CHAPTER SIX The Calculus of Residues
§6-1 Singularities and Zeroes
♣ Laurent Series ∞ n fz()=−∑ cn ( z z0 ) n=−∞ −−21 1 ⇒ fz()()= ""+−+−++−+ czz−−20 czz 10 () c 010 czz ()
♣ Some Definitions: 1) Principal part and analytic part
The portion of the Laurent series containing only the negative powers of ()zz− 0 is called principal part; the remainder of the series ⎯ the summation of the terms with zero and positive powers ⎯ is known as the analytic part. 2) Pole of order N
A function whose Laurent expansion about a singular point z0 has a principal part, in which the most
negative power of ()zz− 0 is −N , is said to have a pole of order N at z0 . ♦ A function has a pole of order one, at some point, is said to have a simple pole at that point.
1. Singular Point (Singularity)
A point z0 is called a singular point of the function f (or called a singularity of the function f), if f
is not analytic at z = z0 , but every neighborhood of z0 contains at least one point at which f is analytic. y
=
||zz− 0 < = z0
x 0
2. Isolated Singularity (Isolated Singular Point).
A point z0 is an isolated singularity (or an isolated singular point) of the function f, if f is not
analytic at z = z0 but is analytic in a deleted neighborhood of z0 . That is, f is analytic in
ℜ′(,){|0|zzzzz00====<−< | , and ∈ C } ♣ Nonisolated singularity
A singular point z = z0 of a function f ()z is nomisolated if every neighborhood of z0 contains
at least one singularity of f ()z other than z0 .
Example 1 The function z fz()= ⇒ zi= ±2 both are isolated singularities z2 + 4 In other words, f ()z is analytic in the deleted neighborhoods 0|< zi−< 2|1 and 0|<+zi 2|1 <. The function f ()zz= Ln ⇒ The branch point z = 0 is a non-isolated singularity of f ()z since every neighborhood of
148 z = 0 contains points on the negative real axis.
Example 2 The function 1 fz()= sin(π /z ) has the singular points z = 0 and znn= 1/ ,=± 1, ± 2, " , all lying on the segment of the real axis from z =−1 to z = 1. Each singular point except z = 0 is isolated. The singular point z = 0 is not singular because every deleted = neighborhood of the origin contains other singular points of the function f ()z .
3. Removable Singularity Let f (z) be analytic in the punctured disk
0|<−zz0 | <= and let
lim(zzfz−=0 ) ( ) 0 n→∞
Then f (z) has a removable singularity at z = z0 .
In other words, f (z) can be assigned a value f ()z at z0 so that the resulting function is
analytic in ||zz− 0 < = . The proof of this theorem can be seen in the textbook " Complex Variable, Levinson / Redheffer" from p.154 to p.154.
Example 3 Consider the function sin z fz()= z 1) Because 0/0 is undefined, the point z = 0 is singular. 2) Since sinzzz=−35 / 3! + z / 5! −+" , we have sin zzz24 = 1−+−" ⇒ Taylor series on the right represents an analytic function z 3! 5! 3) By defining f (0)= 1, we obtain a function ⎧sin z ⎪ ,0z ≠ fz()= ⎨ z ⎩⎪1,z = 0 which is analytic for all z . The singularity of f ()z at z = 0 has been removed by an appropriate definition of f (0) . ⇒ z = 0 is a removable singular point. ♣ z = 0 is also a removable singular point of the function fz()= eLn z .
4. Pole and Essential Singularity 1) If the Laurent Series ∞ ∞ n bn f (z) = ∑ an (z − z0 ) + ∑ n n=0 n=1 (z − z0 )
In the neighborhood of an isolated singular point z = z0 contains only a finite number of negative
powers of z − z0 , then z = z0 is called a pole of f (z) . −m If (z − z0 ) is the highest negative power in the series (expansion), the pole is said to be of order m and the series m b b b b n 1 2 " m ∑ n = + 2 + + m n=1 (z − z0 ) z − z0 (z − z0 ) (z − z0 )
is called the principal part of f (z) at z = z0 .
149 Then, the Laurent series is rewritten as ∞ m n bn f (z) = ∑ an (z − z0 ) + ∑ n n=0 n=1 (z − z0 )
2) If the Laurent series of f (z) in the neighborhood of an isolated singular point z = z0 contains
infinitely many negative powers of z − z0 , then
z = z0 is called an essential singularity of f (z) . ∞ ∞ n bn ⇒ f (z) = ∑ an (z − z0 ) + ∑ n n=0 n=1 (z − z0 ) Example 4 The function 1 fz()= zz(1)− 2 has singularities at z = 0 and z = 1. 1) Laurent series about z = 0 : fz( )= z−12++ 2 3 z + 4 z + ...,0 < z < 1 ⇒ z = 0 is a pole of order 1. 2) Laurent series about z = 1: fz( )= ( z−+−+−−+−+ 1)−−21 ( z 1) 1 ( z 1) ( z 1) 2" , 0 <−< z 1 1 ⇒ z = 1 is a pole of order 2.
Example 5 The function 1 zz−−35 sin= zz−1 −++" , ≠ 0 z 3! 5! ⇒ z = 0 is an essential singularity
Example 6 Consider the function e1/(z− 1)
(1)z − 2 u2 Using euu =++1, +" and putting uz=−(1)−1 , we find that 2! ez1/(z−− 1)(1)− 4 =−(1)(1)zz−−23 +− + +" , z ≠ 1 (1)z − 2 2! ⇒ z = 1 is an essential singularity
♣ Three different kinds of principal part: 1) A principal part with a positive finite number of nonzero terms. −−−−NN(1) 1 Ex. czz−−−−NN()−+0(1)0 c () zz − ++−" czz 10 (), where c− N ≠ 0 2) There are an infinite number of nonzero terms in the principal part.
3) There are functions possessing an isolated singular point z0 such that a sought-after Laurent
expansion about z0 will be found to have no term in the principal part, that is, all the coefficients
involving negative powers of ()zz− 0 are zero.
5. Some Examples about Pole and Singularity 1) Pole If m a) lim (zz− 0 ) fz ( ) exist and zz→ 0
150 m b) lim (zz−≠0 ) fz ( ) 0 zz→ 0
⇒ z = z0 is a pole of order m of f (z) . * A pole of order one is a simple pole. 2) The order of the pole for a function of the form f ()zgzhz= ()/():
If g()z and hz() are analytic at z0 , with gz()00 ≠ and hz()00 = , then f ()z will have an
isolated singularity at z0 . Let NN+1 hz()=−+ aNN ( z z010 ) a+ ( z − z ) +" , where aN ≠ 0 and N ≥ 1 . Using the above equation, we have gz()
fz()= NN+1 azzNN()−+010 a+ () zz − +"
To show that this expression has a pole of order N at z0 , consider N N ()()zz− 0 gz lim⎡⎤ (zz−=0 ) fz ( ) lim NN+1 zz→→⎣⎦ zz 00azz(− )+−+ a ( zz ) ... NN010+ gz() gz ( ) ==lim0 . zz→ 0 aazzNN+−++10( ) ... a N Since this limit is finite and nonzero, we may conclude the following rule: Rule I (Quotients):
If f ()zgzhz= ()/(), where g()z and hz() are analytic at z0 , and if hz()00 = and gz()00 ≠ ,
then the order of the pole of f ()z at z0 is identical to the order of the zero of hz()at this point.
♣ For the case of hz()00 = and gz()00 = , if limfz ( )==∞ lim[ gz ( ) / hz ( )] zz→→00 zz
then f ()z has a pole at z0 , whereas if limfz ( )== lim[ gz ( ) / hz ( )] finite zz→→00 zz
f ()z has a removable singularity at z0 . L’Hôspital rule is often useful in finding the limit. Rule II (Quotients):
The order of the pole of f ()zgzhz= ()/() at z0 is the order of the zero of g()z at this point less the order of the zero of g()z at the same point.
Example 7 The function f defined by 2z + 3 f (z) = (z −1)(z − 2)3 Since 23z + lim (zfz−= 1) ( ) lim = −5 ≠ 0 zz→→11(2)z − 3 ⇒ z = 1 is a simple pole of f (z) . and 23z + lim (zfz−= 2)3 ( ) lim = 7 ≠ 0 zz→→22z −1 ⇒ z = 2 is a pole of order 3 of f (z) . ♣ 在遇到三角函數時，不可直接冒然判斷其 order。
Example 8 The function f defined by sin z f (z) = for all | z |< 0 z 4
151 Hence limzfz4 ( )== lim sin z 0 zz→→00 The limit of limzfz4 ( ) exists, but equal to 0. Therefore, we can not say that z = 0 is a pole of order z→0 four of the function f ()z . However, if we take sin z limzfz3 ( )==≠ lim 1 0 zz→→00z ⇒ z = 0 is a pole of order 3 of the function f (z) . We can also use following method to determine the order of z = 0 ： z 3 z 5 z 7 z − + − +" sin z 3! 5! 7! f (z) = = z 4 z 4 1 1 1 z z3 = − + − + " z3 3! z 5! 7! 1 因為在 此項以前之項的係數均為 0，故知 z 3 ⇒ z = 0 is a pole of order 3 of f (z) .
Example 9 Discuss the singularity of zzcos fz()= (1)(1)(32).zz−+222 zz ++
Example 10 Discuss the singularity of ez fz()= . sin z
Example 11 Find the order of the pole of (1)/(1)ze2 ++z at zi= π .
152
Note that c0 = 0 because at hi()0π = . Since d ce=+=−≠(1)10z 1 dz zi= π We see that hz() has a zero of order 1 at zi= π . Thus, by Rule I, f ()z has a pole of order 1 at zi= π .
Example 12 Find the order of the pole of sinh z fz()= at z= 0. sin5 z
Example 13 Find the poles and establish their order for the function 1 fz()= . (Lnzi−−π )( z1/2 1)
This cut plane is the domain of analyticity of f ()z . If Lnzi−=π 0 , we have Ln zi= π or ze==−iπ 1 This condition can not occur in the domain. Alternatively, we can say that z = −1 is not an isolated singular point of f ()z since it lies on the branch cut containing all nonisolated singular points of f ()z . Consider z1/2 −=10, or z1/2 = 1 Squaring, we get z = 1 . Since the principal value of 11/2 is 1, we see that f ()z has an isolated singular point at z = 1. Now ∞ 1/2 n zcz−=1(1).∑ n − n=0
153 We readily find that c = 0 and cz= [(1/ 2) /1/2 ] 0 1 z=1 This shows that z1/2 −1 has a zero of order 1 at z = 1 . Since ⎛⎞1 ⎜⎟ Ln zi− π fz()= ⎝⎠ z1/2 −1 and the numerator of this expansion is analytic at z = 1 while the denominator z1/2 −1 has a zero of order 1 at the same point, the given f ()z must, by Rule I, have a simple pole at z = 1 .
♣ Comments on the term “pole” 1) If f ()z has a pole at zz= , then limfz ( ) = ∞ . 0 zz→ 0
2) Equivalently, fz( )→ unbounded as zz→ 0 .
3) Suppose f ()z has a pole of order N at zz= 0 . Then −−−NN(1) fz()()=−+ c−−−NN z z0(1)0 c () z − z +++−+"" c 010 cz (), z NN+1 ⇒ ()()zz−=+−++−+0(1)000 fzc−−−NN c () zz"" czz () N Dividing both sides of this equation by ()zz− 0 , we have this representation of f ()z in a
neighborhood of zz= 0 : N fz()=−ψ ()/( z z z0 ) where
ψ ()zc= −−−NN+−+−+ c(1) ( zzc 0 ) −− (2) N ( zz 0 ) "
and c− N ≠ 0 .
Thus, as zz→ 0 , we have
ψ ()zc0 −N fz()≈=NN , ()()zz−−00 zz and so
c− N fz()≈ N . ()zz− 0 We see that the higher the order N of the pole, the more steeply will be the surface depicting f ()z
rise as zz→ 0 .
Example 14 Using MATLAB, obtain a plot of fz()=− 1/[( zz 2)]2 and make comments on the poles of f ()z .
% plot for f(z) x=linspace(-3,3,200); y=linspace(-1,1,200); [X,Y]=meshgrid(x,y); z=X+i*Y; f=z.*(z-2).^2; f=1./f; f=abs(f);mesh(X,Y,f);AXIS([-1.0 3.5 -1 1 0 10]); view(10,15)
154
H.W. 1 (a) Using MATLAB, obtain a plot similar to that of Example 14 but use the function 1 fz()= (2)(2)zzz+−22 which has two poles of order 2, and a simple pole. (b) We have observed that the magnitude of a function does not have a limit of infinity at an essential singularity. Illustrate this by generating a MATLAB plot of the magnitude of e1/z in the region 01<≤z . For a useful plot, it is best not to let z to get too close to zero ⎯ as you will see after some experimentation. 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 7, Exercise 6.2, Pearson Education, Inc., 2005.】
H.W. 2 A nonisolated essential singular point of a function is a singular point whose every neighborhood contains an infinite number of isolated singular points of the function. An example is f ()zz= 1/sin(1/), whose nonisolated essential singular point is at z = 0 . (a) Show that in the domain z < = (0)= > , there are plots at zn=±1/,1/(1),1/(2),π ± n +ππ ± n + " where n is an integer such that n > 1/(π = ) . (b) Is there a Laurent expansion of f ()z in a deleted neighborhood of z = 0 ? (c) Find a different function with nonisolated essential singular point at z = 0 . Prove that is has an infinite number of isolated singular points inside z = = . 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 36, Exercise 6.2, Pearson Education, Inc., 2005.】
§6-2 Residue Theorem
♣ Definition of Residue Let f ()z be analytic on a simple closed contour C and at all points interior to C except for the point
z0 . Then, the residue of f ()z at z0 , written Resf (z ) or Res[ f (zz ), 0 ] , is defined by zz= 0 1 Resfz ( )== Res[] fz ( ), z0 fzdz ( ) >∫ C zz= 0 2πi
♣ Connection between Resf (z ) and the Laurent series of f ()z zz= 0 1) Laurent expansion of f ()z : −−21 1 fz()()= ""+−+−++−+ czz−−20 czz 10 () c 010 czz (), 0 <−zz0 < R
2) Residue of f ()z at zz= 0 :
155 1 Res[]fz ( ), z0 = fzdz ( ) y 2πi >∫ C ∞ 1 n =−czn () z0 dz >∫ zz−= r∑ 2πi 0 n=−∞ ∞ R 1 n =−czzdzn ()0 ∑ >∫ zz−= r 2πi n=−∞ 0 Note that z0 n ⎧ 0,n ≠ − 1 ()zz−=0 dz r >∫ zz−= r ⎨ 0 ⎩2,πin= − 1 Thus, we have x Res[ f (zz ), 01] = c−
♣ Theorem −1 The residue of function f ()z at the isolated singular point z0 is the coefficient of ()zz− 0 in the
Laurent series representing f ()z in an annulus 0 < zz−<0 R.
Example 1 y Let C 1 fz()= zz(1)− z −=11 Using the above theorem, find 1 fzdz() 2πi >∫ C where C is the contour shown in the left figure. x
Example 2
156 1 Find zzdzsin(1/ ) integrated around z = 2 . 2πi >∫ C
1. Three Theorems for finding Residue 1) A formula for the residue in the case of a simple pole.
If z = z0 is a simple pole of f (z)
⇒ Resf (zzzfz )=− lim (0 ) ( ) zz= 0 zz→ 0
2 b1 = a0 + a1 (z − z0 ) + a2 (z − z0 ) +"+ (z − z0 )
where b1 ≠ 0 .
Multiplying both sides of the above equation by z − z0 , we have 2 (z − z0 ) f (z) = b1 + (z − z0 )[a0 + a1 (z − z0 ) + a2 (z − z0 ) +"]
If we let z approach z0 , then
lim (zzfz− 0 ) ( ) zz→ 0 ∞ ⎡ n+1 ⎤ =−+lim⎢∑azn ( z01 ) b⎥ n=0 zz→ 0 ⎣ ⎦
==bfz1 Res ( ) zz= 0 2) Let the function f be defined by p(z) f (z)= q(z)
where the function p(z) and q(z) are both analytic at z = z0 , and p(z0 ) ≠ 0 .
Then, f has a pole of order m at z = z0 if and if only z0 is a pole of order m of q.
In particular, if zz= 0 is a simple pole of f ()z , we have pz() ⇒ Resfz ( ) = zz= 0 qz′()
⇒ q(z0 ) = 0 and q′(z0 ) ≠ 0 Consequently, q(z) can be expanded in a Taylor's series of the form：
157 qz′′() qz()=+ qz () q′ ()( z z −+ z )0 ( z −+ z )2 "" 0002！ 0
qz′′() =−+−+qz′()()() z z0 z z 2 "" 002！ 0 Then, we can see that
Resfz ( )=− lim ( z z0 ) fz ( ) zz= 0 zz→ 0 pz() =−lim (zz0 ) zz→ 0 qz′′() qzzz′()()()−+0 zz −2 +"" 002！ 0 pz() = 0 qz()0
3) Suppose that function f is analytic in a deleted neighborhood of z0 , and if z = z0 is a pole of order m of f (z) . Then, we have m−1 1 d m Resfz ( )=− limm−1 [( z z0 ) fz ( )] zz= 0 zz→ 0 (1)mdz− ！
where bm ≠ 0 , and the series converges in some neighborhood of z = z0 , except at the point
z0 itself. m By multiply both sides of the above equation by (z − z0 ) , we obtain m (z − z0 ) f (z) ∞ n+m m−1 m−2 = ∑ an (z − z0 ) + b1 (z − z0 ) + b2 (z − z0 ) + " + bm n=0 m−1 This shows that the residue b1 of f (z) at z = z0 is now the coefficient of (z − z0 ) in m the series of (z − z0 ) f (z) . Then, we have d m−1 [(z − z ) m f (z)] dz m−1 0
∞ n+1 = ∑ an (n + m)(n + m −1)"(n + 2)(z − z0 ) + (m −1！) b1 n=0 Hence, we obtain that m−1 ⎡⎤d m lim⎢⎥m−1 [(zz− 0 ) fz ( )] zz→ 0 ⎣⎦dz
=−(1)mb！ 1
=−(1)Res()mbfz！ 1 zz= 0 That is, m−1 1 d m Resfz ( )=− lim [( z z0 ) fz ( )] zz→ m−1 zz= 0 (1)mdz− ！ 0
Example 1 Find the residue of the function 4 − 3z 4 − 3z f (z) = = z 2 − z z(z −1)
158 43− z 43− z ⇒ Resfz ( )==− 4 and Resfz ( )= = 1 z=0 21z − z = 0 z=1 21z − z = 1
Example 2 ez Find the residue of fz()= at all poles. (1)zz22+
Example 3 tan z Find the residue of fz()= at all singularities of tan z . zz2 ++1
By expanding cos z in a Taylor series about the point zk0 = π /2+ π , we have 2 coszazz=−+−+1020 ( ) azz ( ) ", where ak1 = −≠cosπ 0
Since cos z has a zero of order 1 at zz= 0 , f ()z must have a pole of order 1 there. Taking gz()=++ sin/( z z2 z 1) and hz()= cos z, we have hz′()= − sin z. Thus, 1 Resfz ( )=+=− Res[] fz ( ),ππ / 2 k 2 z0 zz++1 ππ/2+k −1 ==±±,0,1,2,k " (/2)(/2)1ππππ++kk2 ++ ♣ There are also poles of f ()z , for z satisfying zz2 + +=10, i.e.,
zi1 =−1/2 + 3/2 and zi2 =−1/2 − 3/2
Residue at z1 : Resfz ( )=−+=−+ Res⎡⎤ fz ( ), 1/ 2 i 3 / 2 tan( 1/ 2 i 3 / 2) /( i 3) z1 ⎣⎦
Residue at z2 : Resfz ()=−−=−−− Res⎡⎤ fz (), 1/2 i 3/2 tan(1/2 i 3/2)/( i 3) z2 ⎣⎦
Example 4 z1/2 Find the residue of fz()= at all poles. Use the principal branch of z1/2 . zzz32−+44
159
Example 5 e1/ z Find the residue of fz()= at all singularities. 1− z
−−11⎡⎤11 cz−1 =+++1 " z ⎣⎦⎢⎥2! 3! Thus, we see that 11 Resf (zfzc )===+++=− Res[] ( ),0−1 1" e 1 z=0 2! 3! 111 where recalling the definition e1 =+1 + + +" . 1! 2! 3!
Example 6 ez −1 Find the residue of fz()= at z = 0 . sin3 z
160 Since the numerator has a zero of order 1 and the denominator has a zero of order 3, the quotient has a pole of order 2. Reside of f ()z at z = 0 : Method I: Successive applications of L’Hôspital’s Rule, we see that 1 Resfz ( )== Res[] fz ( ),0 z=0 2 Method II: Using long division, we have z−1 z−2 ++" 2 zzz523 zz3 −+"" +++ 22!3! 1 ⇒ Resfz ( )== Res[] fz ( ),0 coefficient of z−1 = z=0 2
H.W. 1 For each of the following functions state the location and order of each pole and find the corresponding residue. sin zz− (a) fz()= zzsinh 1 (b) fz()= sin z2 cos(1/z ) (c) fz()= sin z 1 (d) fz()= ee2zz++1 1 (e) fz()= []Ln(ze / )− 1 2 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 8, 10, 11, 13, and 14, Exercise 6.3, Pearson Education, Inc., 2005.】
H.W. 2 Find the residue of the following functions at the indicated point: 1 (a) fz()= , at z = 0 ⎛⎞π z cos⎜⎟ez+ sin ⎝⎠2 1 (b) fz()= , at z = 0 ⎡⎤z sin⎣⎦ze()− 1 cos(z − 1) 2 (c) fz()=+, at z = 1 and z = 0 zz10 −1 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 22-25, Exercise 6.3, Pearson Education, Inc., 2005.】
H.W. 3 (a) Consider the analytic function f ()zgzhz= ()/(), having a pole at z0 . Let gz()00 ≠ ,
hz()00== h′ ()0 z , hz′′()00 ≠ . Thus, f ()z has a pole of second order at zz= 0 . Show that
2()g′′′zgzhz0002 ()() Resfz ( )==− Res[] fz ( ), z0 2 zz= 0 hz′() 3 0 []hz′()0 (b) Use the formula of part (a) to obtain ⎡⎤cos z Res⎢⎥2 , e ⎣⎦⎢⎥()Lnz − 1 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 15, Exercise 6.3, Pearson Education, Inc., 2005.】
H.W. 4 Consider the function f ()zz= 1/sin . Here we use residues to obtain
161 ∞ 1 n = ∑ czn for π <
cdefnnnn=++ where ⎡⎤1 ⎡ 1 ⎤ ⎡ 1 ⎤ dn = Res , 0 , en = Res , π , fn = Res , −π ⎣⎦⎢⎥zzn+1 sin ⎣⎢ zzn+1 sin ⎦⎥ ⎣⎢ zzn+1 sin ⎦⎥
(c) Show that for n ≤−2 , dn = 0 . n+1 (d) Show that efnn+ = 0 when n is even, and efnn+=−2/π for n odd. 2 4 (e) Show that cn = 0 when n is even, and that c−1 = −1 , c1 =−2/π + 1/6, c3 =−2/π + 7/360, n+1 and cnn =−2/π for n ≤−3 . 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 41, Exercise 6.3, Pearson Education, Inc., 2005.】
H.W. 5 If a function f ()z is analytic everywhere in the finite plane except for a finite number of singular points interior to a positively oriented simple closed contour C, show that ⎡⎤11⎛⎞ fzdz()= 2π i Res f ⎢⎥2 ⎜⎟ >∫ C z=0 ⎣⎦zz⎝⎠ 【本題摘自：James Ward Brown and Ruel V. Churchill, Complex Variables and Applications, 7rd ed., Sec. 64, pp. 228-229, McGraw Hill, Inc., 2003.】
H.W. 6 (a) Let n ≥ 1 be an integer. Show that the n poles of 1
zznn++++−−12 z n" 1 are at cos(2knπ /(++ 1)) i sin(2 knπ /( + 1)) , kn= 1, 2," , . (b) Show that the poles are simple. (c) Show that the residue at cos(2knπ /(+ 1))++ i sin(2 knπ /( 1)) is cos(2knπ /(++ 1)) i sin(2 knπ /( +− 1)) 1
(nknniknn++++ 1)[] cos(2ππ /( 1)) sin(2 /( 1)) 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 26, Exercise 6.3, Pearson Education, Inc., 2005.】
§6-3 Cauchy's Residue Theorem
1. Cauchy's Residue Theorem If 1) C is a simple closed curve (or contour) in a domain D.
2) f (z) is analytic in D except at finite many points z1 , z2 , z3 ,", zn in the interior of C. Then, we have n ⇒ fzdz()= 2π i∑Res fz () ∮C j=1 zz= j
Let CCC123,,,," Cn be n circles having centers at z1 , z2 , z3 ,", zn respectively such that they all lie interior to C and exterior to each other, as shown in above Figure. Hence, we have
162 y
C2 z2 C D C5 z C 5 3 z3 z1 C 1 Cn C zn 4 z4 x 0
fzdz() ∮C =+++∮∮f() z dz f () z dz" ∮ f () z dz CC12 Cn =+++2Res()2Res()ππifzifz" 2Res() π ifz zz==12 zz zz =n ∞ = 2Res()π ifz∑ j=1 zz= j
Example 1 52z − Find ∮ dz , Cz：||3= C zz(1)− where the integral along C is taken in the positive direction. y 52z −
Example 2 Let us show that sin z π ∮ 4 dz=− i C z 3 where Cz：||1= , described in the positive directions.
163 Method II： sin z 1 z 3 z 5 z 7 Since = [z − + − +"] z 4 z 4 3! 5! 7! 1 1 1 z z 3 = − + − +" z 3 3! z 5! 7! sinzzz⎡⎤ 1 1 1 3 ⇒ Res43=−+−+ Res ⎢⎥" zz==00zzz⎣⎦3! 5! 7! 1 =− 3!
1 =− 6 Thus, we have sinzz sin ∮ 44dz= 2Resπ i C zzz=0 ⎛⎞1 =−2π i ⎜⎟ ⎝⎠6 π =− i 3 Method III： sin z Since z = 0 is a pole of order 3 of f (z) = z 4 2 sinzdz 1⎡⎤3 sin ⇒ Res424=⋅ lim⎢⎥ (z ) z=0 zdzz2！z→0 ⎣⎦ 1sindz2 = lim 2！z→0 dz2 z 12sin2cossin⎡⎤zzz =−−lim 2 z→0 ⎣⎦⎢⎥zzz323 12sin2cossin⎡⎤zz−− zzz2 = lim ⎢⎥3 2 z→0 ⎣⎦z Then, we use L'Hôspital's Rule, sinzzzzzzzzz 1⎡⎤ 2cos−+ 2cos 2 sin − 2 sin −2 cos ⇒ Res43= lim ⎢⎥ z=0 zz23z→0 ⎣⎦ 1cos⎡⎤z =−lim 23z→0 ⎣⎦⎢⎥ 11⎛⎞ =−⎜⎟ 23⎝⎠ 1 =− 6 Hence, we have z−4 sin zdz ∮C sin z = 2Resπ i z=0 z4
⎛⎞1 π =⋅−=−2π ii⎜⎟ ⎝⎠63
Example 3 Let us show that (1+ zz5 ) sinh π i ∮ 6 dz =− C z 60
164 where C is the unit circle | z |= 1 described in the positive direction.
2π i π i = − ⋅1 = 120 60 Method II： Since (1+ z 5 )sinh z
z 6 1+ zzzz5357⎡⎤ =6 ⎢⎥z ++−+" z ⎣⎦3! 5! 7!
−2 55⎡⎤− zz1 =+(1zz ) ⎢⎥ + + − +" ⎣⎦3! 5! 7! Then, we have (1+ zz5 ) sinh Res z=0 z6 −2 55⎡⎤− zz11 =+Res(1zz ) ⎢⎥ ++−+=" z=0 ⎣⎦3! 5! 7! 5！ Hence (1++zz55 ) sinh (1 zz ) sinh ∮ 66dz= 2Resπ i C zzz=0
1 π i =⋅=−2π i 560！ y Example 4 Cz：||2 = Let us show that z2 + 4 dz = 0 i ∮C ()()zizi−+ x 0 where C is the circle | z |= 2 described in the positive directions. −i z2 + 4
165 ⎡ zz22++44⎤ =+2Resπ i ⎢ Res ⎥ ⎣ zi==−()()zizi−+ z i ()() zizi −+⎦ zz22++44 =+2π i zi+−zi==− zi z i ⎡⎤33 =+2π i ⎢⎥ ⎣⎦22ii− =⋅=10π i 0 0
H.W. 1 Evaluate the following integrals by using the method of residue: ∞ 1 (a) dz around zi− = 3 >∫ ∑n=−5 ()(6)!zi++n n ⎛⎞1 (b) zdzsin ⎜⎟ around z = 2 >∫ ⎝⎠z −1 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 5 and 7, Exercise 6.1, Pearson Education, Inc., 2005.】
H.W.2 Use residue calculus to show that if n ≥ 1 is an integer, then ⎧ 2!πin ,oddn 1 ⎪ ⎛⎞⎛⎞nn−+11 ⎛⎞ ⎜⎟⎜⎟!! zdz+=⎝⎠⎝⎠22 >∫ C ⎜⎟⎨ ⎝⎠z ⎪ ⎩0,n even where C is any simple closed contour encircling the origin.
H.W.3 Use residues to evaluate the following integrals. Use the principal branch of multivalued functions. 1 (a) dz around Cz:165− = >∫ C ⎣⎦⎡⎤Ln() Lnz − 1 dz (b) around Cz:0=> a . Note that the integrand is not analytic. Consider ab> >∫ C zb− and ab< .
2. Special Topics Poles and Zeros of Meromorphic Functions Here, we want to discuss Meromorphic functions. Before we discuss Meromorphic functions, we must define it first. Definition： poles of f (z) A function f which is analytic in a domain D, except at some points of D, at which it has poles, is said to be D meromorphic in D. * We know that the non-analytic points is called singularities, it involves： A) Essential Singularities, it Analytic defines that f (z) can be expansion region as
166 n n f (z) = ∑ an (z − a) n=−∞ B) Poles, it defines that f (z) can be expanded by n n f (z) = ∑ an (z − a) n=−m * A meromorphic function may have an essential singularity at z = ∞ . Theorem： Let f (z) be a meromorphic function in a simply connected domain D, in which contains a simple closed contour C. Suppose that f (z) has no zeros on C. Let N be the number of zeros and P be the number of poles of the function f (z) in the interior of C, where a multiple zero or pole is counted according to its multiplicity. Then, 1()fz′ ∮ dz=− N P 2()π ifzC where the integral is taken in the positive direction.
D N
C
poles of f (z)
f ′(z) poles of P f (z)
First of all, observe that the number of zeros and poles of f (z) interior of C is finite. Suppose that z = α is a zero of order K of f ()z . Then, we can express f ()z as f (z) = (z −α) K λ(z) where λ(z) is analytic at z = α and λ()z ≠ 0. ** Here, if λ(z) = 0 , it implies that there is another zero in f ()z , then z = α is a zero of order K +1 of f ()z . Hence, we have f ′(z) = (z −α) K λ′(z) + K(z −α) K −1 λ(z) f ′(z) K λ′(z) ⇒ = + f (z) z −α λ(z) λ′(z) f ′(z) Since is analytic at z = α , it means that has a simple pole at z = α , which is the λ(z) f (z) f ′(z) only one pole that can't be reproduced and is with residue K. f (z) Suppose that z = β is a pole of order m of f (z) , then 1 f (z) = v(z) (z − β ) m where v(z) is analytic at z = β , and v(β ) ≠ 0 . Since −m 1 f ′(z) = v(z) + v′(z) (z − β ) m+1 (z − β ) m we have
167 f ′(z) −m v′(z) = + f (z) z − β v(z) Since v(z) is analytic, then v′(z) is analytic, too. v′(z) This implies that is analytic. v(z) v′(z) f ′(z) Since is analytic at z = β , hence has a simple pole at z = β with residue (−m) . v(z) f (z) f ′(z) Moreover, the only pole of interior to C is precisely the pole or zero of f ()z . f (z)
If α j and β j are the zeros and poles of f (z) interior to C, and K j and m j are their orders, respectively. Then, by Cauchy's Residue Theorem, we have ′ fz() ⎡⎤ ∮ dz=−2π i∑∑ Kjj m C fz() ⎣⎦ =−2(π iN P ) 1()fz′ ⇒ ∮ dz=− N P 2()π ifzC
§6-4 The Evaluation of the Integrals of Certain Periodic Functions Taken Between the limits 0 and 2π ── By complex variable analysis
1. 利用複變分析法解實變數積分時，通常有底下幾種型式： 2π 1) F(cosθ ,sinθ )dθ F(cosθ ,sinθ ) is ∫0 π rational function of 2) F(cosθ ,sinθ )dθ cosθ and sinθ ∫−π π 1 π 3) F(cosθ ,sinθ )dθ = F(cosθ ,sinθ )dθ ∫0 2 ∫−π where F(cosθ ,sinθ ) is even function of θ. In dealing with the above problems, we can change them (variable; real variable) into complex variable. Hence, we consider a unit circle | z |= 1 as an integral path described in the positive direction. Thus, we have z(θ ) = cosθ + i sinθ , −π ≤ θ ≤ π 1 1 ⇒ = z cosθ + i sinθ = cosθ − i sinθ Hence, ⎧ 11⎛⎞ ⎪cosθ =+⎜⎟z ⎪ 2 ⎝⎠z ⎨ ⎪ 11⎛⎞ sinθ =−⎜⎟z ⎩⎪ 2iz⎝⎠ and dz dz = i eiθ dθ = izdθ ⇒ dθ = iz Similarly, z 2 = (cosθ + i sinθ ) 2 = cos 2θ + i sin 2θ 1 = cos 2θ − i sin 2θ z 2
168 hence we have
⎧ 11⎛⎞2 cos2θ =+z ⎪ ⎜⎟2 ⎪ 2 ⎝⎠z ⎨ ⎪ 11⎛⎞2 sin 2θ =−⎜⎟z 2 ⎩⎪ 2iz⎝⎠ Using the same method, we obtain 1 z 3 = cos3θ + i sin 3θ and = cos3θ − i sin 3θ z 3 that is
⎧ 11⎛⎞3 cos3θ =+z ⎪ ⎜⎟3 ⎪ 2 ⎝⎠z ⎨ ⎪ 11⎛⎞3 sin 2θ =−⎜⎟z 3 ⎩⎪ 2iz⎝⎠ 以下依此類推，將上列之結果代入 1), 2) 及 3)式即可使其變成複變數分析型式。
2. Some Examples Example 1 Let us show that 2π dθ 2π = , ab> || ∫0 a + bsinθ a 2 − b 2
∫0 a + bsinθ dz iz =∮ C 11⎛⎞ ab+⋅⎜⎟ z − 2iz⎝⎠ 2 =∮ 2 dz C bz+−2 aiz b 2dz =∮ C ⎛⎞⎛⎞−+aab22 − −− aab 22 − bz⎜⎟⎜⎟−− i z i ⎜⎟⎜⎟ ⎝⎠⎝⎠bb ⎡⎤ ⎢⎥ ⎢⎥2 = 2Resπ i ⎢⎥ −+bab22 − ⎛⎞⎛⎞22 22 z= ⎢⎥−+aab − −− aab − b bz⎜⎟⎜⎟−− i z i ⎢⎥⎜⎟⎜⎟bb ⎣⎦⎝⎠⎝⎠ ⎡⎤ ⎢⎥ ⎛⎞21 = 2π i ⎜⎟⎢⎥ ⎝⎠b ⎢⎥−+aab22 − −aab+−22 ⎢⎥z − z = ⎣⎦b b 41π i =⋅ b 2 ab22− i b 2π = ab22−
169
♣ Similar integral 2π dθπ2 Ik==>,1 ∫0 k + sinθ k 2 −1
Example 2 Find 2π cos2θdθ I = ∫0 54sin− θ
11⎛⎞22 1 ⎛⎞ 1 dz cos2θθθ=+⎜⎟zzd22 , sin 2 = ⎜⎟ − , = 22⎝⎠ziziz ⎝⎠ we have 22− zz+ 4 ⎛⎞dz()zdz+1 ⎛⎞ dz I ==2 >>∫∫2 ⎜⎟22 ⎜⎟ zz==115−+zz−1 ⎝⎠iz22iz() iz+− 52 z i ⎝⎠ iz i () There is a second-order pole at z = 0 . Solving 2520iz2 + z−= i , we find simple poles at zi= /2 and zi= 2 . The pole at zi= 2 is outside the circle z = 1 and can be ignored. Thus, we have ⎡⎤(zdz4 +1) Ii= 2Resπ ⎢⎥, at z = 0 and zi= /2 ∑ ⎢⎥22iz22 iz+− 52 z i ⎣⎦() From previous sections, we find the residue at z = 0 , 4 15d ⎡⎤(z +1) − ⎢⎥= i 28idz⎢⎥()252iz2 +− z i ⎣⎦z=0 and the residue at zi= /2, 4 ⎡⎤⎛⎞i ⎢⎥⎜⎟+1 1172 ⎢⎥⎝⎠ = i 2 ⎢⎥d ⎛⎞i 252iz2 +− z i 24 (2i )⎜⎟⎢⎥() 2 dz ⎝⎠⎣⎦zi= /2 Thus, we obtain 2π cos2θdθπ⎛⎞− 5 17 Iiii==+=−2π ⎜⎟ ∫0 54sin− θ ⎝⎠ 8 24 6
Example 3 Let us show that 2π 2π ecosθ ⋅ cos(sinθ − nθ )dθ = , a < | b | ∫0 n！ where n = 0, 1, 2, 3, ".
2π =⋅⋅eecosθθθiin sin edθ ∫ 0 2π =eedcosθθθ+−iin sin ⋅()θ −−−−−− (1) ∫ 0
170 Let Cz： ()θθθ=+ cos i sin = eiθ , where −π ≤ θ ≤ π . Then, equation (1) becomes dz ⇒ ezzn⋅⋅− ∮C iz −iez = dz ∮C zn+1 1 =⋅−⋅⋅2()π ii ez n！ z=0 2π = n！ Since Im = Im , Re = Re , we have 2π ⇒ ecosθ ⋅ cos(sinθ − nθ )dθ ∫0 ⎡2π ⎤ 2π = Re⎢ ⎥ = ⎣ n！⎦ n！ and 2π cosθ ⎡2π ⎤ e ⋅ cos(sinθ − nθ )dθ = Im⎢ ⎥ = 0 ∫0 ⎣ n！⎦
Example 4 Show that 2π dθ 2π 2 = 2 , ∫0 1− 2a cosθ + a 1− a where | a | < 1 .
H.W. 1 Using residues, establish the following identities: 3/2π cosθπ 2⎡⎤ 1 (a) dabθ =−⎢⎥1,for0 >> ∫ 22 −π /2 ab+ cosθ b⎢⎥⎣⎦ab−
171 2π m 2!π m (b) cosθθdm=≥m 2 , for 0 even. Show that the preceding integral is zero ∫0 2 ⎡⎤⎛⎞m ⎢⎥⎜⎟! ⎣⎦⎝⎠2 when m is odd. 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 4 and 6, Exercise 6.4, Pearson Education, Inc., 2005.】
H.W. 2 Using residues, establish the following identities: 2π daθπ2 (a) =>≥,forab 0 ∫0 22 ()ab+ sinθ ()ab22− 2π dθπ2 (b) =>,fora 0 ∫0 aaa++sin2 θ ( 1) 2π cosθθd 2 π (c) =>, foraa real, 1 ∫0 12cos−+aaaaθ 22 ( − 1) 2π cosndθθ 2 π (− 1) e−na (d) =≥,na 0 is an integer, > 0 ∫0 coshaa+ cosθ sinh 2π dθπ2 (e) = , forab , real, ab> 0 ∫0 aba22sinθθ++ 2 cos 2 2ab 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 7-9 and 11-12, Exercise 6.4, Pearson Education, Inc., 2005.】
§6-5 The Evaluation of Certain Types of Integrals Taken Between The Limits −∞ and ∞
♣ Improper integrals 1. The improper integral is an integral in which one or both limits are infinite, that is expressions of the form ∞ k ∞ fxdx() , fxdx() , fxdx() , ∫k ∫−∞ ∫−∞ where f ()x is a real function of x, and k is a real constant.
2. The improper integral ∞ f (x)dx ∫−∞ can be defined as following form: ∞ 0 t fxdx()=+ lim fxdx () lim2 fxdx () ∫∫∫−∞ t 0 tt12→−∞1 →∞ and we let ∞ I = f (x)dx ∫−∞ 0 I1 = limfxdx ( ) ∫ t t1 →−∞ 1
t2 I2 = limfxdx ( ) ∫ 0 t2 →∞
If both of I1 and I 2 exist, ∞ ⇒ f (x)dx is convergent. ∫−∞ *** In advanced calculus, we know that: if the function f (x) is continuous in [,a ∞ ), 1) limxf ( x )=≠ A 0 x→∞
172 ∞ ⇒ fxdx () is divergent. ∫ a 2) limx P fx ( ) = A x→∞ where A is finite and P > 1 , ∞ ⇒ f (x)dx is convergent. ∫a *** In the above two cases, a can be equal to −∞, too.
Example 1 ∞ R 11−−11π π (a) 22dx==−=−lim dx lim() tan R tan 1 exists; however, ∫∫1111++xxRR→∞ →∞ 24 ∞ 1 (b) dx=−lim() ln R ln1 fails to exist, as does ∫1 1+ x2 R→∞ ∞ (c) cosxdx= lim sin R . ∫0 R→∞ In case (b), as x increases, the curve yx= 1/ does not fall to zero fast enough for the area under the curve to approach a finite limit. In case (c), a sketch of yx= cos shows that, along the positive x-axis, the total area under this curve has not meaning.
3. 1) If the integral ∞ f (x)dx is convergent, ∫−∞ then we can define that ∞ R fxdx()= lim fxdx () ∫∫−∞R→∞ −R 2) If the integral ∞ f (x)dx is divergent, and ∫−∞ R limfxdx ( ) exists, R→∞ ∫ −R R then limfxdx ( ) is called the Cauchy's Principal Value of the integral R→∞ ∫ −R ∞ f (x)dx ∫−∞ and is denoted by ∞ P.V.fxdx ( ) ∫ −∞
4. If 1) the function f (x) is rational and is expressed in terms of the following form: p(x) f (x) = y q(x)
where qx()≠ 0, for all real x. CR：| z |= R z 2) the degree of q(x) ≥ the degree of [ p(x) + 2] 2 z5 z1 z4 3) and z j , j = 1, 2, 3, " , k are poles of f (z) in the z3 upper zk half plane. x Then, we can obtain that −R 0 1 R ∞ K f (x)dx = 2π i ∑Res f (z) ∫−∞ j=1 z=z j
173 Let CR denote the upper half of the circle and | z |= R
described in the positive direction, and take R sufficiently large so that all the poles zk of f (z) lie in
its interior. Hence, let C = [−R, R] +C R . This implies that C is a simple closed curve. And C encloses all poles of f (z) in the upper half plane. By Cauchy's Residue Theorem, we then have K fzdz()=⋅⋅ 2π i∑ Res fz () ∮C j=1 zz= j
R K ⇒ f (x)dx + f (z)dz = 2π i ∑Res f (z) ∫−R ∫C j=1 z=z j R *** When we integrate in the interval of [−R, R] , y-component is zero, this means that z = x + i 0 ⇒ d z = d x Hence, we have RR fxdx()= fzdz () ∫∫−−RR Here, we want to prove that 1) f (z)dz → 0 , when R → ∞ , and ∫C R R ∞ 2) fzdz()⇒ fxdx () , when R → ∞ . ∫∫−−∞R First, suppose that p(z) f (z) = q(z)
m m−1 am z + am−1 z + "+ a1 z + a0 = n n−1 bn z + bn−1 z + " + b1 z + b0
where n ≥ m + 2 , and bn ≠ 0 . ⎡ 1 1 1 ⎤ z m a + a ⋅ +"+ a + a ⎢ m m−1 1 m−1 0 m ⎥ ⎣ z z z ⎦ ⇒ f (z) = ⎡ 1 1 1 ⎤ z n b + b ⋅ +"+ b ⋅ + b ⎢ n n−1 1 n−1 0 n ⎥ ⎣ z z z ⎦ But, we know that 111 aa+⋅++" a + a mm−110mm−1 a zzz → m , when | z | → ∞ 111b bb+⋅++⋅+" b b n nn−110zzznn−1 Therefore, when | z | = | R | → ∞ , we obtain m R am |()|fz →⋅n ------(1) R bn And, the ML inequality tells us that | f (z)dz | ≤ MA ∫C ⇒ | f (z) | ≤ M where A is the arc length. Thus, equation (1) still satisfies following form: 1 | f (z) | ≤ ⋅ M R n−m Hence 1 Mπ fzdz() ≤⋅⋅= Mπ R ∫ C nm−−− nm1 R RR Since, n ≥ m + 2 , then
174 n − m −1 ≥ 1 Thus, when R → ∞ , we know that
f (z)dz → 0 ∫C R This means that f (z)dz → 0 , when R → ∞ ∫C R Hence, when R → ∞ , the following equation
R K f (x)dx + f (z)dz = 2π i ∑Res f (z) ∫−R ∫C j=1 z=z j R becomes R K f (x)dx + 0 = 2π i ∑Res f (z) ∫−R j=1 z=z j That is, R K f (x)dx = 2π i ∑Res f (z) z=z ∫−R j=1 j
5. Some Examples
Example 1 Evaluate the integral ∞ x 2 2 2 dx ∫0 (x +1) x 2
Example 2 Show that ∞ x 2 + 3 5 2 2 dx = π ∫−∞ (x +1)(x + 4) 6 ∞ x 2 + 3
⎡ zz22++33⎤ =+2π i ⎢ 22 22 ⎥ ⎣(zz++ 1)( 4)zi== ( zz ++ 1)( 4) z2 i⎦
175 ⎡⎤11 =+2π i ⎢⎥312ii ⎣⎦ ⎡⎤11 55π =+=⋅=22ππ ⎣⎦⎢⎥312 126
Example 3 Evaluate ∞ 1 4 dx ∫−∞ x +1
176 p(z) ** If z = z is a simple pole of f (z) = 0 q(z) pz() ⇒ Resfz ( ) = 0 zz= 0 q′()z0
Example 4 ∞ x2 Find dx using residues. ∫−∞ x4 +1
Example 5 ∞ x2 Find dx . ∫−∞ xx42++1
H.W.1 Evaluate the following integrals by means of residue calculus. Use the Cauchy principal value. ∞ xxx32+++1 ∞ 1 (a) dx , (b) dx , ab, real, b> 0 ∫−∞ x4 +1 ∫−∞ ()xa++22 b
H.W.2 Consider ∞ xx32+ dx ∫−∞ (1)(4)xx22++
177 Does the above-mentioned theorem apply directly to this integral? Evaluate this integral by evaluating the sum of the two Cauchy principal values.
H.W.3 (a) When a and b are positive, prove that ∞ dx π 1 = , for both ab≠ and ab= . ∫−∞ ()()xaxb2222++2(baab+ ) (b) Verify the answer to (a) by performing the integration with Symbolic Math Toolbox in MATLAB. Consider both possible cases. 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 27, Exercise 6.5, Pearson Education, Inc., 2005.】
H.W.4 Show that for a, b, c real, and bac2 < 4 , the following hold. ∞ dx 2π (a) 2 = ∫−∞ ax++ bx c 4ac− b2 ∞ dx4π a (b) = ∫−∞ ()ax22++ bx c 3 ⎡⎤4ac− b2 ⎣⎦⎢⎥ Obtain the result in (b) using residues, but check the answer by differentiating both sides of the equation in (a) with respect to c. You may differentiate under the integral sign. 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 28 and 29, Exercise 6.5, Pearson Education, Inc., 2005.】
∞ dx H.W.5 Find .
∞ dx H.W.6 Find .
∞ H.W.7 (a) Explain why x /(xdx4 + 1) cannot be evaluated through the use of a closed semicircular in the ∫0 upper or lower half-plane. (b) Consider the quarter-circle contour shown in the following figure is the arc of radius R > 1 . Show that R xy0 z⎡ z⎤ dx−+= dy dz2Resπ i at all poles in first quadrant 444∑ ⎢ 4⎥ ∫∫∫0 xy+++111RC1 z⎣ z + 1⎦ ∞ (c) Let R →∞ and show that xx/(4 += 1) dxπ / 4 . y ∫0 【本題摘自：A. David Wunsch, Complex Variable with Applications, CR：1 arc of radius 3rd ed., Problem 36, Exercise 6.5, Pearson Education, Inc., 2005.】
H.W.8 Show that ∞ xm π n dx = x ∫0 x +1 nmnsin[]π (+ 1) / R where n and m are nonnegative integers and nm− ≥ 2 .
178 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 38, Exercise 6.5, Pearson Education, Inc., 2005.】 y
CR：1 arc of radius
2/π n x R H.W.9 (a) Show that ∞ u1/l π du = ∫0 uk +1 kllksin[]π (+ 1) /( ) where k and l are integers, l > 0 , which satisfy lk()2− l ≥ . Take u1/l as nonnegative real function in the interval of integration.
§6-6 Improper Integrals From Fourier Analysis
1. Basic concept Residue theory can be useful in evaluating convergent improper integrals of the form ∞ ∞ ∞ fx()sin axdx, f()cos x axdx , or fxedx()iax (A) ∫−∞ ∫−∞ ∫−∞ where a denotes a positive constant and f ()x is a rational function of x. Assume that f ()xpxqx= ()/(), where px() and qx() are polynomials with real coefficients and no factors in common. Also, qz() has no real zeros. Integrals of type (A) occur in the theory and application of the Fourier integral. The method described in previous section cannot be applied directly here since sinaz2 =+ sin22 ax sinh ay and cosaz2 =+ cos22 ax sinh ay More precisely, since eeay− − ay sinh ay = 2 the moduli sin az and cosaz increase like eay as y tends to infinity. The modification illustrated in the example below is suggested by the fact that RRR f()cos x axdx+= i f ()sin x axdx f () x eiax dx ∫∫∫−−−RRR together with the fact that the modulus eeiaz=== ia() x+− iy eee ay iax − ay is bounded in the upper half plane y ≥ 0 .
Example 1 To give some insight into the method discussed in this section, we try to evaluate +∞ cos(3x ) dx ∫−∞ (1)1x −+2
179 using the technique of the preceding section.
Closed y
contour C Cz：||= R 1
i
x
−R 0 1 R
−iz3 Assuming we can argue that the integral over arc C1 vanishes as R →∞ (the troublesome e no longer appears), we have, in this limit, +∞ ee33ix⎡ iz ⎤ dx= 2Reπ i s in u.h.p. 22∑ ⎢ ⎥ ∫−∞ (1)1xz− +−+⎣ (1)1⎦ Putting exixix3 =+cos3 sin 3 and rewriting the integral on the left as two separate expressions, we have +∞cos3xxe +∞ sin3 3iz dx+= i dx2π i in u.h.p. ∫∫−∞22 −∞ ∑ 2 (1)1xz− +−+−+ (1)1res (1)1 z When we equate corresponding parts (reals and imaginaries) in this equation, the values of two real integrals are obtained: +∞ cos3xe⎡⎤⎡ 3iz ⎤ dx= Re⎢⎥ 2π i Res at all poles in u.h.p. (2) ∫ 22∑ ⎢ ⎥ −∞ (1)1xz−+⎣⎦⎢⎥⎣ (1)1 −+⎦ +∞ sin 3xe⎡⎤⎡ 3iz ⎤ dx= Im⎢⎥ 2π i Res at all poles in u.h.p. (3) ∫ 22∑ ⎢ ⎥ −∞ (1)1xz−+⎢⎥⎣⎦⎣ (1)1 −+⎦ Equation (2) contains the result being sought, while the integral in Eq. (3) has been evaluated as a bonus. Solving the equation (1)1z −=−2 and finding that zi= 1± , we see that on the right sides of Eqs. (2) and (3), we must evaluate a residue at the simple pole zi= 1+ . From the previous section, we obtain ⎛⎞ee33iz iz 2ππis Re ,1+= i 2 i lim ⎜⎟2 ⎝⎠(1)1z −+ zi→+(1 ) 2(z − 1) ==ππee−+33i − 3[]cos3 + i sin 3
180 Using the result in Eqs. (2) and (3), we have, finally, +∞cos3xx +∞ sin 3 dx==ππ e−−33cos3 and dx e sin 3 ∫∫−∞(1)1xx−+22 −∞ (1)1 −+
2. Theorem
Let f ()z have the following property in the half-plane Imz ≥ 0 . There exist constants, kR> 0,0 , and µ such that µ fz( )≤ k for all z≥ R0 z iθ Then if C1, is the semicircular arc Re , 0 ≤ θ ≤ π , and RR> 0 , we have limfzedz ( )ivz = 0 when v > 0 (4) R→∞ ∫ C1 When v < 0 , there is a corresponding theorem that applies in the lower half-plane. Notice that when the factor eivz is not present, we require k > 1, whereas the validity of Eq. (4) requires the less-restrictive condition k > 0 . To prove Eq. (4), we rewrite the integral on the left, which we call I, in terms of polar coordinates; taking zRe= iθ , dz= Reiθ idθ , we have
π iθ I ==∫∫fzedz()ivz fR ( e iθθ ) e ivRe iR e i dθ (5) C1 0 Recall now the inequality bb g()θ dgdθθθ≤ () ∫∫aa Applying this to Eq. (5) and resulting that eiθ = 1, we have
π iθ I ≤ RfRe(e)iivRθ e dθ (6) ∫0 We see that iθ eeivRe== ivR (cosθ +− i sinθθθ ) ee vR sin ivR cos Now eivR cosθ = 1 and since e−vRsinθ > 0 , we find that iθ eeivResin= − vR θ Rewriting Eq. (6) with the aid of the previous equation, we have π I ≤ RfRe(e)ivRθθ− sin dθ ∫0 With the assumption fz()=≤ fRe (ikθ )µ / R, it should be clear that
ππ µµ−−vRsinθθ vR sin I ≤=Redkkθ −1 edθ (7) ∫∫00RR Since sinθ is symmetric about θ = π /2 (see the y figure shown below), we can perform the integration on the right in Eq. (7) from 0 to π /2 and the double sinθ the result. 2θ Hence n 2µ π /2 I ≤ ed−vRsinθ θ (8) Rk−1 ∫0 The above figure also shows that over the interval x 0/2≤≤θ π , we have π/2 π µπµπ /2 Iede≤=−−−vRθπ2/ θ ⎡1 vR ⎤ kk−1 ⎣ ⎦ RvR∫0
181 With R →∞, the right-hand side of his equation becomes zero, which implies I → 0 in the same limit. Thus limfze ( )ivz dz= 0 R→∞ ∫ C1
♣ Jordan’s Lemma Pz() limedzivz => 0 if v0, degree Q −degree P ≥1 (9) R→∞ ∫ C1 Qz()
3. Improper Integrals From Fourier Analysis Evaluation for ∫ ePzQzdzivz ()/() : All zeros of Qz() in u.h.p. are assumed enclosed by the contour, and we also assume Qx()≠ 0 for all real values of x. Therefore, +∞ Px()ivx Pz () ivz⎡ Pz () ivz ⎤ ∫∫edx+= edz2Resπ i∑ ⎢ e⎥ at all poles in u.h.p. (10) −∞ Qx()C1 Qz ()⎣ Qz () ⎦ Now, provided the degrees of Q and P are as described in Eq. (9), we put R →∞ in Eq. (10) and discard the integral over C1 in this equation by invoking Jordan’s lemma. We obtain the following: +∞ Px()ivx⎡ Pz () ivz ⎤ edx= 2Resπ i∑ ⎢ e⎥ (11) ∫−∞ Qx()⎣ Qz () ⎦ The derivation of Eq. (11) requires that v > 0 , Qx()≠ 0 for −∞
4. If 1) function f is rational, and is in the following form as p(x) f (x) = q(x) where q(x) ≠ 0 , for all real x. 2) the degree of q(x) ≥ the degree of [ p(x) +1] imz 3) z j , j = 1, 2, 3, " , k are the poles of f (z)e located in the upper half plane 4) and m > 0 , then we obtain that K ∞ fxedx()imz= 2π i∑ Res() fze imz ∫ −∞ j=1 zz= j
5. Some Examples
Example 2 Evaluate ∞ cos x 2 2 dx ∫−∞ (x + 4)(x + 9)
182
Example 3 Evaluate ∞ sin x 2 2 dx ∫−∞ (x + 4)(x + 9) ∞ sin x
Example 4 Evaluate ∞ xsin x 2 2 dx ∫0 (x +1)(x + 4) xsin x
183 ∞ xsin x 1 ∞ xsin x 2 2 dx = 2 2 dx ∫0 (x +1)(x + 4) 2 ∫−∞ (x +1)(x + 4) 1 ⎪⎪⎧⎫⎡⎤zeiz ze iz =+Im⎨⎬ 2π i ⎢⎥ Res22 Res 22 2⎩⎭⎪⎪⎣⎦zi== (1)(4)(1)(4)zz++ z2 i zz ++ ⎪⎪⎧⎫⎡⎤zeiz ze iz =+Im ⎨⎬π i ⎢⎥22 22 (zz++ 1)( 4)zi== ( zz ++ 1)( 4) z2 i ⎩⎭⎪⎪⎣⎦ ⎡⎤⎛⎞ee−−12 =−Im ⎢⎥π i ⎜⎟ ⎣⎦⎝⎠66 π ⎛⎞11 =−⎜⎟2 6 ⎝⎠ee
Example 5 Evaluate +∞ xeixω +∞ x cosωx +∞ xsinωx dx , dx , and dx for ω > 0 . ∫−∞ x4 +1 ∫−∞ x4 +1 ∫−∞ x4 +1
6. Jordan's Inequality The cosθ is a strictly decreasing function for 0 ≤ θ ≤ π 2 . Therefore, the mean (or average) coordinate of the graph y = cos x over the range 0 ≤ x ≤ θ also decreasing steadily. This means that the coordinate is given by 1sinθ θ cos xdx= θ ∫ 0 θ Hence, for 0 ≤ θ ≤ π 2 , we have 2sinθ ≤≤1 πθ Thus, we have 2θ π ≤ sinθ ≤ θ , where 0 ≤ θ ≤ (A) π 2 and π π ed−msinθ θ < , for m > 0 ∫0 m which is known as Jordan's Inequality.
184 Observe that π ππ −−−mmmsinθθθ2 sin sin edθ =+ edθθπ ed ∫∫∫00 2 where m is a real number. Let θ = π − φ , substitute it into the last integral above. We can obtain that π 0 ed−−−mmsinθπφθ =− e sin( ) dθ ∫∫ππ 22 π = 2 ed−msinθ θ ∫ 0 Thus, we have π π ed−−mmsinθθθ = 2 2 ed sin θ ∫∫00 From Eq. (a), if m > 0 ee−−mmsinθ ≤ 2θπ / when 0 ≤ θ ≤ π 2 and so ππ π 22ed−−mmsinθθπθθ≤=− e 2 / d()1 e − m ∫∫002m Hence, π π 2 ed−msinθ θ ≤ , m > 0 (B) ∫ 0 2m Combining Eqs.(A) and (B) yields π π ed−msinθ θ < , for m > 0 ∫0 m
H.W. 1 Evaluate the following integrals by residue calculus. Use Cauchy principal values and take advantage of even and odd symmetries where appropriate. ∞ ()cos(2)xx32+ x ∞ xeix /3 (a) dx (b) dx ∫−∞ x4 +1 ∫−∞ ()4xi−+2 ∞ ()sin(/2)xxx32++ x ∞ cos x (c) dx (d) dx ∫−∞ (1)(4)xx22++ ∫0 (4)x22+ 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problems 7, 8, 12, and 13, Exercise 6.6, Pearson Education, Inc., 2005.】
H.W.2 Show that ∞ sinmx sin nx π dx= e−ma sinh na , for mn≥≥0 . Assume a is positive. ∫0 xa22+ 2a
H.W.3 The following is an extension of the work used in deriving Eq.(11) and deals with integrands having removable singularities. In parts (b) and (c), take mn,0> . eeimz− inz (a) Show that has a removable singularity at z = 0 . z ∞ sinmx− sin nx (b) Use the above to prove that dx = 0 . ∫0 x ∞ sinmx− sin nx π (c) Show that dx=−⎡ e−−na e ma ⎤ , where a > 0 . ∫−∞ xx()22+ a a 2⎣ ⎦
185 ∞ 2 cos (ππx / 2) −π (d) Show that 4 dx=−()1 + e . ∫−∞ x −1 4 1+ eizπ
H.W.4 To establish the well-known result ∞ sin x π dx = , ∫0 x 2 we proceed as follows: (a) Show that eiz −1 fz()= z Has a removable singularity at z = 0 . How should f (0) be defined to remove the singularity? (b) Using the contour shown below, prove that y Closed contour C Cz：1 ||= R
i
x 1 −R 0 R R eeix−−11 iz ∫∫dx+= dz 0 −RCxz1 and also RRcosxx− 1 sin 1 eiz ∫∫∫∫dx+=− i dx dz dz −−RRCCxxzz11 iθ (c) Evaluate the first integral on the above right by using the polar representation of C1: zRe= , 0 ≤≤θ π . Pass to the limit R →∞ and explain why the second integral on the right goes to zero. Thus prove that ∞ cosx − 1 dx = 0 ∫−∞ x and ∞ sin x dx = π ∫−∞ x and finally that ∞ sin x π ∫ dx = Closed y 0 x 2 contour C Cz：||= R 【本題摘自：A. David Wunsch, Complex Variable with 1 Applications, 3rd ed., Problem 24, Exercise 6.6, Pearson Education, Inc., 2005.】 i
H.W.5 Consider the problem of evaluating x 1 ∞∞cos xeix −R 0 R I ==dxRe dx ∫∫−∞coshx −∞ cosh x If we try to evaluate the preceding integral by employing the contour shown above and the methods leading to Eq. (12) we get into difficult because the function eziz /cosh has an infinite number of poles in the upper half-plane, i.e., at zin= (π +=π / 2), n 0, 1, 2, ... , which are the zeros of cosh z . Thus the Theorem in Point 2, in page 182 is inapplicable. However, we can determine I with the aid of the contour C shown below. (a) Using residues show that
186 eiz dz= 2π e−π /2 >∫ C cosh z
Closed y
contour C CIII
π
C IV π/2 CII
x −R CI 0 R
(b) Using the appropriate values of z integrals along the top and bottom portions of the rectangle show that RReeix i() x+ iπ ee iz iz ∫∫dx+++= dx ∫∫ dz dz2π e−π /2 −−RRcoshxxizz cosh(+ π ) CCII cosh IV cosh (c) Combine the first two integrals on the left in the preceding equation to show that R eeeix iz iz ∫∫∫dx(1++ e−−ππ ) dz + dz = 2π e /2 −RCCcoshxzzII cosh IV cosh
(d) Let R →∞ in the preceding. Using the ML inequality argue that the integrals along CII and CIV are zero in this limit.
Thus on CII and CIV we have eiz 1 ≤ coshzR sinh Prove finally that ∞ cosxe 2ππ−π /2 dx == ∫−∞ coshx 1+ e−π cosh(π / 2) and explain why ∞ cos x π dx = ∫0 coshx 2cosh(π / 2) 【本題摘自：A. David Wunsch, Complex Variable with Applications, 3rd ed., Problem 26, Exercise 6.6, Pearson Education, Inc., 2005.】
H.W.6 Using a technique similar to that employed in H.W.5 prove the following: ∞ cosh x π (a) dx = for a > 1 ∫0 coshax a cos(π / 2 a )
integrand, and prove that the integrations along segments CII and CIV each go to zero as R →∞. ∞ ex π (b) dx = for a > 1 ∫−∞ 1+ eax aasin(π / )
H.W.7 The result
∞ 22 π edx−ax = , a > 0 ∫−∞ a is derived in many standard texts on real calculus. Use this identity to show that
∞ 22π 2 2 ebxdxe−−mxcos = b/(4 m ) ∫−∞ m where b is a real number and m > 0 .
187 2
§6-7 Integrals Involving Indented Contour y 1. Theorem Suppose that
(1) a function f ()z has a simple pole at a point zx= 0 on the real axis, with a Laurent series representation in a C ρ ρ punctured disk 0 < zx−<02 R (see Fig.A) and with x residue B0 ; x0 (2) Cρ denotes the upper half of a circle zx− 0 = ρ , where R2 ρ0 ρ < R2 and the clockwise direction is taken. Then
limfzdz ( ) =− B0π i (1) ρ→0 ∫ Cρ Fig. A
Now the function g()z is continuous when zx− 02< R, according to the convergence theorem of
power series. Hence if we choose a number ρ0 such that ρ < ρ02< R (see Fig. A), it must be bounded
on the closed disc zx− 00< ρ , according to the section of discussing continuity. That is, there is a nonnegative constant M such that
g()zM≤ where zx− 00< ρ ;
and, since the length L of the path Cρ is L = πρ , it follows that ∫ gzdz() ≤= ML Mπρ Cρ Consequently, limgzdz ( )= 0 (3) ρ→0 ∫ Cρ
Inasmuch as the semicircle ⎯ −Cρ has parametric representation iθ zx=+0 ρ e (0 ≤≤θ π ) the second integral on the right in Eq. (2) has the value ππ 111iθ dz=− dz =−iθ ρ ie dθθπ =− i d =− i ∫∫CCρρzx−−00− zx ∫00ρe ∫ Thus
188 1 lim dz=−π i (4) ρ→0 ∫Cρ zx− 0 Limit (1) now follows by letting ρ tend to zero on each side of equation (2) and referring to limits (3) and (4).
♣ Concept review
1) Let the function f ()z possesses a simple pole at the point z0 , and let C be a circle of radius r
centered at z0 .
2) Suppose c−1 is the residue of f ()z at z0 . 3) Assuming that f ()z has no other singularities on and inside C, we know that
fzdz()== 2ππ ic−1 2 i Res() fz (1) >∫ z C 0
2. Theorem Let the function f ()z have a simple pole at the y Fig. B point z0 . An arc C0 of radius r is constructed θ2
using z0 as its center. The arc subtends an angle C0 α at z0 (see Fig. B). Then ⎡⎤α α limfzdz ()= 2π i⎢⎥ Res[] fz (), z0 (2) r→0 ∫C0 ⎣⎦2π where the integration is done in the r θ1 counterclockwise direction. (For clockwise z integration, a factor of −1 is placed on the right 0 of Eq. (2).) ♦ If C0 is a circle, then α = 2π . x
We first expand f ()z in a Laurent series about z0 .
Because of the simple pole at z0 , the series assumes the form cc∞ f ()zczzgz=+−=+−−11 ( )n () ()zz−−∑ n 0 () zz 00n=0 where ∞ n gz()=−≡∑ cn ( z z0 ) Taylorseries is analytic at z0 n=0 and cfz−1 = Res ( ) z0
We now integrate the series expansion of f ()z along C0 in Fig.B. Thus c f() z dz=+−1 dz g () z dz (3) ∫∫CC00()zz− 0 ∫ C 0 Applying ML inequality to the second integral on the right in Eq.(3), we have ∫ g()zdz≤ Mrα (4) C0 where
g(zM )≤≡ upper bound of gz ( ) in a neighborhood of z0
rα ≡ the length of contour C0 From eq. (4), we see that limgzdz ( )= 0 (5) r→0 ∫ C0 iθ iθ With zz=+0 re, dz= ire dθ , and the limits on θ indicated in Fig.B, we have
189 θα+ iθ ccired−−111 θα dz===iθ c−11απ i2 i c− (6) ∫∫C01()zz− 0 θ re 2π
+1311π /2 1 dx,,and dx dx ∫∫−−11x (2)xx− ∫π /2 sin
13. 課本§22-4 利用 Residue Theorem 求級數和的方法.
190