= 푎2= 푎3=…..=푎푛 . 46 m n m 2 2 a 1 . a 2 . . . a n K K n a 1 . a 2 . . . a n K 1 n a 1 . a 2 . . . a n K 1 a a . . . a n 1 2 n or a 1 . a 2 . . . a n n with equality iff 푎1= 푎2= 푎3=…..=푎푛 . Combing this result with the earlier one, we come to the conclusion that if 푎1, 푎2, 푎3, … … 푎푛 are n positive numbers, then 1 a a . . . a 1 2 n n a 1 . a 2 . . . a n n or 퐴 ≥ 퐺 . Corollary. If 푎1, 푎2, 푎3, … … . , 푎푛 are n positive real numbers , then 퐺 ≥ 퐻. Proof. We know that for any positive integer n, 1 푎1+푎2+⋯푎푛 ≥ (푎 푎 푎 … … . 푎 )푛 where 푎 > 0 ∀ 푖, 1 ≤ 푖 ≤ 푛 . 푛 1 2 3 푛 푖 1 1 1 Thus result holds for , , … … . 푎1 푎2 푎푛 Therefore 1 1 1 + +⋯……………… 1 푎 푎 푎 1 2 푛 ≥ (푎 푎 푎 … … . 푎 )푛 . 푛 1 2 3 푛 47 1 푛 1 1 1 1 n 1 1 1 ≤ . . . . . + +⋯+ 푎 푎 푎 a a a a 1 2 푛 1 2 3 n 퐻 ≤ 퐺 . Equality holds if and only if 푎1 = 푎2 = ⋯ . = 푎푛 . Corollary. Since 퐴 ≥ 퐺 and 퐺 ≥ 퐻 , thus 퐴 ≥ 퐺 ≥ 퐻 . 푎1+푎2+⋯푎푛 푛 This gives ≥ 1 1 1 . 푛 + +⋯……………… 푎1 푎2 푎푛 푛 푛 1 2 Therefore, ∑푖=1 푎푖 ∑푖=1 ≥ 푛 . 푎푖 Theorem Cauchy Schwarz Inequality 3.17. If 푎1, 푎2, 푎3, … … . , 푎푛 and 푏1, 푏2, 푏3, … … . , 푏푛are real numbers, then n 1 1 푛 푛 a b ≤ (∑ 2 ) 2 ( ∑ 2 2 . i i 푖=1 푎푖 ) 푖=1 푎푖 ) i 1 Proof. Let be a real number, then 2 ( a i b i ) 0 with equality iff a i b i 0 for i 1,2,3,. .. n. 2 2 2 This implies a i 2 a i b i b i 0 for i 1,2,3,. .. Adding for i=1, 2, 3. . . , we get n n n 2 2 2 a 2 ( a b ) b i 0 i i i i 1 i 1 i 1 i.e., f A 2 2 B C 0 for every real 48 n n n 2 2 A a i B a b b . 푤ℎ푒푟푒 i i 푎푛푑 C = i i 1 i 1 i 1 Now obviously A o . If A=0, then there is nothing to prove because both sides of the proposed inequality reduce to zero. So, let A o . We claim that f o for every , implies B 2 AC . If this is not true, then B 2 AC , so 2 2 B AB _ B B AC f 2 B C 0 . 2 A A A A B Thus, for , f 0 which contradicts the fact that A f 0 for every real . Hence our supposition that B 2 AC is wrong. Therefore B 2 AC must be true. 2 n n 2 2 a b a i b i i i i 1 i 1 i 1 1 1 푛 푛 2 푛 2 or |∑푖=1 푎푖푏푖| ≤ (∑푖=1 푎푖 )2(∑푖=1 푏푖 )2 Hence we have Cauchy’s - Schwarz Inequality, with equality iff a i b i 0 for i 1,2,3,. .. n. 49 Convex Function. Let 푦 = 푓(푥) be well defined in some interval and 푙푒푡 푥1 ≠ 푥2 . 푆푎푦 푥1 < 푥2 be the abscissa the graph of 푦 = 푓(푥) so that the point 푃 and 푄 are [푥1, 푓(푥1)] 푎푛푑 [푥2, 푓(푥2)] respectively. Y 푄(푋2, 푓(푋2)) 푦 = 푓(푥) 푋 + 푋 푋 1 2 푋 1 2 2 O X If the graph of f(x) between P and Q lies below the cord PQ,then the function is said to be convex downwards or simply convex. The equation of the straight line through PQ is said to be convex downwards or simply convex. The equation of the straight line 푃푄 is 푦−푓(푥 ) 푓(푥 )−푓(푥 ) 1 = 2 1 푥−푥1 푥2−푥1 푥−푥1 That is 푦 = 푓(푥1) + [푓(푥2) − 푓(푥1)]. 푥2−푋1 50 푥 +푥 Since the point 1 2 lies between 푥 and 푥 and the curve is convex, 2 1 2 the y coordinate of the curve must be less than the y coordinate of the cord i.e., we must have 푥 + 푥 푓(푥 ) + 푓(푥 ) 푓 ( 1 2) ≤ 1 2 2 2 푥 +푥 푓(푥 )+푓(푥 ) Thus, 푓 ( 1 2) ≤ 1 2 for convex function f. 2 2 Theorem 3.16. Prove that if 푓 is convex function, then 푥 + 푥 + ⋯ 푥 푓(푥 ) + 푓(푥) + ⋯ 푓(푥 ) 푓 ( 1 2 푛) ≤ 1 푛 . 푛 푛 Proof. For 푛 = 1, there is nothing to prove. We will first prove the result for all positive integer 푚 = 2푛 by using induction on 푛. 푥 +푥 푓(푥 )+푓(푥 ) For 푛 = 1 , then result holds, because 푓 ( 2 1) ≤ 1 2 . 2 2 For a convex function, suppose result holds for 2푘. That is 푥 +푥 +⋯푥 푓(푥 )+푓(푥 )+⋯푓(푥 ) 푓 ( 1 2 2푘) ≤ 1 2 2푘 (24) 2푘 2푘 We will show that result holds for2푘+1, we have 푥 +푥 +⋯푥 푥 +⋯푥 1 2 2푘 1+2푘 2푘+1 푥1+푥2+⋯푥 +푥 +⋯푥 + 푓 ( 2푘 1+2푘 2푘+1) ≤ 푓 ( 2푘 2푘 ) 2푘 2 푥1 + 푥2 + ⋯ 푥2푘 푥1+2푘 + ⋯ 푥2푘+1 푓 ( 푘 ) + 푓 ( 푘 ) ≤ 2 2 2 because result holds 2 terms. This implies that 51 푓(푥1)+푓(푥2)+⋯푓(푥 푘) 푓(푥 푘)+⋯푓(푥 푘+1) 푥 +푥 +⋯푥 2 + 1+2 2 푓 ( 1 2 2푘+1) ≤ 2푘 2푘 by ( 24) . 2푘+1 2 Therefore 푥 +푥 +⋯푥 푓(푥1)+푓(푥2)+⋯푓(푥 ) 푓 ( 1 2 2푘+1) ≤ 2푘+1 2푘+1 2.2푘 푓(푥 ) + 푓(푥 ) + ⋯ 푓(푥 푘+1) = 1 2 2 . 2푘+1 Thus result holds for 2푘+1 . Hence by principal of Mathematical induction result holds for all integers of the form 2푛 , 푛 ≥ 1. Now, let 푛 be any positive integer. Choose positive integer 푚, 푠푢푐ℎ 푡ℎ푎푡 2푚 > 푛 푥 +푥 +⋯푥 Let 1 2 푛 = 퐾 (25) 푛 푥푛+1 = 푥푛+2 = ⋯ = 푥2푚 = 퐾 (26) 푓( 푥1) + 푓( 푥2) + ⋯ + 푓( 푥푛) = 퐾1 (27) We have 푥 +푥 +⋯푥 +⋯푥 푓( 푥 )+푓( 푥 )+⋯+푓( 푥 )+푓( 푥 )+⋯푓( 푥 ) 푓 ( 1 2 푛 2푚) ≤ 1 2 푛 푛+1 2푚 . 2푚 2푚 This implies that 푛퐾+(2푚−푛)퐾 퐾 +(2푚−푛)푓(푘) 푓 ( ) ≤ 1 2푚 2푚 퐾 +2푚푓(퐾)−푛푓(퐾) 푓(퐾) ≤ 1 2푚 푚 푚 2 푓(퐾) ≤ 퐾1 + 2 푓(퐾) − 푛푓(퐾) 0 ≤ 퐾1 − 푛푓(퐾) 퐾 This implies 푛푓(퐾) ≤ 퐾 or 푓(퐾) ≤ 1 . 1 푛 52 푥 +푥 +⋯푥 푓( 푥 )+푓( 푥 )+⋯+푓( 푥 ) Therefore, 푓 ( 1 2 푛) ≤ 1 2 푛 for all n, not of the form 2푚. 푛 푛 푥 +푥 +⋯푥 푓( 푥 )+푓( 푥 )+⋯+푓( 푥 ) Hence 푓 ( 1 2 푛) ≤ 1 2 푛 for all 푛 ≥ 1, if f is convex. 푛 푛 Proposition . Suppose 푓 is defined 푖푛 [푎, 푏] and 푓′′(푥) exists, 푓′′(푥) ≥ 0 in this interval. Then 푓 is convex . Proof. Let 푡1 ≠ 푡2, say 푡1 < 푡2 i.e., any two points in [푎, 푏] = 퐼 , so that 푡 +푡 푡 < 1 2 < 푡 . 1 2 2 푡 +푡 푡 −푡 푡 +푡 1 푡 −푡 2 Then , 푓(푡 ) = 푓 ( 1 2) + ( 1 2) 푓′ ( 1 2) + ( 1 2) 푓′′(퐶 ) 1 2 2 2 2 2 1 t +t where t < C < 1 2 . 1 1 2 t +t t −t t +t 1 t −t 2 Also, f(t ) = f ( 1 2) + ( 2 1) f ′ ( 1 2) + ( 2 1) f ′′(C ) 2 2 2 2 2 2 2 t +t where 1 2 < C < t . 2 2 2 Adding these two equations, we get 푡 +푡 1 푡 −푡 2 푓(푡 ) + 푓(푡 ) = 2푓 ( 1 2) + 퐸 푤ℎ푒푟푒 퐸 = ( 2 1) [푓′′(퐶 ) + 푓′′(퐶 )]≥ 0. 1 2 2 2 2 1 2 푡 +푡 Therefore, 푓(푡 ) + 푓(푡 ) ≥ 2푓 ( 1 2). 1 2 2 푡 +푡 푓(푡 )+푓(푡 ) This implies that 푓 ( 1 2) ≤ 1 2 . 2 2 Therefore f is convex. Proposition. If 푓 is convex and 푓′′ exist in [푎, 푏], then 푓′′ ≥ 0 푖푛 [푎, 푏] . Proof. Let ℎ > 0 Take 푥 − ℎ = 푥1, 푥 + ℎ = 푥2 . Since 푓 is convex. 53 푥 +푥 (푥 )+푓(푥 ) Therefore 푓 ( 2 1) ≤ 1 2 . 2 2 This implies that 푥−ℎ+푥+ℎ 푓(푥−ℎ)+푓(푥+ℎ) 푓 ( ) ≤ 2 2 푓(푥−ℎ)+푓(푥+ℎ) 푓(푥) ≤ . 2 This gives 푓(푥 − ℎ) + 푓(푥 + ℎ) − 2푓(푥) ≥ 0 (28) Since 푓′′(푥) exists , we have 푓(푥 − ℎ) + 푓(푥 + ℎ) − 2푓(푥) 푙푖푚 = 푓′′(푥) . ℎ→0 ℎ2 Therefore by (28) , we have 푓′′(푥) ≥ 0. In the year 1906 Jenson obtained some considerable extensions of the AM-GM inequality. These extensions were based on the theory of convex functions, founded by Jenson himself. Theorem Jensen’s Inequality 3.18. ′′ Suppose 푓 is convex and 푓 (푥) exists finitely in [a, b] and 푥1, 푥2, … . 푥푛 are n-points in this interval. Further let 푎1, 푎2, … . 푎푛 be n positive numbers. Then 푎 푥 +푎 푥 +⋯푎 푥 푎 푓(푥 )+푎 푓(푥 )+⋯푎 푓(푥 ) 푓 ( 1 1 2 2 푛 푛) ≤ 1 1 2 2 푛 푛 . 푎1+푎2+⋯푎푛 푎1+푎2+⋯푎푛 Proof. Since f is convex and 푓′′(푥) exists finitely in [a, b], therefore 푓′′(푥) ≥ 0. 푛 푎1푥1+푎2푥2+⋯푎푛푥푛 ∑푖=1 푎푖푥푖 Let β = = 푛 . 푎1+푎2+⋯푎푛 ∑푖=1 푎푖 54 푛 푛 This implies that ∑푖=1 푎푖푥푖 − 훽 ∑푖=1 푎푖 = 0 (29) By Taylors theorem on f(x) defined in [a,b], we have (푥 −훽)2 푓(푥 ) = 푓(훽) + (푥 − 훽)푓′(훽) + 푖 푓′′(퐶 ) 푖 푖 2 푖 푤here β <퐶푖 ≤ 푥푖 푓표푟 푎푙푙 푖 = 1,2,3, … . . , 푛 . Multiplying both sides by ai, 푤푒 푔푒푡 푎 (푥 −훽)2 푎 푓(푥 ) = 푎 푓(훽) + (푎 푥 − 푎 훽)푓′(훽) + 푖 푖 푓′′(퐶 ) for 푖 = 1,2,3, … 푛 . 푖 푖 푖 푖 푖 푖 2! 푖 푎 (푥 −훽)2 Since 푖 푖 푓′′(퐶 ) ≥ 0. 2! 푖 ′ Therefore, 푎푖 푓(푥푖) ≥ 푎푖푓(훽) + (푎푖푥푖 − 푎푖훽)푓 (훽). By adding, we get 푛 푛 푛 ′ ∑푖=1 푎푖 푓(푥푖) ≥ 푓(훽) ∑푖=1 푎푖 + (∑푖=1(푎푖푥푖 − 푎푖훽)푓 (훽)) . By (29) coefficient 푓′(푥) is zero. 푛 푛 Therefore ∑푖=1 푎푖 푓(푥푖) ≥ 푓(훽) ∑푖=1 푎푖 . 푛 ∑푖=1 푎푖 푓(푥푖) This implies that 푓(훽) ≤ 푛 . ∑푖=1 푎푖 푛 푛 ∑푖=1 푎푖푥푖 ∑푖=1 푎푖 푓(푥푖) This gives 푓 ( 푛 ) ≤ 푛 which is required Inequality. ∑푖=1 푎푖 ∑푖=1 푎푖 Deduction from Jensen’s Inequality. Consider the function 푓(푥) = −푙표푔 푥 , 푥 > 0 . 1 Then 푓′′(푥) = > 0, 푥 > 0 . 푥2 Therefore f is convex function for all positive x. Let 푡1, 푡2, … . , 푡푛 be positive numbers and 푎1, 푎2, … . 푎푛 be positive numbers. Then by Jensen’s Inequality 55 푎 푡 +푎 푡 +⋯푎 푡 푎 푓(푡 )+푎 푓(푡 )+⋯푎 푓(푡 ) 푓 ( 1 1 2 2 푛 푛) ≤ 1 1 2 2 푛 푛 푎1+푎2+⋯푎푛 푎1+푎2+⋯푎푛 This implies that 푎 푡 +푎 푡 +⋯푎 푡 −[푎 푙표푔푡 +푎 푙표푔푡 +⋯+푎 푙표푔 푡 ] – 푙표푔 푥 [ 1 1 2 2 푛 푛] ≤ 1 1 2 2 푛 푛 푎1+푎2+⋯푎푛 푎1+푎2+⋯푎푛 This implies 1 푎 푡 +푎 푡 +⋯푎 푡 1 1 2 2 푛 푛 푎1 푎2 푎푛 푎 +푎 +⋯푎 푙표푔 푥 [ ] ≥ 푙표푔(푡1 . 푡2 … . 푡푛 ) 1 2 푛 푎1+푎2+⋯푎푛 1 푎 푡 +푎 푡 +⋯푎 푡 1 1 2 2 푛 푛 푎1 푎2 푎푛 푎 +푎 +⋯푎 This implies that ≥ (푡1 . 푡2 … . 푡푛 ) 1 2 푛 (30) 푎1+푎2+⋯푎푛 Set 푎1 = 푎2 = ⋯ 푎푛 = 퐼 ,so we get from (30) 1 푡1+푡2+⋯푡푛 ≥ (푡 , 푡 , … . , 푡 )푛 . 푛 1 2 푛 퐴. 푀 ≥ 퐺. 푀. Holder’s Inequality and Minkowski’s Inequality 3.19. 1 1 If 1 < 푝 < ∞ and + = 1 and 푎 , 푏 , 푗 = 1,2,3, … . , 푛 are real numbers , then 푝 푞 푗 푗 1 1 푛 푛 푝 푝 푛 푞 푞 ∑푗=1|푎푗푏푗| ≤ (∑푗=1|푎푗| ) (∑푗=1|푏푗| ) (a) Proof. For the proof, we first prove Lemma. 1 1 Lemma. If 1 < 푝 < ∞ and q be such that + = 1 , then for any non- 푝 푞 negative real numbers a and b, we have 푎푝 푏푞 푎푏 ≤ + . 푝 푞 Proof of lemma. If 푏 = 0 , then there is nothing to prove. Let then b > 0, then for 푡 ∈ 푅 . 1 1 1 Let 푓(푡) = + ( ) 푡 − 푡푝 . 푞 푝 56 1 1 −1 Then 푓′(푡) = [1 − 푡푝 ] 푝 1 1 − = [1 − 푡 푞] 푝 so that for 푡 < 1 , 푓′(푡) < 0 푎푛푑 푓표푟 푡 > 1 , 푓′(푡) > 0. Also 푓′(푡) = 0 . 1 1 1 −1 This implies 푡 = 1 푎푛푑 푓′′(푡) = [ 푡푞 ] 푝 푞 1 1 − 1 = 푡 푝 = 푓표푟 푡 = 1 . 푝푞 푝푞 That is 푓′′(푡) > 0, 푓표푟 푡 = 1 . f(t) has minimum value at 푡 = 1 . 1 1 1 Thus 푓(푡) ≥ 푓(1) , ∀ 푡 ∈ 푅 this implies that + 푡 − 푡푝 ≥ 0 . 푞 푝 1 1 1 푡푝 ≤ + ( ) 푡. 푞 푝 푎푝 Letting 푡 = , we get 푏푞 1 푎푝 1 1 푎푝 ( )푝 ≤ + ( ) . 푏푞 푞 푝 푏푞 푎 1 1 푎푝 푞 ≤ + ( ) 푞 푝 푏푞 푏푝 푎푏푞 푏푞 푎푝 푞 ≤ + 푞 푝 푏푝 푞 푞− 푏푞 푎푝 푎푏 푝 ≤ + 푞 푝 푏푞 푎푝 1 1 푞 푎푏′ ≤ + (because + = 1 . Implies that + 1 = 푞) . 푞 푝 푝 푞 푝 57 푏푞 푎푝 Therefore, 푎푏 ≤ + . (31) 푞 푝 This completes the lemma. Proof of the theorem. 1 푛 푝 푝 Let α = (∑푗=1|푎푗| ) 1 푛 푞 푞 β = (∑푗=1|푏푗| ) (32) If either 훼 = 0 or 훽 = 0 , then both sides of the Inequality (a) are zero. Let then 훼 ≠ 0 , 훽 ≠ 표 f표푟 푗 = 1,2,3, … . 푛. |푎 | |푏 | Letting 푎 = 푗 and 푏 = 푗 , then from (31), we have 훼 훽 푃 푞 |푎 | |푏 | 1 |푎 | 1 |푏 | ( 푗 ) ( 푗 ) ≤ ( 푗 ) ( 푗 ) . 훼 훽 푃 훼 푞 훽 Taking summation on both sides, we get 푝 1 ∑푛 푛 푞 푗=1|푎푗| 1 ∑ |푏푗| ∑푛 |푎 | |푏 | ≤ αβ [푝 + 푗=1 ]. 푗=1 푗 푗 훼푝 푞 훽푞 This implies that 1 훼푝 1 훽푞 ∑푛 |푎 | |푏 | ≤ 훼훽 [ . + . ] by (32) gives 푗=1 푗 푗 푝 훼푝 푞 훽푞 1 1 ∑푛 |푎 | |푏 | ≤ 훼훽 ( + ) = 훼훽 푗=1 푗 푗 푝 푞 1 1 푛 푝 푝 푛 푞 푞 = (∑푗=1|푎푗| ) (∑푗=1|푏푗| ) . This completes the Holder’s Inequality. Minkowski’s Inequality 3.20. 퐼푓 푎푗, 푏푗, 푗 = 1,2,3, … , 푛 are real numbers, and 1 < 푝 < ∞, 1 < 푞 < ∞ , 1 1 such that + = 1 , 푡ℎ푒푛 푝 푞 58 1 1 1 푛 푝 푝 푛 푝 푝 푛 푞 푞 (∑푗=1|푎푗 + 푏푗| ) ≤ (∑푗=1|푎푗| ) +(∑푗=1|푏푗| ) . 푛 푝 Proof. Let Y = ∑푗=1(|푎푗| + |푏푗|) 푛 푝−1 = ∑푗=1(|푎푗| + |푏푗|) (|푎푗| + |푏푗|) 푛 푝−1 푛 푝−1 = ∑푗=1|푎푗| (|푎푗| + |푏푗|) + ∑푗=1|푏푗| (|푎푗| + |푏푗|) 1 1 푛 푝 푝 푛 (푝−1)푞 푞 ≤ (∑푗=1|푎푗| ) . (∑푗=1(|푎푗| + |푏푗|) ) + 1 1 푛 푝 푝 푛 (푝−1)푞 푞 + (∑푗=1|푏푗| ) . (∑푗=1(|푎푗| + |푏푗|) ) . By Holder’s Inequality 1 1 푝 푝 1 푛 푝 푛 푝 푞 = ((∑푗=1|푎푗| ) + (∑푗=1|푏푗| ) ) 푌 (33) as (푝 − 1)푞 = 푝 Thus 1 1 1 푝 1− 푛 푝 푝 푞 (∑푗=1|푎푗 + 푏푗| ) ≤ 푌 = 푌 1 1 1 − 푝 푝 ′ 푞 푛 푝 푛 푝 = 푌 . 푌 ≤ ((∑푗=1|푎푗| ) + (∑푗=1|푏푗| ) ) 1 1 − = 푌푞 . 푌 푞 푏푦 (33). Therefore 1 1 1 푛 푝 푝 푛 푝 푝 푛 푞 푞 (∑푗=1|푎푗 + 푏푗| ) ≤ (∑푗=1|푎푗| ) +(∑푗=1|푏푗| ) . This completes the proof. Holder’s Inequality from Jensen’s Inequality 3.21. Consider a function 푓(푥) = 푥푞 , 푥 > 0, 푞 > 1 , 푡ℎ푒푛 푓′′(푥) = 푞(푞 − 1)푥푞−2 > 0 . Therefore f is convex functions. ′ Let 훼1, 훼2, … , 훼푛 and 훽1, 훽2, … 훽푛 be all positive , then by Jensen s Inequality 59 훼 훽 +훼 훽 +⋯훼 훽 훼 푓(훽 )+훼 푓(훽 )+⋯+훼 푓(훽 ) 푓 ( 1 1 2 2 푛 푛) ≤ 1 1 2 2 푛 푛 훼1+훼2+⋯+훼푛 훼1+훼2+⋯+훼푛 푛 ∑푗=1 훼푗훽푗 훼푗푓(훽푗) This implies that 푓 [ 푛 ] ≤ ( 푛 ) . ∑푗=1 훼푗 ∑푗=1 훼푗 푛 푞 푞 ∑푗=1 훼푗훽푗 훼푗훽푗 [ 푛 ] ≤ 푛 ∑푗=1 훼푗 ∑푗=1 훼푗 1 ∑푛 푞 푗=1 훼푗훽푗 푛 훼푗훽푗 푞 푛 ≤ [∑푗=1 ( 푛 )] ∑푗=1 훼푗 ∑푗=1 훼푗 1 − 푛 푛 푛 푞 푛 푞 Thus ∑푗=1 훼푗훽푗 ≤ ∑푗=1 훼푗 (∑푗=1 훼푗) .∑푗=1 훼푗훽푗 . This implies that 1/푝 1 1 ∑푛 훼 훽 ≤ (∑푛 훼 ) . ∑푛 훼 훽 푞 where + =1 . 푗=1 푗 푗 푗=1 푗 푗=1 푗 푗 푝 푞 푝 푞 푞 Set 훼푗 = 푎푗 , and 훼푗훽푗 = 푏푗 ,so that 푞 푞 푏푗 훽푗 = 푝 . 푎푗 1 푞 푝 푏푗 푞 Therefore 훼푗훽푗 = 푎푗 . ( 푝) 푎푗 푝 푝− 푞 = 푎푗 푏푗 = 푎푗푏푗 푛 푛 푛 푝 푝 푛 푞 푞 Therefore ∑푗=1 훼푗훽푗 = ∑푗=1 푎푗푏푗 ≤ (∑푗=1 푎푗 ) (∑푗=1 푏푗 ) 1 1 where + = 1 푝 푞 which is Holder’s Inequality . Problem. If 푎, 푏, 푐, are positive and 푎 + 푏 + 푐 = 1 , then prove that ퟏ 1 1 ( − 1) ( − 1) ( − 1) ≥ 8 , when does equality hold. 풂 푏 푐 60 퐒퐨퐥퐮퐭퐢퐨퐧. We have ퟏ 1 1 (1−푎)(1−푏)(1−푐) ( − 1) ( − 1) ( − 1) = 풂 푏 푐 푎푏푐 푏+풄 푐+푎 푎+푏 = . . 푎 푏 푐 2√푏푐. 2√푎푐. 2√푎푏 ≥ 푎푏푐 8푎푏푐 푥 + 푦 = ( as ≥ √푥푦 푎푏푐 2 1 1 1 Thus ( − 1) ( − 1) ( − 1) ≥ 8 and equality holds if and only if 푎 푏 푐 푎 = 푏 = 푐. 1 But then ( − 1)3 = 8 implis 푎 1 ( − 1) = 2 푎 1 푎 = . 3 Thus equality holds if and only if 푎 = 푏 = 푐 . Problem. Prove that the volume of the maximum rectangular parallelepiped which can be inscribed in the ellipsoid x 2 y 2 z 2 8 abc 1 is given by . 2 2 2 a b c 3 3 Solution. Let the semi edges of the rectangular box be x, y, z then its volume is 8xyz.Thus, V= 8xyz . We have to maximize V subject to the condition that x 2 y 2 z 2 1 . 2 2 2 a b c 61 We have 2 2 2 1 x y z 2 2 2 3 x y z a 2 b 2 c 2 1 . 2 2 2 a b c 3 1 x 2 y 2 z 2 3 1 2 2 2 Thus a b c 3 2 x . y . z 3 1 or a .b .c 3 1 x . y . z 1 2 or a.b .c 27 abc or x y z . 3 3 8 abc Hence the required maximum volume is given by . 3 3 62 Functions of several variables. We already know about the functions of a single independent variable and their related concepts with regard to their limits, continuity, differentiability etc. In this unit we will be discussing the functions of several variables and their characteristic properties. Definitions 4.1. Consider the set of n independent variables x 1 , x 2 , x 3 , . . . x n and one dependent variable u, then the equation u f x 1 , x 2 . . . . x n denotes the functional relation and is known as a function of several variables. In this case are n arbitrary assigned variables, the corresponding values of the dependent variable u is determined by the function relation. The function represented above is an explicit function but where several variables are concerned it is rarely possible to obtain an equation expressing one of the variables explicitly in terms of the others. Thus most of the functions of more than one variable are implicit functions, that is to say we are given a functional relation x 1 , x 2 . . . . x n 0 connecting the n variables, it is not in general possible to solve this equation to find an explicit function which expresses one of these variable say x, in terms of the other n-1 variables. Limits and Continuity of functions of two or more variables. Let u=f(x, y) be a function of two independent variables x and y which is defined in some domain D R 2 . Let a , b D . We say limit f x , y exists as x , y a , b and is equal to l and we write lim f x , y l . x , y a ,b 63 If given 0 , howsoever small we can find a positive number , such that f x, y l when ever x l , y l We should note that x , y can tend to a , b in any manner i.e., along any path and the value of f(x, y) is independent of the path choosen joining the point x , y to the po int a , b . 2 xy Example 4.1. Let f x , y as x , y 0 , 0 , find f x , y . 2 2 lim x y x , y a , b Solution. We approach the origin (0, 0) along the path y=mx. Then, y 0 as x 0 and we have along this path lim f x , y lim f x , mx x , y 0 , 0 x 0 2 x mx lim 2 2 2 x 0 x m x 2 m 2 m which depends on m and is therefore different values of m . lim 2 2 x 0 1 m 1 m Hence we conclude that lim f x , y does not exist. x , y a , b x 2 y Example 4.2. Let f x , y as x , y 0, 0 , find f x , y . 4 2 lim x y x , y a , b Solution. We approach the origin (0, 0) along the path y mx 2 . Then 64 2 lim f x , y lim f x , mx x , y 0 , 0 x 0 x mx 2 lim 4 2 4 x 0 x m x m 2 m which depends on m and is therefore different values of m . lim 2 2 x 0 1 m 1 m Hence we conclude that lim f x , y does not exist. x , y a , b 2 xy 2 Example 4.3. Let f x , y as x , y 0, 0 , find f x , y . 2 4 lim x y x , y a , b Solution. We approach the origin (0, 0) along the path y mx . Then y 0 as x 0 and we have along this path lim f x , y lim f x , mx x , y 0 , 0 x 0 2 x mx lim 2 2 2 x 0 x m x 2 x 2 m 2 m which depends on m and is therefore different values of m . lim 2 2 2 x 0 x (1 m ) 1 m Hence we conclude that does not exist. 2 xy Example 4.4. Let f x , y as x , y 0, 0 , find f x , y . 2 2 lim x y x , y a , b Solution. We approach the origin (0, 0) along the path y mx . 65 Then y 0 as x 0 and we have along this path lim f x , y lim f x , mx x , y 0 , 0 x 0 2 x mx lim 2 2 2 x 0 x m x 2 x 2 m 2 xm 0 . lim 2 2 lim 2 x 0 x 1 m x 0 1 m . This shows that lim f x , y exists and is equal to 0. We now show x , y a ,b that lim f x , y 0 by showing that x , y a ,b f x, y 0 for x 0 and y 0 i.e., we show that f x, y for x and y , we have 2 x y 0 for x , y real or x 2 y 2 2 xy 0 2 xy x 2 y 2 2 xy 1 x 2 y 2 2 xy x 2 y 2 x 2 y 2 2 xy Hence f x , y x 2 y 2 x 2 y 2 i x 2 y 2 We choose x , y . 2 2 Now, by equation (i),we have 66 2 2 f x , y 0 x 2 y 2 2 . 2 2 Thus, f x , y 0 for x 0 and y 0 . 2 2 Therefore, lim f x , y 0 . x , y a ,b Definition 4.2. The limit lim lim f x , y ; lim lim f x , y are x a y b y b x a known as repeated limits whereas the limit . The lim f x , y is known as the simultaneous limit or a double limit. x , y a ,b Example 4.5. Show that for the following functions the two repeated limits exist at (0, 0) and are unequal but simultaneous limit or double limit does not exists. x y (i) f x , y x , y 0 , 0 . x y x 2 y 2 (ii) f x , y x , y 0, 0 . x 2 y 2 Solution. (i) We have x y lim lim f x , y lim lim x 0 y 0 x 0 y 0 x y x y lim x 0 x y lim 1 1 . x 0 67 Again, x y lim lim f x , y lim lim y 0 x 0 y 0 x 0 x y x y lim y 0 x y lim 1 1 . y 0 Hence lim lim f x , y lim lim f x , y . x a y b y b x a Hence, the two repeated limits exist but are not equal. To see that the simultaneous limit exists or not we approach the origin along the path y mx . Then y 0 as x 0 and we have along this path lim f x , y lim f x , mx x , y 0 , 0 x 0 x mx lim x 0 x mx 1 m 1 m lim x 0 1 m 1 m which depends on m and is therefore different values of m . Hence the simultaneous limit does not exist. (ii) This is left to the student as an exercise. Continuity of functions of two or more variables. Let u=f(x, y) be a function of two independent variables x and y which is defined in some domain D R 2 . Let a , b D . 68 A function f x , y is said to be continuiou s at a po int a , b if given 0 , howsoever small we can find a positive number , such that f x, y f a , b for x a , y b In other words the function f x , y is said to be continuiou s at a po int a , b D if lim f x , y f a , b . The function f x , y is said to be continuiou s in D if it is x , y a ,b continuous at all points of D. Example 4.6. Discuss the continuity of the function xy x 2 y 2 f x , y as x , y 0 , 0 x 2 y 2 0 as x , y 0 , 0 . Solution. We show the function f x , y is continuiou s at a po int 0 ,0 by showing that lim f x , y f 0 , 0 0 . x , y a ,b In fact we show that f x, y f 0 , 0 for x 0 , y 0 or f x, y f 0 , 0 f x , y for x , y 2 2 xy x 2 y 2 xy x y We have f x , y 2 2 x y x 2 y 2 2 2 xy x y x 2 y 2 x y x 2 y 2 x y x 2 y 2 69 Thus, f x, y x y . We choose . f x , y x y for x and y . Therefore f x, y f 0 , 0 for x 0 , y 0 . This shows that lim f x , y f 0 , 0 0 . x , y 0 , 0 Hence the function f x , y is continuiou s at a po int 0 ,0 . Example 4.7. Discuss the continuity of the function x 2 y 2 f x , y x , y 0 , 0 x 4 y 4 x 2 y 2 0 x , y 0 , 0 . Solution. Here f 0 , 0 0 . To see whether lim f x , y exists or not, we approach the origin x , y 0 , 0 along a path y mx , then y 0 as x 0 and we have along this path lim f x , y lim f x , mx x , y 0 , 0 x 0 x 2 m 2 x 2 lim 4 4 4 4 2 x 0 x m x x m m 2 lim 4 2 x 0 1 m m which depends on m and is therefore different values of m . 70 Therefore, lim f x , y does not exist. x , y 0 , 0 Hence the function f x , y is discontinuous at the origin. Example 4.8. Discuss the continuity of the following function at the origin 1 1 f x , y x sin y sin ; x , y 0 , 0 y x 0 ; x , y 0 , 0 . Solution. To see whether exists or not, we approach the origin along a path y mx , then y 0 as x 0 and we have along this path lim f x , y lim f x , mx x , y 0 , 0 x 0 1 1 lim x sin m x sin x 0 mx x 0 . Therefore exists as x , y 0 , 0 and is equal to zero. Now 1 1 f x , y f 0 , 0 f x , y 0 f x , y x sin y sin y x x y . We choose x and y . 2 2 Therefore the function is continuous at the origin. Partial differential of function of two or more variables. Let u f x , y be a function of two independent variables x and y which is defined in some domain D R 2 , then partial derivative of 71 f x , y with respect x at a point a , b where a , b D which is denoted f by a , b or f x a , b is defined as x f f a h , b f a , b a , b or f a , b x lim x h 0 h Similarly the partial derivative of with respect y at a point f where which is denoted by a , b or f y a , b is defined y as f f a , b k f a , b a , b or f a , b y lim y k 0 k More generally if f x 1 , x 2 , . . . x n is a function of n independent variables. Then the partial derivative of f with respect x i i 1 , 2 , 3, . . .n at the point x 1 , x 2 , . . . x n is defined by f a , a , a , . . . a or f a , a , a , . . . a and is defined as 1 2 3 n x i 1 2 3 n x i f a , a . . . a h . . . a f a , a , a , . . . a lim 1 2 i n 1 2 3 n h 0 h for i 1 , 2 , . . . n . x 3 y 3 f x , y x , y 0 , 0 Example 4.9. A function x y 0 x , y 0 , 0 Show that first order partial derivative of f x , y at a po int 0 , 0 exists but the function f x , y is discontinu ious at a po int 0 ,0 . 72 Solution. We have f f 0 h , 0 f 0 , 0 0 ,0 lim x h 0 h f h , 0 f 0 , 0 lim h 0 h h 3 0 h lim h 0 h lim h 0 . h 0 Similarly f f h , 0 k f 0 , 0 0 ,0 lim x k 0 k f 0 , k f 0 , 0 lim k 0 k k 2 0 lim k 0 k lim k 0 . k 0 Hence, the first order partial derivative exist at (0,0) and are both equal to zero. To see whether lim f x , y exists or not we approach the origin x , y 0 , 0 along the curve y x mx 3 , then y 0 as x 0 , and we have along this curve 3 lim f x , y lim f x , mx mx x , y 0 , 0 x 0 3 x 3 x m x 2 lim 3 x 0 x x m x 2 1 1 m x 2 lim x 0 m m which depends on m and is therefore different values of m . 73 Therefore, lim f x , y does not exist and hence the function f x , y is x , y 0 , 0 discontinuous at the origin. Total differentiation. Let u f x , y be a function of two independent variables x and y, then u u du dx dy x y f f i.e., df dx dy x y f f where and are continuous functions of x an y . x y Proof. We have u f x , y . Let x and y receive simultaneous increments x and y respectively. Let the corresponding increments of u be u ,then we have u u f x x , y y u f x x , y y f x , y f x x , y y f x , y y f x , y y f x , y i Thus we have expressed u as a sum of two differences in which for the first difference y remains constant at the value y y and x varies from x to x x . In the second difference x remains constant at the value x and y varies from y to . By Mean value theorem, we have f x x , y y f x , y x f x x 1 x , y y where 0 1 1 74 and f x , y y f x , y y f y x , y 2 y where 0 2 1 Using this in equation (i), we get u x f x x 1 x , y y y f y x , y 2 y Since f x and f y are continuous functions of x and y in the domain considered. f x x , y y f x , y lim x 1 x x , y 0 , 0 and f x , y y f x , y lim y 2 y . x , y 0 , 0 Hence, we have f x x 1 x , y y f x x , y 1 where 1 0 as x , y 0 , 0 and f y x , y 2 y f y x , y 2 where 2 0 as x , y 0 , 0 Using this equation in (ii), we get f u x [ f x 1 ] y [ f y 2 ] f f x y x 1 y 2 x y where 1 0 , 2 0 as x , y 0 , 0 . The principle part in the increment is called the total differential of u with respect to x and y and is denoted by du. Hence, we have f f d u x y ii x y 75 u u If in particular, we take u=x, then 1 , 0 and so du x . Also x y du dx , since u=x. Hence dx x . Similarly we can have dy y by putting u=y. Hence (ii) becomes f f du dx dy x y f f or df dx dy x y Remark. If u f x , y is a function of two variables in x and y and these variables are functions of a variable t, then du u dx u dy dt x dt y dt and if x t , s and y t , s , then du u x u y dt x t y t du u x u y . ds x s y s f f Mean Value Theorem. If f x , y , , are all continuous x y functions in a circular domain D of Centre a , b and radius large enough for the point a h , b k to be within D , then f a h , b k f a , b h f x a h , b k k f y a h , b k where 0 1 . Proof. Consider a function g t F a th , b tk where 0 t 1 . Then g t is a continuous function in the closed interval [0, 1] and differential in the open interval (0, 1), therefore by Lagrange’s Mean Value Theorem, we have 76 g 1 g 0 1 0 g / where 0 t 1 . or g 1 g 0 g / where 0 t 1 . But, g 1 F a h , b k Thus F a h , b k F a , b g / where 0 t 1 i Now g t F a t h , b t k F x , y x a t h where y a t k dx dy Thus h and k dt dt Therefore dg f dx f dy dt x dt y dt f x x , y h f y x , y k h f x a th , b tk k f y a th , b tk Using this in eq. (i), we get F a h , b k F a , b h f x a h , b k k f y a h , b k where 0 1 . u u Theorem 4.1. If u f x , y then du dx dy x y holds no matter what the independent variables be. In other words holds regardless of whether x, y are independent or dependent variables. 77 Proof. Let u f x , y be a function of two variables x and y. If x, y are independent variables, then we have already proved that u u du dx dy . x y Now suppose x, y are dependent variables, say x r , s y r , s u f x , y f r , s , r , s Then f r , s u u du dr ds where r , s are independen t var iables . r s We show u u u s dx dy dr ds x y r s Now , u u x u y . . r x r y r u u x u y and . . s x s y s Substituting the values, we get u u u x u y u x u y dr ds . . dr . . ds r s x r y r x s y s 78 u x x u y y . dr .ds dr ds x r s y r s u u dx dy x y x x because dx dr ds r s y y and dy dr ds r s u u Thus du dx dy x y and the theorem is completely proved. Differentiability of functions of two variables. Let u f x , y be a function of two variables in x and y which is defined in some domain D R 2 . Let a , b D , then f f f a h , b k f a , b h a , b k a , b h 2 k 2 h , k x y where h , k 0 , 0 as h , k 0 .0 . Example 4.10. Discuss the continuity and differentiability of the function xy f x , y as x , y 0 , 0 x 2 y 2 0 as x y 0 at the origin . Solution. We first show that the function f x , y is continuous at 0 , 0 . 79 We approach the origin along a path y mx , then y 0 as x 0 and we have along this path lim f x , y lim f x , mx x , y 0 , 0 x 0 x mx lim 2 x 0 x 1 m mx 0 . lim 2 x 0 1 m This shows that lim f x , y as x , y 0 , 0 exists along the path x , y 0 , 0 and is equal to 0. We now show that lim f x , y f 0 , 0 0 . x , y 0 , 0 We have f x , y f 0 , 0 f x , y 0 f x , y xy i x 2 y 2 2 Now x y 0 for x , y real or x 2 y 2 2 xy 0 x 2 y 2 2 xy x 2 y 2 xy 2 x 2 y 2 xy 2 80 2 2 xy x y x 2 y 2 2 We choose 2 for x 2 and y 2 Now from eq. (i), we have x 2 y 2 2 2 2 2 f x , y . 2 2 Hence f x , y f 0 , 0 for x 0 2 and y 0 2 We now show that f x , y is not differentiable at (0,0). f x , y will be differentiable at (0,0) if f f f 0 h ,0 k f 0 , 0 h 0 , 0 k 0 , 0 h 2 k 2 h , k x y where h , k 0 , 0 as h , k 0 .0 f f or f h , k f 0 , 0 h 0 , 0 k 0 , 0 h 2 k 2 h , k ii x y where h , k 0 , 0 as h , k 0 .0 Now f f 0 h , 0 f 0 , 0 0 ,0 lim x h 0 h f h , 0 f 0 , 0 lim h 0 h 0 0 lim 0 . h 0 h Similarly 81 f f 0 , 0 k f 0 , 0 0 ,0 lim y h 0 k f 0 , k f 0 , 0 lim h 0 k 0 0 lim 0 . h 0 k Hence from eq. (ii), f x , y will be differentiable at (0, 0) if hk 0 h . 0 k . 0 h 2 k 2 h , k h 2 k 2 h , k h 2 k 2 hk where h , k 0 as h , k 0 , 0 and hence h , k . h 2 k 2 Here we shall approach the origin along the path k mh then k 0 as h 0 , we get lim h , k lim h , mh h , k 0 , 0 x 0 h mh lim 2 2 x 0 h k m 1 m 2 which depends on m and is different for different values of m. Hence lim h , k does not exist and as such f x , y is not differentiable at h , k 0 , 0 (0, 0). Second and higher order partial derivatives. Let u f x , y be a function of two variables in x and y which is defined in some domain D R 2 , then 82 2 f 2 f 2 f 2 f f , f , f , f x x x y y x y y x 2 xy y x y 2 are called second order partial derivatives of the function f. 2 f 2 f In general . x y y x f f a h , b f a , b We have a , b lim f x a , b x h 0 h f f a , b k f a , b a , b lim f y a , b y k 0 k 2 f f f y a h , b f y a , b Now f y a , b lim x y x h 0 h 1 f a h , b k f a h , b f a , b k f a , b lim { lim lim } h 0 h k 0 k k 0 k 1 lim lim f a h , b k f a h , b f a , b k f a , b h 0 k 0 hk 2 f 2 f h , k Thus f xy lim lim x y h 0 k 0 hk where 2 f h , k f a h , b k f a h , b f a , b k f a , b . 2 f f f x a , b k f x a , b Similarly f x a , b lim y x y h 0 k 1 f a h , b k f a , b k f a h , b f a , b lim { lim lim } h 0 k k 0 h k 0 h 1 lim lim f a h , b k f a , b k f a h , b f a , b k 0 h 0 hk 2 f 2 f h , k Thus f yx lim lim . y x k 0 h 0 hk 83 Since, in general 2 f h , k 2 f h , k lim lim lim lim . h 0 k 0 hk k 0 h 0 hk Therefore in general f xy f yx . xy x 2 y 2 f x , y as x , y 0 , 0 Example 4.11. Let x 2 y 2 0 as x y 0 . Show that f xy f yx at 0 , 0 . Solution. We have for x , y 0 , 0 . x 3 y xy 3 f x , y 2 2 x x x y x 2 y 2 3 x 2 y y 3 x 3 y xy 3 2 x 2 x 2 y 2 x 3 y xy 3 f x , y and 2 2 y y x y x 2 y 2 x 3 3 xy 2 x 3 y xy 3 2 y 2 x 2 y 2 f 0 h , 0 f 0 , 0 Also f x 0 , 0 lim h 0 h 0 0 lim 0 . h 0 h f 0 , 0 k f 0 , 0 and f y 0 , 0 lim k 0 k 0 0 lim 0 . k 0 k 84 f y 0 h , 0 f y 0 , 0 Therefore f x x 0 , 0 f y 0 , 0 lim x h 0 h h 3 0 3 h 2 h lim lim 1 . 3 h 0 h h 0 h f x 0 , 0 k f x 0 , 0 Also f y y 0 , 0 f x 0 , 0 lim y k 0 k k 3 0 3 k 2 k lim lim 1 . 3 h 0 k h 0 k Hence f xy 0 , 0 f yx 0 , 0 . Sufficient conditions for the equality of f xy and f yx . We now give two theorems which we show that under what conditions f xy and f yx are equal at a certain point which are known as sufficient conditions for equality of Theorem 4.2 (Young’s Theorem). If f x and f y exist in the neighborhood of the point (a, b) and are differentiable at the point (a, b), then f xy f yx at a , b . Proof. We shall prove this result by taking equal increment h for both x and y and calculate 2 f in two different ways where 2 f f a h , b h f a h , b f a , b h f a , b h k 85 Now, consider the function H x f x , b h f x , b Then H a h f a h , b h f a h , b so that H a h H a f a h , b h f a h , b f a , b h f a , b 2 f . Thus 2 f H a h H a . Since f x exists in the neighborhood of the point (a, b); we apply Lagrange’s mean value theorem to H(x) in the interval (a, a +h), we get H a h H a h H / a h where 0 1 = h f x a h , b h f x a h , b i Since is differentiable at (a,b), then 2 2 f x a h , b h f x a , b h f x x h f y x h 1 where 1 0 as h 0 and f x a h , b f x a , b h f x x h 2 where 2 0 as h 0 Hence 2 f x a h , b h f x a h , b h f y x h [ 1 1 2 ] 2 h f y x h where 1 1 2 and 0 as h 0 . Using this in eq. (i), we get 2 2 f h [ f y x ] where 0 as h 0 . and therefore 86 2 f lim f at a , b ii 2 y x h 0 h We now consider the function K y f a h , y f a , y then K b h f a h , b h f a , b h , so that K b h K b f a h , b h f a , b h f a h , b f a , b 2 f By the same procedure as done earlier, we get 2 f lim f at a , b iii 2 x y h 0 h From eqs. (ii) and (iii), we get f x y f y x at a , b . This proves the Theorem. Theorem 4.3 (Schwartz Theorem). If f x and f y and f y x all exist in the neighborhood of the point (a, b) and f y x is continuous at the point (a, b). Then f x y also exists and f xy f yx at a , b . Proof. We have 2 f h , k f x y lim lim h 0 k 0 hk 2 f h , k and f y x lim lim . k 0 h 0 hk 87 where 2 f h , k f a h , b k f a h , b f a , b k f a , b Now, consider the function H a f a , b k f a , b Then H a h f a h , b k f a h , b so that H a h H a f a h , b k f a h , b f a , b k f a , b 2 f h , k . Thus 2 f h , k H a h H a . Since f x exists in the neighborhood of the point (a, b); we apply Lagrange’s mean value theorem to H(x) in the interval (a, a +h), we get 2 f h , k H a h H a h H / a h where 0 1 = h f x a h , b k f x a h , b i Since f y x exists in the neighborhood of the point (a, b); we apply Lagrange’s mean value theorem again to the R.H.S of eq.(i) and get = h k f y x a 1 h , b 2 k } where 0 1 1 & 0 2 1 ii Since is continuous at (a,b), then lim f y x a 1 h , b 2 k f y x a , b h , k 0 , 0 and hence f y x a 1 h , b 2 k f y x a , b where 0 as h , k 0,0 . Using this in eq. (ii), we get 88 2 f h , k h k [ f y x ] where 0 as h , k 0 ,0 . 2 f h , k or f y x where 0 as h , k 0 ,0 . hk We first take limits when k 0 and then when h 0 , we get 2 f h , k lim lim f y x h 0 k 0 hk 2 f h , k But f x y lim lim . h 0 k 0 hk and thus f x y f y x at a , b . This proves the result. Change of variables. Let u f x , y be a function of two variables in x and y, then u u du dx dy x y 2 u u d u d du d dx dy x y u u d dx d dy x y u u u u 2 2 d dx d x d dy d y x x y y 2 2 2 2 u 2 u u u u 2 u dx dy dx d 2 x dx dy dy d 2 y x 2 y x x x y y 2 y 89 Thus 2 2 2 u 2 u u u 2 u d 2 u dx 2 dx dy d 2 x dy d 2 y x 2 x y x y 2 y Now, if x and y are independent variables, then dx, dy are constants so that d 2 x 0 and d 2 y 0 , so that 2 2 2 u 2 u u 2 d 2 u dx 2 dx dy dy . x 2 x y y 2 2 u 2 u Example 4.12. Prove that is invariant for change of x 2 y 2 rectangular axes. Solution. Let the axes turn through an angle , then x x / cos y / sin y x / sin y / cos where x , y are the co-ordinates of a point with respect to XOY and x / , y / are the co-ordinates of a point with respect to X / OY / . Thus, we have 2 2 2 u 2 u u u 2 u d 2 u dx 2 dx dy d 2 x dy d 2 y i x 2 x y x y 2 y 2 2 2 u 2 u u 2 d 2 u dx / 2 dx / dy / dy / ii 2 / / 2 x / x y y / x / , y / are independent variables so that d 2 x / 0 and d 2 y / 0 . Now 90 y x / sin y / cos dx dx / cos dy / sin dy d x / sin d y / cos and d 2 x 0 , d 2 y 0 . Putting these values in eq.(i), we get 2 2 u 2 u d 2 u dx / cos dy / sin 2 dx / cos dy / sin dx / sin dy / cos x 2 x y 2 u dx / sin dy / cos y 2 2 2 2 2 2 2 u u u 2 u u u 2 cos 2 2 cos sin sin 2 dx / sin 2 2 cos sin cos 2 dy / 2 2 2 2 x x y y x x y y B dx / dy / iii where B is the coefficient of dx / dy / . Now comparing eq. 9ii) and (iii), we get 2 2 2 2 u u u u 2 cos 2 2 cos sin sin 2 dx / iv / 2 2 2 x x x y y 2 2 2 2 u u u u 2 sin 2 2 cos sin cos 2 dy / v / 2 2 2 y x x y y Adding eq’s (iv) and (v), we have 91 2 2 u u 2 u 2 u = . 2 2 2 2 x / y / x y Thus is invariant for the change of rectangular axes. Maxima and Minima of function of two or more variables. Let u f x , y be a function of two variables in x and y which is defined in some domain D R 2 . Let a , b D . If f x , y f a , b for all points x , y belonging to a neighborhood of the point a , b , then f x , y is said to have a relative or local maximum at and if f x , y f a , b for all points belonging to a neighborhood of the point , then is said to have a relative or local minimum at . At a stationary point i.e., at an extreme point, we have f f 0 and 0 y x 2 2 2 f 2 f f 2 and d 2 f dx 2 dx dy dy . x 2 x y y 2 f f At a stationary point 0 . x yx 2 f 2 f dx 2 x y 2 x Now d f dx dy . 2 2 f f dy 2 x y y 92 If a , b is a stationary point of x , y then f x , y will have a relative minima at if d 2 f is positive i.e., d 2 f 0 . i.e., d 2 f is a positive definite quadratic form, which is possible if 2 2 2 2 2 f f f f 0 and . 0 . 2 2 2 x x y x y Now, will have a relative maxima at if is negative i.e., d 2 f 0 which is possible if 2 2 2 2 2 f f f f 0 and . 0 . 2 2 2 x x y x y Example 4.13. Find the maxima and minima of the function Z x 2 3 y 2 6 y . Solution. At a stationary point Z Z 2 x 0 and 6 y 6 0 y 1 . x y Thus x=0 and y= 1. 2 Z 2 Z 2 Z Now 2 , 0 and 6 . x 2 x y y 2 2 2 f 2 f 2 f Thus . 2 . 6 0 12 0 . 2 2 x y x y Hence Z x 2 3 y 2 6 y has a minima at the point (0, 1). Restricted Maxima and Minima. Lagrange’s method of undetermined multipliers for maxima and minima. 93 Let u f x 1 , x 2 , . . . , x n be a function of n variables which are connected with m equations of the form f 1 x 1 , x 2 , . . . , x n 0 f 2 x 1 , x 2 , . . . , x n 0 (I) . f m x 1 , x 2 , . . . , x n 0 The problem is to find the stationary values of subject to m given conditions. Lagrange’s method of undetermined multipliers consists of the following F x 1 , x 2 , x n f x 1 , x 2 , . . . , x n 1 f 1 2 f 2 . . . m f m where 1 , 2 , . . . m are multipliers. The stationary point of f may be found by determining the stationary points of F. At a stationary point of F, we have F F F 0 , 0 , . . . , 0 x 1 x 2 x n which gives 94 f f f f 1 2 . . . m 0 1 2 n x x x x 1 1 1 1 f f f f 1 2 . . . m 0 1 2 n x x x x 2 2 2 2 . II . . f f f f 1 2 m 1 2 . . . n 0 x x x x n n n n (II) are n equations out of which we shall find the value of m multipliers 1 , 2 , . . . m are put these values in the remaining (n-m) equations of the system (II). These (n-m) equations will be free of ' s . These (n-m) equations taken together with m equations of the system (I) or in all n-m+m=n equations, which are sufficient to determine the values of x 1 , x 2 , . . . , x n which will give rise to the stationary values of f. Example 4.14. Find the volume of the greatest rectangular parallelopiped that can be inscribed in the ellipsoid x 2 y 2 z 2 1 . a 2 b 2 c 2 Solution. Let x, y, z be the half of the sides of a required parallelepiped that can be inscribed in an ellipsoid Then the volume of parallelepiped is V 2 x. 2 y. 2 z 8 xyz . We have to find the maximum value V 8 xyz subject to the condition 95 x 2 y 2 z 2 1 . a 2 b 2 c 2 We consider the function x 2 y 2 z 2 F x , y , z 8 xyz 1 2 2 2 a b c At a stationary point of F, we have F 2 x 8 yz 0 i x a 2 F 2 y 8 xz 0 ii y b 2 F 2 z 8 xy 0 iii z c 2 Multiplying eq. (i) by x, eq. (ii) by y and eq. (iii) by z and then adding the result, we get x 2 y 2 z 2 24 xyz 2 0 2 2 2 a b c 24 xyz 2 1 0 or 12 xyz Putting this value of in eq. (i), we get 2 x 12 xyz 8 yz 0 a 2 a x 3 96 b c Similarly y and z . 3 3 Hence Volume (V) = 8xyz = 8 abc . 3 3 We show this volume is maximum by showing 2 f is negative. a b c i.e., 2 f 0 at , , . 3 3 3 a b c At a stationary point of x , y , z , , . 3 3 3 2 2 f 2 f f 2 f dx 2 dx dy d 2 x 2 x x y x 2 2 dx 2 8 z dx dy 2 a 2 dx c 2 16 dx dy a 3 2 dx 16 2 c dx dy a 3 2 dx 16 2 12 xyz c dx dy a 3 2 8 dx 16 abc c dx dy i 3 a 3 2 x We have 1 2 a 2 x or dx 0 2 a 97 a b c and therefore at , , ,we get 3 3 3 a 3 2 dx 0 2 a 2 dx dx or 0 0 ii 3 a a 2 dx 2 0 a a 2 12 ab Eq.(ii) gives { } a 2 dx dx dy 2 0 a a b 2 c dxdy dx or 2 abc a 2 abc dx c dxdy 2 a Putting this in eq.(i), we get 2 2 2 8 abc dx 16 abc dx d f 3 a 2 3 a 2 2 2 8 abc dx 8 abc dx 16 abc dx 0 3 a 3 a 3 a 8 abc i.e., d 2 f 0 and therefore, maximum volume is given by . 3 3 Jacobians. Let F1 , F 2 , . . . F n denote n differential functions of (n+ p) variables u 1 , u 2 , . . . u n ; x 1 , x 2 , . . . x p , then the functional determinant 98 F F F 1 1 . . . 1 u 1 u 2 u n F F F J 2 2 . . 2 u 1 u 2 u n F F F n n . . n . u 1 u 2 u n is called the Jacobian of the n functions with respect to n variables u 1 , u 2 , . . . u n and is denoted by F , F , . . . , F J 1 2 n . u 1 , u 2 , . . . , u n x , y Example 4.15. Let x u v , y u v , then find J u , v Solution. We have x x u v 1 1 = -2. y y 1 1 u v Example 4.16. Let x u v , y u v , then find Solution. We have 1 1 = u- v. v v 99 x , y Example 4.17. Let x r cos , y r sin , then find J u , v Solution. We have x x u v cos r sin y y sin r cos u v = r cos 2 sin 2 = r. 1 Example 4.18. Prove that if f 0 0 , f / x , then show that 1 x 2 x y f x f y f . 1 xy x y Solution. Let u f x f y and v 1 xy f / x f / y 1 y 2 1 x 2 2 2 1 xy 1 xy 1 1 1 x 2 1 y 2 = 0. 1 y 2 1 x 2 2 2 1 xy 1 xy Thus J = 0. Therefore, there is a fundamental relation between u and v say u v 100 x y i.e., f x f y . 1 xy Put y=0, we get f x f 0 x But f 0 0 Therefore f x x x . or f x y and hence f x f y f . 1 xy Theorem 4.4. If u 1 , u 2 , . . . u n are n differentiable functions of the independent variables x 1 , x 2 , . . . x n and there exists an identical differentiable functional relation f u 1 , u 2 , . . . u n 0 which does not involve F 1 , F 2 , . . . , F n x ' s explicitly then, the Jacobian J provided as a u 1 , u 2 , . . . , u n function of u ' s has no stationary values in the domain considered. Proof. We have u 1 , u 2 , . . . u n 0 . Therefore d 0 which implies that du 1 du 2 . . . du n 0 i u 1 u 2 u n u 1 u 1 u 1 But du 1 dx 1 du 2 . . . du n 0 x 1 x 2 x n u 2 u 2 u 2 du 2 dx 1 du 2 . . . du n 0 x 1 x 2 x n . 101 . u n u n u n du n dx 1 du 2 . . . du n 0 . x 1 x 2 x n Hence from eq. (i), we get u u u u u u 1 dx 1 du . . . 1 du 2 dx 2 du . . . 2 du 1 2 n 1 2 n u x x x u x x x 1 1 2 n 2 1 2 n u u u + n dx n du . . . n du 0 1 2 n u x x x n 1 2 n u u u u or 1 . . . n dx 1 . . . n dx . . . 1 2 u x u x u x u x 1 1 n n 1 2 n 2 u u 1 . . . n dx 0 ii n u x u x 1 n n n Since dx 1 , dx 2 , . . . , dx n are arbitrary differentials of independent variables, it follows from eq. (ii) that u u 1 . . . n 0 u 1 x 1 u n x n u u 1 . . . n 0 u 1 x 2 u n x 2 . . u u 1 . . . n iii u 1 x n u n x n 102 Since has no stationary values in the domain considered, therefore 0 , 0 , . . . , 0 . u 1 u 2 u n Eliminating , , . . . , from the systems of equations (iii), we get u 1 u 2 u n u u u 1 2 . . . n x 1 x 1 x 1 u u u 1 2 . . . n x 2 x 2 x 2 . 0 . u u u 1 2 . . . n x n x n x n u u u 1 1 . . . 1 x 1 x 2 x n u u u 2 2 . . . 2 x 1 x 2 x n or . 0 . u u u n n . . . n x 1 x 2 x n u , u , . . . , u i.e., J 1 2 n 0 . x 1 , x 2 , . . . , x n