7.8 Improper Integrals Improper Integral Type I (On Infinite Intervals) A] Assume That F Is Continuous on [A, ∞). Then ∫

Improper Integral Type I (on infinite intervals)

a] Assume that f is continuous on [a, ). Then ∞ ∞ t f(x) dx := lim f(x) dx. Za t→∞ Za b] Assume that f is continuous on ( ,a]. Then −∞ a a f(x) dx := lim f(x) dx. Z−∞ t→−∞ Zt c] Assume that f is continuous on ( , ). Then −∞ ∞ ∞ a ∞ f(x) dx := f(x) dx + f(x) dx. Z−∞ Z−∞ Za (It can be shown that the value of the integral on ( , ) is independent of the choice of a, we can choose a = 0, for instance.) −∞ ∞ In parts a] or b] we say that the improper integral converges (or the improper integral is convergent) if the limit in the definition exists as a finite real number. In any other cases we say that the improper integral diverges. It can diverge “nicely” to or if the limit in the definition exists as or , but the improper integral can diverge∞ −∞ “due to oscillation” as well (roughly speaking).∞ −∞ In part c] we say that the improper integral converges (or the improper integral is convergent) if both integrals in the right-hand side converges. In any other cases we say that the improper integral diverges. Recall that + L = , + L = , + = , and ( )+( ) = . ∞ ∞ −∞ −∞ ∞ ∞ ∞ −∞ −∞ −∞ E.1. We show that ∞ 1 1 p dx = , p > 1 , Z1 x p 1 − and ∞ 1 p dx = , p 1 . Z1 x ∞ ≤ In other words ∞ 1 p dx Z1 x converges to 1/(p 1) for all p > 1, while it diverges to for all p 1. − ∞ ≤ Using the definition we have

∞ t t −p+1 t 1 1 −p x p dx = lim p dx = lim x dx = lim Z1 x t→∞ Z1 x t→∞ Z1 t→∞ p +1 − 1 t−p+1 1 1 1 = lim = 0 = t→∞ p +1 − p +1 − p +1 p 1 − −1 − − for all p > 1 as p +1 < 0 in the exponent of t tending to . Also, − ∞ ∞ t t −p+1 t 1 1 −p x p dx = lim p dx = lim x dx = lim Z1 x t→∞ Z1 x t→∞ Z1 t→∞ p +1 − 1 t−p+1 1 1 = lim = = t→∞ p +1 − p +1 ∞ − p +1 ∞ − − − for all p < 1 as p +1 > 0 in the exponent of t tending to . Finally for p = 1 we have − ∞ ∞ t t 1 1 −1 t p dx = lim dx = lim x dx = lim [ln x]1 Z1 x t→∞ Z1 x t→∞ Z1 t→∞ = lim (ln t ln1) = 0 = . t→∞ − ∞ − ∞

The above example will be quite useful later in this section when we detect convergence or divergence of improper integrals by using comparison theorems. ∞ 1 E.2. Show that p dx converges for all p > 1, while it diverges to for all p 1. Ze x(ln x) ∞ ≤ Hint: Use a substitution u = ln x to observe that ∞ 1 ∞ 1 p dx = p du, Ze x(ln x) Z1 u and recall the result of E.1. ∞ 1 E.3. Show that p dx converges for all p > 1, while it diverges to for Zee x(ln x)(lnln x) ∞ all p 1. Hint:≤ Use a substitution u = lnln x to observe that ∞ 1 ∞ 1 p dx = p du, Zee x(ln x)(ln ln x) Z1 u and recall the result of E.1. ∞ E.4. Does xe−x dx converge? If it does, compute its value. Z0 Using the deﬁnition we get

∞ t xe−x dx = lim xe−x dx. Z0 t→∞ Z0

The Integration by Parts (IP) formulas with f(x) = x, g′(x) = e−x, f ′(x) = 1 and g(x) = e−x gives −

xe−x dx = x( e−x) 1( e−x) dx = xe−x e−x , Z − − Z − − − 2 hence t −x −x −x t −t −t t +1 xe dx =[ xe e ]0 = te e ( 1)=1 t , Z0 − − − − − − − e and taking the limit with an application of the L’Hospital’s Rule gives

∞ t −x −x t +1 t +1 xe dx = lim xe dx = lim 1 t =1 lim t Z0 t→∞ Z0 t→∞ − e − t→∞ e 1 1 = 1 lim =1 =1 0=1 . − t→∞ et − − ∞ ∞ E.5. Does e−|x| dx converge? If it does, compute its value. Z−∞ Using the deﬁnition in part c] we get

∞ 0 ∞ e−|x| dx = e−|x| dx + e−|x| dx =1+1=2 . Z−∞ Z−∞ Z0

Using the deﬁnition in parts a] and b] here we have

∞ ∞ t −|x| −x −x −x t e dx = e dx = lim e dx = lim [ e ]0 Z0 Z0 t→∞ Z0 t→∞ − = lim ( e−t ( e0))=1 0=1 , t→∞ − − − − and

0 0 0 −|x| x x x 0 e dx = e dx = lim e dx = lim [e ]t Z−∞ Z−∞ t→−∞ Zt t→−∞ = lim (e0 et)=1 0=1 , t→−∞ − −

∞ 4x E.6. Does 4 dx converge? If it does, compute its value. Z0 x +4 If we substitute u = x2/2 with du/dx = x, then we get

4x 4 1 dx = du = du = arctan u = arctan(x2/2)(+C) , Z x4 +4 Z 4u2 +4 Z u2 +1

hence

∞ t 4x 4x 2 t 4 dx = lim 4 dx = lim [arctan(x /2)]0 Z0 x +4 t→∞ Z0 x +4 t→∞ π π = lim (arctan(t2/2) arctan0) = 0 = . t→∞ − 2 − 2 3 ∞ E.7/a. Does sin xdx converge? If it does, compute its value. Z0 The deﬁnition gives ∞ t sin xdx := lim sin xdx Z0 t→∞ Z0 However if n is an even integer, then nπ nπ sin xdx =[ cos x]0 =( 1) ( 1)=0 , Z0 − − − − while if n is an odd integer, then nπ nπ sin xdx =[ cos x]0 = ( 1) ( 1)=1+1=2 . Z0 − − − − − Hence, by recalling the concept of limit of f(t) when t , we see that the limit does t → ∞ not exist as the function f(t) := sin xdx oscillates between 0 and 2 for arbitrarily large Z0 t values of t. So lim sin xdx does not exist (not even as an extended real number), so t→∞ Z0 ∞ we have to say that sin xdx diverges due to oscillation (roughly speaking). Z0 ∞ E.7/b. Show that sin x dx diverges to . Z0 | | ∞ Hint: This follows from the observation that t sin x dx Z0 | | is as large positive number as we wish for all suﬃciently large values of t. ∞ ln x E.8. Does 2 dx converge? If it does, compute its value. Z1 x Recall that the Integration by Parts Formula gives that ln x x−1 ln x dx = (ln x)x−2 dx = (ln x) x−1( x−1) dx = + x−2 dx Z x2 Z 1 − Z − − x Z − ln x x−1 = + . − x 1 − We have ∞ ln x t ln x ln x x−1 t 2 dx = lim 2 dx = lim + Z1 x t→∞ Z1 x t→∞ − x 1 − 1 ln t 1 ln1 1 = lim t→∞ − t − t − − 1 − 1 ln t 1 1/t = 1+ lim − − =1+ lim − =1+0=1 , t→∞ t t→∞ 1 where at the end we used L’Hospital’s Rule to calculate an / type limit. ∞ ∞ 4 Improper Integral Type II (on finite intervals)

a] Assume that f is continuous on (a,b]. Then

b b f(x) dx := lim f(x) dx. + Za t→a Zt b] Assume that f is continuous on [a,b). Then

b t f(x) dx := lim f(x) dx. Za t→b− Za c] Assume that f is continuous on [a,b] except at c [a,b]. Then ∈ b c b f(x) dx := f(x) dx + f(x) dx. Za Za Zc

It can be shown that the value of the integral on [a,b] is independent of the choice of c. In parts a] or b] we say that the improper integral converges (or the improper integral is convergent) if the limit in the definition exists as a finite real number. In any other cases we say that the improper integral diverges. It can diverge “nicely” to or if the limit in the definition exists as or , but the improper integral can diverge∞ −∞ “due to oscillation” as well (roughly speaking).∞ −∞ In part c] we say that the improper integral converges (or the improper integral is convergent) if both integrals in the right-hand side converges. In any other cases we say that the improper integral diverges. Recall that + L = , + L = , + = , and ( )+( ) = . ∞E.9. We show∞ −∞ that −∞ ∞ ∞ ∞ −∞ −∞ −∞ 1 1 1 dx = , p < 1 , Z xp 1 p 0 − and 1 1 p dx = , p 1 . Z0 x ∞ ≥ In other words 1 1 p dx Z0 x converges to 1/(1 p) for all p < 1, while it diverges to for all p 1. − ∞ ≥ Using the definition we have

1 1 1 −p+1 1 1 1 −p x p dx = lim p dx = lim x dx = lim Z0 x t→0+ Z x t→0+ Z t→0+ p +1 t t − t 1 t−p+1 1 1 = lim = 0 = t→0+ p +1 − p +1 1 p − 1 p − −5 − − for all p < 1 as p +1 > 0 in the exponent of t tending to 0+. Also, − 1 1 1 1 1 1 x−p+1 dx = lim dx = lim x−p dx = lim Z xp t→0+ Z xp t→0+ Z t→0+ p +1 0 t t − t 1 t−p+1 1 = lim = ∞ = t→0+ p +1 − p +1 p +1 − p +1 ∞ − − − − for all p > 1 as p +1 < 0 in the exponent of t tending to 0+. Finally for p = 1 we have − 1 1 1 1 1 1 dx = lim dx = lim x−1 dx = lim [ln x] p 0+ 0+ 0+ t Z0 x t→ Zt x t→ Zt t→ = lim (ln1 ln t)=0 ( ) = . t→0+ − − −∞ ∞

The above example will be quite useful later in this section when we detect convergence or divergence of improper integrals by using comparison theorems. Compare this result carefully with the result in E.1. 1 E.10. Does ln xdx converge? If it does, compute its value. Z0 The deﬁnition gives

1 1 ln xdx = lim ln xdx = lim [x ln x x]1 0+ 0+ t Z0 t→ Zt t→ − ln t = lim ( 1 (t ln t t))= lim ( 1 + t) lim t ln t = 1 lim t→0+ − − − t→0+ − − t→0+ − − t→0+ 1/t 1/t = 1 lim = 1+ lim t = 1+0= 1 , − − t→0+ 1/t2 − t→0+ − − − where we used L’Hospital’s Rule to evaluate a / type limit. −∞ ∞ 10 1 E.11. Does 3 dx converge? If it does, compute its value. Z1 √x 9 − What is wrong with the calculation below?

10 10 10 2/3 1 −1/3 (x 9) 3 dx = (x 9) dx = − Z1 √x 9 Z1 − 2/3 − 1 3 3 9 = 12/3 ( 8)2/3 = (1 4) = . 2 − − 2 − −2

The above calculation gives the correct answer but the computation is incorrect. As we cannot divide by 0 the the integrand is not deﬁned at 9, which is inside the interval [1, 10] 6 of the integration. Hence the integral is improper and we are supposed to handle this integral as an improper integral of type II, part c]. The deﬁnition gives

10 1 9 1 10 1 3 dx = 3 dx + 3 dx Z1 √x 9 Z1 √x 9 Z9 √x 9 − t− − 10 = lim (x 9)−1/3 dx + lim (x 9)−1/3 dx 9 9+ t→ − Z1 − t→ Zt − t 10 (x 9)2/3 (x 9)2/3 = lim − + lim − t→9− t→9+ 2/3 1 2/3 t 3 3 3 3 = lim (t 9)2/3 ( 8)2/3 + lim (t 9)2/3 t→9− 2 − − 2 − t→9+ 2 − 2 − 3 9 = 6 + = . − 2 −2 Hence 10 1 9 3 dx = , Z1 √x 9 −2 − that is, the improper integral convergess to 9/2. − 3 1 E.12. We study 2 dx. Z0 (x 1) − What is wrong with the calculation below?

3 3 −1 3 1 −2 (x 1) 1 3 2 dx = (x 1) dx = − = 1 = . Z0 (x 1) Z0 − 1 −2 − −2 − − 0 The above calculation gives an incorrect answer by an incorrect computation. As we cannot divide by 0 the the integrand is not deﬁned at 1, which is inside the interval [0, 3] of the integration. Hence integral is improper and we are supposed to handle this integral as an improper integral of type II, part c]. The deﬁnition gives

3 1 1 1 3 1 2 dx = 2 dx + 2 dx Z0 (x 1) Z0 (x 1) Z1 (x 1) − −t − 3 = lim (x 1)−2 dx + lim (x 1)−2 dx 1 1+ t→ − Z0 − t→ Zt − 3 (x 1)−1 t (x 1)−1 = lim − + lim − t→1− 1 t→1+ 1 − 0 − t (t 1)−1 1 (t 1)−1 = lim − 1 + lim − t→1− 1 − t→1+ −2 − 1 − − 1 1 1 = ( 1) 1 + + =( 1)+( 1/2 + ) − 0 − −2 0+ ∞ − − ∞ − = + = . ∞ ∞ ∞ 7 Hence 3 1 2 dx = , Z0 (x 1) ∞ − that is, the improper integral diverges to . ∞ Comparison Theorems

In this section we are interested in ﬁnding out whether or not the given improper integral converges as the improper integral looks too complicated or impossible to compute. Comparison Theorem 1 (CT 1) (on inﬁnite intervals). Suppose f and g are continuous functions with 0 g(x) f(x) for all x a. ≤ ≤ ≥ ∞ ∞ (a) If f(x) dx converges, then g(x) dx also converges. Za Za ∞ ∞ (b) If g(x) dx diverges, then f(x) dx also diverges. Za Za

Note that the assumption 0 g(x) f(x) for all x a is very important. Do not use Comparison Theorem 1 without≤ checking≤ this assumption.≥ An analogous statement is true on intervals ,a] and , ). −∞ −∞ ∞ Comparison Theorem 2 (CT 2) (on ﬁnite intervals). Suppose f and g are continuous functions with 0 g(x) f(x) for all x (a,b]. ≤ ≤ ∈ b b (a) If f(x) dx converges, then g(x) dx also converges. Za Za b b (b) If g(x) dx diverges, then f(x) dx also diverges. Za Za

Note that the assumption 0 g(x) f(x) for all x (a,b] is very important. Do not use Comparison Theorem 2 without≤ checking≤ this assumption.∈ An analogous statement is true on intervals [a,b) and [a,b] with a discontinuity only at c [a,b]. ∈ ∞ sin2 x E.13. Does 2 dx converge? Z1 x Observe that sin2 x 1 0 , x [1, ) . ≤ x2 ≤ x2 ∈ ∞ Here ∞ 1 2 dx Z1 x 8 converges as the exponent p = 2 in the integrand is greater than 1 (recall the result in E.1). Hence by CT 1 (a) ∞ sin2 x 2 dx Z1 x also converges. ∞ √1 + x−1/3 E.14. Does dx converge? Z1 √x Observe that √1 + x−1/3 1 1 = 0 , x [1, ) . √x ≥ √x x1/2 ≥ ∈ ∞ Here ∞ 1 1/2 dx Z1 x diverges as the exponent p = 1/2 in the integrand is not greater than 1 (recall the result in E.1). Hence by CT 1 (b) ∞ √1 + x−1/3 dx Z1 √x also diverges. 1 1 E.15. Does 2/3 dx converge? Z0 arcsin(x ) Observe that (recall Math 151) 0 < sin x 0 . − Z0 ≤ · As the function arcsin x is strictly increasing on [0, 1] we have 0 < x < arcsin x, x (0, 1] , ∈ and replacing x by x2/3 we get 0

∞ √x3 +3x2 a] 3 8 6 dx Z1 √x +4x

∞ √x3 +3x2 b] dx 3 7 5 Z1 √x +6x

1 √x3 +3x2 c] 3 8 6 dx Z0 √x +4x

1 √x3 +3x2 d] 3 7 5 dx Z0 √x +6x

We claim that the integral in a] converges. Indeed, on the interval [1, ) we have 0 x2 x3 and 0 x6 x8, and hence ∞ ≤ ≤ ≤ ≤ √x3 +3x2 (4x3)1/2 2 0 =2x3/2−8/3 =2x−7/6 = , x [1, ) . ≤ √3 x8 +4x6 ≤ x8/3 x7/6 ∈ ∞

Here ∞ 2 7/6 dx Z1 x converges as the exponent p = 7/6 in the integrand is greater than 1 (recall the result of ∞ √x3 +3x2 E.1). Hence by CT 1 (a) dx also converges. 3 8 6 Z1 √x +4x

We claim that the integral in b] diverges. Indeed, on the interval [1, ) we have 0 x2 x3 and 0 x5 x7, and hence ∞ ≤ ≤ ≤ ≤ √x3 +3x2 (x3)1/2 1 1 1 1 = x3/2−7/3 = x−5/6 = 0 , x [1, ) . √3 x7 +6x5 ≥ (7x7)1/3 71/3 71/3 71/3 x5/6 ≥ ∈ ∞

Here ∞ 1 1 1/3 5/6 dx Z1 7 x diverges as the exponent p = 5/6 in the integrand is not greater than 1 (recall the result of E.1). Hence by CT 1 (b) ∞ √x3 +3x2 3 7 5 dx Z1 √x +6x also diverges. 10 We claim that the integral in c] diverges. Indeed, on the interval (0, 1] we have 0 x3 x2 and 0 x8 x6, and hence ≤ ≤ ≤ ≤ √x3 +3x2 (3x2)1/2 31/2 31/2 31/2 1 = x2/2−6/3 = x−1 = 0 , x (0, 1] . √3 x8 +4x6 ≥ (5x6)1/3 51/3 51/3 51/3 x ≥ ∈ Here 1 31/2 1 1/3 dx Z0 5 x diverges as the exponent p = 1 in the integrand is not less than 1 (recall the result of E.9). Hence by CT 2 (b) 1 √x3 +3x2 3 8 6 dx Z0 √x +4x also diverges.

We claim that the integral in d] converges. Indeed, on the interval (0, 1] we have 0 x3 x2 and 0 x7 x5, and hence ≤ ≤ ≤ ≤ √x3 +3x2 (4x2)1/2 2 2 2 1 0 = x2/2−5/3 = x−2/3 = , x (0, 1] . ≤ √3 x7 +6x5 ≤ (6x5)1/3 61/3 61/3 61/3 x2/3 ∈ Here 1 2 1 1/3 2/3 dx Z0 6 x converges as the exponent p =2/3 in the integrand is less than 1 (recall the result of E.9). Hence by CT 1 (a) 1 √x3 +3x2 3 7 5 dx Z0 √x +6x also converges. ∞ 1 E.17. Does x dx converge? Z0 3√x + e Observe that 1 1 0 = e−x . ≤ 3√x + ex ≤ ex Here ∞ t −x −x −x t −t 0 e dx = lim e dx = lim [ e ]0 = lim ( e ( e )) Z0 t→∞ Z0 t→∞ − t→∞ − − − = lim (1 e−t)=1 0=1 t→∞ − − converges. Hence by CT 1 (a) ∞ 1 x dx Z1 3√x + e also converges. 11 ∞ 1 E.18. Does x dx converge? Z1 3√x + e /x Observe that 1 1 0 = xe−x . ≤ 3√x + ex/x ≤ ex/x Here the result of E.4 shows

∞ ∞ xe−x dx xe−x dx =1 Z1 ≤ Z0 converges. Hence by CT 1 (a) ∞ 1 x dx Z1 3√x + e /x also converges.

My sections 507, 508, and 509 of Math 152 are now open for midterm evaluation in PICA. You can log on to pica.tamu.edu to give feedback. Midterm feedback closes Oct. 4.