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MATH 311: COMPLEX ANALYSIS  CONTOUR INTEGRALS LECTURE
Recall the Residue Theorem: Let be a simple closed loop, traversed counterclockwise. Let f be a function that is analytic on and meromorphic inside . Then f (z) dz = 2i
c inside
Resc (f ).
This writeup shows how the Residue Theorem can be applied to integrals that arise with no reference to complex analysis. 1. Computing Residues Proposition 1.1. Let f have a simple pole at c. Then Resc (f ) = lim (z  c)f (z).
zc
Proposition 1.2. Let f have a pole of order n 1 at c. Define a modified function g(z) = (z  c)n f (z). Then Resc (f ) = 1 lim g (n1) (z). (n  1)! zc
Proof. It suffices to prove the second proposition, since it subsumes the first one. Recall that the residue is the the coefficient a1 of 1/(z  c) in the Laurent series. The proof is merely a matter of inspection: an a1 f (z) = + ··· + + a0 + · · · , n (z  c) zc and so the modified function g is g(z) = an + · · · + a1 (z  c)n1 + a0 (z  c)n + · · · , whose (n  1)st derivative is g (n1) (z) = (n  1)! a1 + n(n  1) · · · 2 a0 (z  c) + · · · . Thus
zc
lim g (n1) (z) = g (n1) (c) = (n  1)! a1 .
This is the desired result. In applying the propositions, we do not go through these calculations again every time. L'Hospital's Rule will let us take the limits without computing the Laurent series. Incidentally, note that a slight rearrangement of the proposition, Resc g(z) (z  c)n+1 = g (n) (c) , n! n 0,
shows that the Residue Theorem subsumes Cauchy's integral representation for derivatives.
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MATH 311: COMPLEX ANALYSIS  CONTOUR INTEGRALS LECTURE
2. Rational Functions Let f (x) = p(x)/q(x) be a rational function of the real variable x, where q(x) = 0 for all x R. Further assume that deg(f ) 2, meaning that deg(q) deg(p) + 2. Then
f (x) dx = 2i
x= Im(c)>0
Resc (f ).
Here is an example. Let f (z) = We want to compute the integral
z2 . z4 + 1
I=
x=
f (x) dx.
To do so, let r be a large positive number, and let consist of two pieces: the segment [r, r] of the real axis and the upper halfcircle r of radius r. Thus
r
f (z) dz =
x=r
f (x) dx +
r
f (z) dz.
On the halfcircle r we have 1 (since deg(f ) = 2), r2 while the length of r is r. Therefore, letting the symbol " " mean "is asymptotically at most," r r f (z) dz  0. r2 r On the other hand, f (z)
r x=r
f (x) dx  I, and so f (z) dz  I.
r
r
But for all large enough values of r, also f (z) dz = 2i
Im(c)>0
Resc (f ).
So to evaluate the integral I we need only compute the sum of the residues. The denominator z 4 +1 of f (z) has simple poles at the fourth roots of 1. These values are 3 5 7 8 , 8 , 8 , 8 , where 8 = e2i/8 .
3 Of the four roots, only 8 and 8 lie in the upper half plane. Compute, using L'Hospital's Rule that
z8
lim
(z  8 )z 2 z 2 + (z  8 )2z z2 1 1 7 = lim = lim = lim = = 8, z8 z8 4z 3 z8 4z z4 + 1 4z 3 48 4
MATH 311: COMPLEX ANALYSIS  CONTOUR INTEGRALS LECTURE
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and similarly
z8
lim3
3 (z  8 )z 2 1 5 = 3 = 8. z4 + 1 48 4
So the sum of the residues is
5 7  2i i 8 + 8 = = . Resc (f ) = 4 4 2 2 Im(c)>0 i I = 2i = . 2 2 2
And consequently the integral is
3. Rational Functions Times Sine or Cosine Consider the integral I= sin x dx. x=0 x To evaluate this real integral using the residue calculus, define the complex function eiz . z This function is meromorphic on C, with its only pole being a simple pole at the origin. Let r be a large positive real number, and let be a small positive real number. Define a contour consisting of four pieces: f (z) = = [r, ] [, r] r , where is the upper halfcircle of radius , traversed clockwise, while r is the upper halfcircle of radius r, traversed counterclockwise. By the Residue Theorem (which subsumes Cauchy's Theorem), f (z) dz = 0.
Note that on we have e0 1 = , z z so that, since is a halfcircle traversed clockwise, f (z) f (z) dz i,
and this approximation tends to equality as shrinks toward 0. Meanwhile, parameterize r by letting z = rei where 0 . On r we have f (z) = f (rei ) = exp(irei )  exp(ir(cos + i sin )) er sin = = , rei r r
and dz = irei d = r d, so that f (z) dz
r =0
er sin d.
The integrand bounded above by 1 (and now matter how large r gets, the integrand always equals 1 for = 0 and = ), but as r grows, the integrand tends to 0
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MATH 311: COMPLEX ANALYSIS  CONTOUR INTEGRALS LECTURE
uniformly on any any compact subset of (0, ), and so the integral goes to 0 as r goes to . The remaining two pieces of the integral are, by a change of variable,
 x=r
eix dx + x
r x=
eix dx = x =
x=r r x=
eix d(x) + x e
r ix
r x=
eix dx x
= 2i
x= +
e dx x sin x dx. x
ix
Now let 0 and let r . Our calculation has shown that 2iI  i = 0, i.e., I= . 2
4. Rational Functions of Cosine and Sine Consider the integral
2
I=
=0
d , a + cos
a > 1.
This integral is not improper, i.e., its limits of integration are finite. The distinguishing characteristic here is that the integrand is a rational function of cos and sin , integrated from 0 to 2. Thus we may set z = ei , 0 2,
and view the integral as a contour integral over the unit circle. On the unit circle we have z + z 1 z  z 1 dz cos = , sin = , d = . 2 2i iz Thus the integral becomes the integral of a rational function of z over the unit circle, and the new integral can be computed by the residue calculus. For the particular integral in question, the calculation is I=
z=1
1 a+
z+z 1 2
·
dz 2 = iz i
z=1
z2
dz . + 2az + 1
Analyze the denominator as follows: z 2 + 2az + 1 = (z  r1 )(z  r2 ), r1 + r2 = 2a, r1 r2 = 1.
Neither root lies on the unit circle since the condition a > 1 ensures that the original denominator a + cos is never zero. Let r1 be the root inside the circle and r2 be the root outside it. Thus the integral is I= 2 · 2iResr1 i 1 (z  r1 )(z  r2 ) = 4 lim
zr1
4 (z  r1 ) = . (z  r1 )(z  r2 ) (r1  r2 )
But the roots of the quadratic polynomial z 2 + 2az + 1 are 2a ± 4a2  4 r1 , r2 = = a ± a2  1, 2
MATH 311: COMPLEX ANALYSIS  CONTOUR INTEGRALS LECTURE
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and so I= 4 2 = . (r1  r2 ) a2  1
5. Integration Around a Branch Point Consider the integral xs dx , 0 < s < 1. x=0 x + 1 For positive real values x we have the formula I= xs = exp(s ln x). Let be a small positive real number and r be a large positive real number. Define a contour consisting of four pieces: = + r  , where · + is the the real axis traversed from up to r, viewing its points as having argument 0, · r is the circle of radius r traversed counterclockwise, viewing its points as having argument increasing from 0 to 2, ·  is the the real axis traversed from r down to , viewing its points as having argument 2, · and is the circle of radius traversed clockwise, viewing its points as having argument decreasing from 2 to 0. On r we have z s  =  exp(s ln r  is) = rs , so that, since z + 1 r on r , z s z+1 Also, r has length 2r, and so
r
rs . r
r
z s dz z+1
2rs  0.
Here it is relevant that s > 0. A similar analysis, using the condition 1  s > 0 and the fact that z + 1 1 on shows that also
z s dz z+1
21s  0.
0+
For points z = x on + we have z s = xs , but for points z = x on  , where we are viewing the argument as 2 rather than 0, we have z s = exp(s log z) = exp(s ln z  2is) = xs e2is . And so r z s xs + dz = (1  e2is ) dx z+1 +  x= x + 1
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MATH 311: COMPLEX ANALYSIS  CONTOUR INTEGRALS LECTURE
The coefficient in front of the integral is 1  e2is = (eis  eis )eis = 2i sin(s) eis , and so as 0+ and r , we have
z s dz  2i sin(s)eis I. z+1 z s z+1
But also,
z s dz = 2i Res1 z+1 = 2i(1)s
= 2i exp(s ln   1  is) = 2ieis . In sum, 2i sin(s)eis I = 2ieis , so that . sin s A similar calculation, using the same contour, shows that I=
x=0
ln x dx = 0. x2 + 1
6. The Riemann Zeta Function for Even Integers The Riemann zeta function is (s) = 1 , ms m=1
Re(s) > 1.
To evaluate (k) where k 2 is an even integer, we use the meromorphic function f (z) = cot z. This function has a simple pole with residue 1 at z = 0 because for z near 0, 1 f (z) . z Thus, by Zperiodicity, f has a simple pole with residue 1 at each integer. Let n be a positive integer. Let be the rectangle with vertical sides at ±(n+1/2) and with horizontal sides at ±in. For any even integer k 2 we have cot z dz = 2i Res0 zk cot z zk +2 1 mk m=1
n
.
But
e2iz + 1 2i = i + 2iz , e2iz  1 e 1 while from the homework we know that cot z = i 2iz = e2iz  1
j=0
Bj (2iz)j , j!
MATH 311: COMPLEX ANALYSIS  CONTOUR INTEGRALS LECTURE
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with Bk = 0 for all odd k except for B1 = 1/2. And so
cot z =
j=0
Bj (2i)j z j1 , j!
summing only over even j.
Therefore, Res0 cot z zk = (2i)k Bk k! for even k 2.
Summarizing so far, the integral is cot z dz = 2i zk (2i)k Bk 1 +2 k! mk m=1
n n
 2i
(2i)k Bk + 2(k) . k!
But also, the integral tends to 0 as n gets large. To see this, estimate the integrand cot z/z k on , by returning to the formula cot z = i + If z = ±(n + 1/2) + iy then e2iz  1 = e2y  1 < 1, and so the formula shows that (small exercise) if z = ±(n + 1/2) + iy then  cot z < . On the other hand, if z = x ± in and n is large then e2iz  = e±2n is either very large or very close to 0, and in either case if z = x ± in then  cot z . Thus two conditions hold on ,  cot z and 1 zk 1 . n2 2i . e2iz  1
Since has length roughly 8n, it follows that cot z n dz  0. k z
By the explicit formula for the integral, it follows that 2(k) =  That is, (2) = and so on. 2 , 6 (4) = 4 , 90 (6) = 6 , 945 (2i)k Bk k! for even k 2.
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MATH 311: COMPLEX ANALYSIS  CONTOUR INTEGRALS LECTURE
7. The Fourier Transform of the Gaussian The onedimensional Gaussian function is the function f : R  R,
f (x) = ex .
2
An exercise in multivariable calculus shows that f is normalized, f (x) dx = 1.

Its Fourier transform is the function
f : R  R,
f () =

f (x)e2ix dx.
(The Fourier transform of the Gaussian is realvalued because the Gaussian is even and the sine function is odd, so that the imaginary part of the integral vanishes.) ^ We use contour integration to show that f = f . Compute that the integrand is f (x)e2ix = e(x
2
+2ix 2 )  2
e
= e(x+i) f (),
2
and so the Fourier transform of f is in fact
f () = f ()

e(x+i) dx.
2
It suffices to show the integral in the previous display is 1, and to show this, it suffices to show that the integral is the integral of the original Gaussian, since the original Gaussian is normalized. To show that the integral in the previous display is the integral of the Gaussian, let be the rectangular contour that traverses from r to r along the real axis, then up to r + i, then horizontally back to r + i, and finally back down to r. Let 2 f (z) = ez . Since f is entire, the Residue Theorem says that f (z) dz = 0.
Also, the integrals along the sides of the rectangle go to 0 as r gets large. This is because if z = ±r + iy for any y between 0 and then f (z) = e(±r+iy)  = e(r
2 2
±2iryy 2 )
 = e(r
2
y 2 )
,
and so as soon as r , the integrand is uniformly small as y varies from 0 to . Since the total integral vanishes and the side integrals go to 0 as r grows, the top and bottom integrals agree in the limit as r . But the top integral is the integral that we want to equal 1,
+i +i 
ez dz =
 
2
e(x+i) dx,
2
while the bottom integral is the original Gausssian integral, which does equal 1, ez dz =
2
ex dx = 1.
2
Thus the Fourier transform equals the original Gaussian, as claimed. That is, f =f for f (x) = ex .
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