MATH 3: 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 = 2πi Res c (f). c inside This writeup shows how the Residue Theorem can be applied to integrals that arise with no reference to complex analysis.. Computing Residues Proposition.. Let f have a simple pole at c. Then Res c (f) = lim z c (z c)f(z). Proposition.2. Let f have a pole of order n at c. Define a modified function g(z) = (z c) n f(z). Then Res c (f) = (n )! lim z c g(n ) (z). Proof. It suffices to prove the second proposition, since it subsumes the first one. Recall that the residue is the the coefficient a of /(z c) in the Laurent series. The proof is merely a matter of inspection: f(z) = a n (z c) n + + a z c + a 0 +, and so the modified function g is whose (n )st derivative is g(z) = a n + + a (z c) n + a 0 (z c) n +, g (n ) (z) = (n )! a + n(n ) 2 a 0 (z c) +. Thus lim z c g(n ) (z) = g (n ) (c) = (n )! a. 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, ( ) g(z) Res c (z c) n+ = g(n) (c), n 0, n! shows that the Residue Theorem subsumes Cauchy s integral representation for derivatives.
2 MATH 3: 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 meaning that deg(q) deg(p) + 2. Then x= Here is an example. Let We want to compute the integral deg(f) 2, f(x) dx = 2πi f(z) = x= Im(c)>0 z2 z 4 +. f(x) dx. Res c (f). 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 half-circle r of radius r. Thus r f(z) dz = f(x) dx + f(z) dz. x= r r On the half-circle r we have f(z) r 2 (since deg(f) = 2), while the length of r is πr. Therefore, letting the symbol mean is asymptotically at most, f(z) dz πr r r r 2 0. On the other hand, and so r x= r But for all large enough values of r, also f(z) dz = 2πi f(x) dx r I, f(z) dz r I. Im(c)>0 Res c (f). So to evaluate the integral I we need only compute the sum of the residues. The denominator z 4 + of f(z) has simple poles at the fourth roots of. These values are ζ 8, ζ 3 8, ζ 5 8, ζ 7 8, where ζ 8 = e 2πi/8. Of the four roots, only ζ 8 and ζ8 3 lie in the upper half plane. Compute, using L Hospital s Rule that (z ζ 8 )z 2 z 2 + (z ζ 8 )2z z 2 lim z ζ 8 z 4 = lim + z ζ 8 4z 3 = lim z ζ 8 4z 3 = lim z ζ 8 4z = = ζ7 8 4ζ 8 4,
MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE 3 and similarly (z ζ lim 8)z 3 2 z ζ8 3 z 4 = + 4ζ8 3 So the sum of the residues is Im(c)>0 And consequently the integral is Res c (f) = ζ5 8 + ζ 7 8 4 = ζ5 8 4. = 2 i 4 2πi i 2 2 = π 2. = i 2 2. 3. Rational Functions Times Sine or Cosine Consider the integral sin x x=0 x dx. To evaluate this real integral using the residue calculus, define the complex function f(z) = 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 ε be a small positive real number. Define a contour consisting of four pieces: = [ r, ε] ε [ε, r] r, where ε is the upper half-circle of radius ε, traversed clockwise, while r is the upper half-circle of radius r, traversed counterclockwise. By the Residue Theorem (which subsumes Cauchy s Theorem), f(z) dz = 0. Note that on ɛ we have f(z) e0 z = z, so that, since ɛ is a half-circle traversed clockwise, ε f(z) dz πi, and this approximation tends to equality as ε shrinks toward 0. Meanwhile, parameterize r by letting z = re iθ where 0 θ π. On r we have f(z) = f(re iθ ) = exp(ire iθ ) sin θ exp(ir(cos θ + i sin θ)) e r re iθ = =, r r and dz = ire iθ dθ = r dθ, so that π f(z) dz r θ=0 e r sin θ dθ. The integrand bounded above by (and now matter how large r gets, the integrand always equals for θ = 0 and θ = π), but as r grows, the integrand tends to 0
4 MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE uniformly on any any compact subset of (0, π), and so the integral goes to 0 as r goes to. The remaining two pieces of the integral are, by a change of variable, ε x= r e ix x dx + r x=ε e ix x dx = = ε x=r r = 2i x=ε r x=ε e ix x d( x) + e ix e ix dx x sin x x dx. r x=ε e ix x dx Now let ε 0 + and let r. Our calculation has shown that 2iI πi = 0, i.e., π 2. Consider the integral 4. Rational Functions of Cosine and Sine 2π θ=0 dθ a + cos θ, a >. 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 θ and sin θ, integrated from 0 to 2π. Thus we may set z = e iθ, 0 θ 2π, and view the integral as a contour integral over the unit circle. On the unit circle we have cos θ = z + z, sin θ = z z 2 2i 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 z = a + z+z 2 Analyze the denominator as follows: dz iz = 2 i z =, dθ = dz iz. dz z 2 + 2az +. z 2 + 2az + = (z r )(z r 2 ), r + r 2 = 2a, r r 2 =. Neither root lies on the unit circle since the condition a > ensures that the original denominator a + cos θ is never zero. Let r be the root inside the circle and r 2 be the root outside it. Thus the integral is 2 ( ) i 2πiRes r (z r )(z r 2 ) But the roots of the quadratic polynomial z 2 + 2az + are r, r 2 = 2a ± 4a 2 4 2 (z r ) = 4π lim z r (z r )(z r 2 ) = 4π (r r 2 ). = a ± a 2,
MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE 5 and so 4π (r r 2 ) = 2π a2. Consider the integral 5. Integration Around a Branch Point x=0 x s dx, 0 < s <. x + For positive real values x we have the formula x s = exp( s ln x). Let ε be a small positive real number and r be a large positive real number. Define a contour consisting of four pieces: = + r ε, where + is the the real axis traversed from ε 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 2π, is the the real axis traversed from r down to ε, viewing its points as having argument 2π, and ε is the circle of radius ε traversed clockwise, viewing its points as having argument decreasing from 2π to 0. On r we have z s = exp( s ln r isθ) = r s, so that, since z + r on r, z s z + r s r. Also, r has length 2πr, and so z r s z + dz 2πr s r 0. Here it is relevant that s > 0. A similar analysis, using the condition s > 0 and the fact that z + on ε shows that also z ε s z + dz 2πε s ε 0 + 0. For points z = x on + we have z s = x s, but for points z = x on, where we are viewing the argument as 2π rather than 0, we have z s = exp( s log z) = exp( s ln z 2πis) = x s e 2πis. And so ( + + ) z s r z + dz = ( x s e 2πis ) x=ε x + dx
6 MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE The coefficient in front of the integral is e 2πis = (e πis e πis )e πis = 2i sin(πs) e πis, and so as ε 0 + and r, we have z s z + dz 2i sin(πs)e πis I. But also, In sum, z s ( ) z s z + dz = 2πi Res z + = 2πi( ) s = 2πi exp( s ln isπ) = 2πie πis. 2i sin(πs)e πis 2πie πis, so that π sin πs. A similar calculation, using the same contour, shows that x=0 ln x x 2 dx = 0. + 6. The Riemann Zeta Function for Even Integers The Riemann zeta function is ζ(s) = m=, Re(s) >. ms To evaluate ζ(k) where k 2 is an even integer, we use the meromorphic function f(z) = π cot πz. This function has a simple pole with residue at z = 0 because for z near 0, f(z) z. Thus, by Z-periodicity, f has a simple pole with residue at each integer. Let n be a positive integer. Let be the rectangle with vertical sides at ±(n+/2) and with horizontal sides at ±in. For any even integer k 2 we have π cot πz z k dz = 2πi ( Res 0 ( π cot πz z k ) + 2 n m= But π cot πz = πi e2πiz + e 2πiz = πi + 2πi e 2πiz, while from the homework we know that 2πiz e 2πiz = j=0 B j j! (2πiz)j, m k ).
MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE 7 with B k = 0 for all odd k except for B = /2. And so Therefore, π cot πz = j=0 B j j! (2πi)j z j, summing only over even j. ( ) π cot πz Res 0 z k = (2πi)k B k k! Summarizing so far, the integral is ( π cot πz (2πi) k B k z k dz = 2πi + 2 k! n m= for even k 2. ) ( n (2πi) k ) B k m k 2πi + 2ζ(k). k! But also, the integral tends to 0 as n gets large. To see this, estimate the integrand π cot πz/z k on, by returning to the formula If z = ±(n + /2) + iy then π cot πz = πi + 2πi e 2πiz. e 2πiz = e 2πy <, and so the formula shows that (small exercise) if z = ±(n + /2) + iy then π cot πz < π. On the other hand, if z = x ± in and n is large then e 2πiz = e ±2πn is either very large or very close to 0, and in either case if z = x ± in then π cot πz π. Thus two conditions hold on, π cot πz π and n 2. z k Since has length roughly 8n, it follows that π cot πz z k dz n 0. By the explicit formula for the integral, it follows that 2ζ(k) = (2πi)k B k k! for even k 2. That is, and so on. ζ(2) = π2 6 π4 π6, ζ(4) =, ζ(6) = 90 945,
8 MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE 7. The Fourier Transform of the Gaussian The one-dimensional Gaussian function is the function f : R R, f(x) = e πx2. An exercise in multivariable calculus shows that f is normalized, Its Fourier transform is the function f : R R, f(x) dx =. f(η) = f(x)e 2πiηx dx. (The Fourier transform of the Gaussian is real-valued 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 ˆf = f. Compute that the integrand is f(x)e 2πiηx = e π(x2 +2iηx η 2) e πη2 = e π(x+iη)2 f(η), and so the Fourier transform of f is in fact f(η) = f(η) e π(x+iη)2 dx. It suffices to show the integral in the previous display is, 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 + iη, then horizontally back to r + iη, and finally back down to r. Let f(z) = e πz2. 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 η then f(z) = e π(±r+iy)2 = e π(r2 ±2iry y 2) = e π(r2 y 2), and so as soon as r η, the integrand is uniformly small as y varies from 0 to η. 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, +iη +iη e πz2 dz = e π(x+iη)2 dx, while the bottom integral is the original Gausssian integral, which does equal, e πz2 dz = e πx2 dx =. Thus the Fourier transform equals the original Gaussian, as claimed. That is, f = f for f(x) = e πx2.