Computing residues

\[ \int_\gamma z^k \dz = \int_0^{2\pi} e^{ik\theta} ie^{i\theta } \dtheta = i\int_0^{2\pi} e^{i(k+1)\theta \dtheta } = \begin{cases} 2\pi i & k=-1 \\ 0 & \text{else}. \end{cases} \]

For the first integral: \[ \int_\gamma{1\over z^2-1}\dz &= {1\over 2}\int_\gamma {1\over z-1} + {1\over z+1}\dz \\ &= {1\over 2}\int_\gamma {1\over z-1} \\ &= {1\over 2}\cdot 2\pi i \Res_{z=1}f(z) \\ &= {1\over 2}\cdot 2\pi i \cdot 1 \\ &= \pi i ,\] using that \(f(z) = {1\over z-1}\) is already an expansion of \(f\) about \(z=1\) since it is a Laurent series in \((z-1)^k\), so the residue is \(1\).

For the second integral: attempt to define a primitive \[ F(z) \da {1\over 2}\log\qty{z-1\over z+1} \implies F'(z) = {1\over z^2-1} .\] If this primitive is well-defined on \(\gamma\), the integral will vanish because this is a closed curve. Choose the branch cut \(\CC\sm(-\infty, 0]\) and define \(g(z) = {z-1\over z+1}\), so that \(F(z) = {1\over 2}\log(g(z))\). Then then \(g(z)\in (-\infty, 0] \iff z\in [-1, 1]\), and this is well-defined on \(\gamma\) since \([-1, 1]\) does not intersect \(\gamma\).

Thus for \(\gamma\) any parameterization of \(\abs{z} = 2\), \[ \int_{\abs{z} = 2} f(z) \dz = \int_{\abs{z} = 2} F'(z)\dz = F(\gamma(0)) - F(\gamma(1)) = 0 .\]

Write \(f(z) = 2\sinh(z) = e^{z} - e^{-z}\) and apply the generalized Cauchy formula: \[ f^{(n-1)}(0) &= {(n-1)! \over 2\pi i} \int_{S^1} {f(z) \over (z-0)^n}\dz \\ \\ \implies \int_{S^1}{f(z)\over z^n}\dz &= {2\pi i \over (n-1)!} f^{(n-1)}(0) \\ &= {2\pi i\over (n-1)!} 2\cdot \qty{e^z - (-1)^{n-1} e^z \over 2}\evalfrom_{z=0} \\ &= {2\pi i\over (n-1)!} 2\cdot \qty{1 + (-1)^{n} \over 2} \\ &= \begin{cases} {4\pi i \over (n-1)!} & n \text{ even } \\ 0 & n \text{ odd }. \end{cases} .\]

For the smaller circle, use Cauchy’s formula \[ \int_{\abs{z} = 1} f(z) \dz &= \int_{\abs{z} = 1} { {z^2 + 1 \over z^2 + 4} \over z} \dz \\ &= 2\pi i \qty{z^2 + 1 \over z^2 + 4} \evalfrom_{z=0} \\ &= {i\pi \over 2} .\].

For the larger circle, break into principal parts. Write \[ {z^2 + 1 \over z(z^2+4)} = {a\over z} + {b\over z+2i} + {c\over z-2i} ,\] and use PFD/cover-up method/finding residues:

• \(a = \qty{z^2+1\over z^2+4}\evalfrom_{z=0} = {1\over 4}\)
• \(b= \qty{z^2 + 1 \over z(z-2i)}\evalfrom_{z=-2i} = {-4+1\over -2i(-4i)} = {3\over 8}\)
• \(c = \qty{z^2 + 1 \over z(z+2i)}\evalfrom_{z=2i} = {-4+1 \over 2i(4i)} = {3\over 8}\)

Thus \[ \int_\gamma f(z) \dz &= \int_\gamma \qty{ {1/4\over z} + {3/8\over z+2i} + {3/8 \over z-2i} } \dz \\ &= 2\pi i\qty{ {1\over 4} + {3\over 8} + {3\over 8}} .\]

Use polynomial long division to write \[ z^3 = z(z^2+1) - z \implies {z^3 \over z^2 + 1} = z - {z\over z^2 + 1} .\] Factor the latter part: \[ {z\over z^2 + 1} = {a\over z+i} + {b\over z-i} \implies a(z-i) + b(z+i) = z ,\] evaluate at \(z=i\) to get \(b=1/2\), and at \(z=-i\) to get \(a=1/2\). Thus \[ f(z) = z - {1/2 \over z+i} - {1/2 \over z-i} = P_\infty + P_{-i} + P_{i} ,\] yielding poles at \(\pm i\) with residues \[ \Res_{z=\infty} f(z) &= 1 \\ \Res_{z = i} f(z) &= -1/2 \\ \Res_{z = -i} f(z) &= -1/2 \\ .\]

Indeed, \[ f(z)=z-{1\over z}+O(z^{-3}) \qquad(z\to\infty), \] so the coefficient of \(z^{-1}\) is \(-1\) and therefore \(\Res_\infty f=-(-1)=1\). Equivalently, it is the negative of the sum of the finite residues.

Expand: \[ {1\over z - \sin(z)} &= \inverseof{z} \inverseof{\qty{1 - \inverseof{z} \sin(z) }} \\ &= \inverseof{z} \inverseof{\qty{1 - \inverseof{z}\qty{ z - {1\over 3!}z^3 + {1\over 5!} z^5 - \cdots}}}\\ &= \inverseof{z} \inverseof{\qty{1 - \qty{ 1 - {1\over 3!}z^2 + {1\over 5!} z^4 - \cdots}}} \\ &= \inverseof{z} \inverseof{\qty{{1\over 3!}z^2 - {1\over 5!}z^4 + \cdots}} \\ &= \inverseof{z} \cdot 3! z^{-2} \inverseof{\qty{1 - {1\over 5!/3!}z^2 + \cdots}} \\ &= {3!\over z^3} \qty{1 \over 1 - \qty{ {1\over 5\cdot 4}z^2 + \cdots}} \\ &= {3!\over z^3}\qty{1 + \qty{{1\over 5\cdot 4}z^2} + \qty{{1\over 5\cdot 4}z^2}^2 + \cdots} \\ &= 3! z^{-3} + {3!\over 5\cdot 4}\inverseof{z} + O(z) \\ .\]

First expand \(\inverseof{(\sin(z))}\): \[ {1\over \sin(z)} &= \inverseof{\qty{z - {1\over 3!}z^3 + {1\over 5!}z^5 -\cdots }} \\ &= \inverseof{z} \inverseof{\qty{1 - {1\over 3!}z^2 + {1\over 5!}z^4 - \cdots }} \\ &= \inverseof{z} \qty{1 + \qty{{1\over 3!}z^2 - {1\over 5!} z^4 + \cdots} + \qty{{1\over 3!}z^2 - \cdots}^2 + \cdots } \\ &= \inverseof{z}\qty{1 + {1\over 3!}z^2 + O(z^4) } ,\] using that \(\inverseof{(1-x)} = 1 + x + x^2 + \cdots\) with \(x={1\over 3!}z^2 - {1\over 5!} z^4 + \cdots\).

Thus \[ z^{-2}\inverseof{\qty{\sin(z)}} &= z^{-2} \cdot \inverseof{z}\qty{1 + {1\over 3!}z^2 + O(z^4) } \\ &= z^{-3} + {1\over 3!}\inverseof{z} + O(z) ,\] and the residue is the coefficient of \(\inverseof{z}\): \[ \Res_{z=0} {1\over z^2 \sin(z)}=\boxed{1\over 6} .\]

Applying the rational function formula: \[ \Res_{z=z_0}{1\over 1+z^2} &= {1\over 2z}\evalfrom_{z= z_0} \implies\\ \Res_{z=i}f(z) &= {1\over 2i} = -{i\over 2} \\ \Res_{z=-i}f(z) &= -{1\over 2i} = {i\over 2} .\]

Check that \(\dd{}{z} z^n+1 = nz^{n-1}\neq 0\) for \(z\neq 0\), so this has no repeated roots since \(z=0\) is not a root. Thus all of the poles are simple, and the residue of \(1/q\) at a simple zero \(z_0\) of \(q\) is \(1/q'(z_0)\): \[ \Res_{z=\omega_n} {1\over z^n + 1} &= {1 \over nz^{n-1}}\evalfrom_{z=\omega_n} \\ &= {1\over n\omega_n^{n-1}} \\ &= {\omega_n^{1-n}\over n} \\ &= -{\omega_n \over n} ,\] which follows from expanding \[ \omega_n^{1-n} = e^{i\pi (1-n) \over n} = e^{i\pi\over n}e^{-i\pi n \over n} = e^{i\pi \over n}\cdot (-1) = -\omega_n .\].

Recall \[ \Res_{z=\infty}f(z) =-\Res_{w=0}\frac{f(1/w)}{w^2}. \]

For \(f(z)=e^z\), \[ \frac{e^{1/w}}{w^2} =\sum_{k=0}^\infty {w^{-k-2}\over k!}. \] There is no \(w^{-1}\) term, hence \[ \Res_{z=\infty}e^z=0. \]

For \[ f(z)={z-1\over z+1}, \] we have \[ \frac{f(1/w)}{w^2} ={1\over w^2}{1-w\over1+w} ={1\over w^2}\qty{1-2w+2w^2-2w^3+\cdots}. \] Thus \[ \Res_{w=0}\frac{f(1/w)}{w^2}=-2, \] so \[ \Res_{z=\infty}{z-1\over z+1}=2. \] Equivalently, the only finite pole is at \(z=-1\) with residue \(-2\), and the sum of all residues on \(\CP^1\) is zero.

Goal: Show that the complex zeros of \(f(z) = \sin(\pi z)\) are exactly \(\ZZ\), each of order 1, and compute the residue of \(1/\sin(\pi z)\) at \(z = n \in \ZZ\).

<1>1. The zeros of \(\sin(\pi z)\) in \(\CC\) are exactly the integers.

<1>2. Each zero \(z = n\) is simple.

<1>3. \(1/\sin(\pi z)\) has a simple pole at each \(z = n \in \ZZ\).

<1>4. The residue of \(1/\sin(\pi z)\) at \(z = n\) is \((-1)^n/\pi\).

<1>5. Q.E.D.

• Factor \((1+z^2)^2 = ((z-i)(z+i))^2\), so \(f\) has poles at \(\pm i\) of order 2.
• Take a semicircular contour \(\gamma \da I_R \union D_R\), then \(f(z) \approx 1/z^4 \to 0\) for large \(R\) and \(\int_{D_R} f \to 0\).
• Note \(\int_{I_R} f \to \int_\RR f\), so \(\int_\gamma f \to \int_\RR f\).
• \(\int_\gamma f = 2\pi i \sum_{z_0} \Res_{z=z_0} f\), and \(z_0 = i\) is the only pole in this region.
• Compute \[ \Res_{z=i} f &= \lim_{z\to i} {1\over (2-1)!} \dd{}{z} (z-i)^2 f(z) \\ &= \lim_{z\to i} \dd{}{z} {1\over (z+i)^2 } \\ &= \lim_{z\to i} {-2 \over (z+i)^3 } \\ &= -{2 \over (2i)^3 } \\ &= {1\over 4i} \\ \\ \implies \int_\gamma f &= {2\pi i \over 4i} = \pi/2 ,\]

Let \(\log\) be the branch with \(0<\arg z<2\pi\) on \(\CC\setminus[0,\infty)\), and let \[ f(z)={\log^2 z\over(1+z^2)^2},\qquad (1+z^2)^2 = (z+i)^2(z-i)^2 .\] Integrate \(f\) over a keyhole contour with outer radius \(R\) and inner radius \(\rho\), slit along \([0,\infty)\):

Write \(I\) for the requested integral and \(A\coloneqq\int_0^\infty{dx\over(1+x^2)^2}={\pi\over4}\).

<1>1. The circles of radii \(R\) and \(\rho\) contribute \(0\) in the limit.

<1>2. The two edges of the slit contribute \(-4\pi i I+4\pi^2A\).

<1>3. \(\Res_{z=i}f=-{\pi\over4}+{i\pi^2\over16}\) and \(\Res_{z=-i}f={3\pi\over4}-{9i\pi^2\over16}\).

<1>4. \(I=\boxed{-\pi/4}\).

Take an indented semicircle. Let \(I\) be the original integral, then \[ I = {1\over 2i} \int_\RR {e^{iz} - 1 \over z } \dz .\]

Goal. Evaluate \(\oint_{|z|=1} \frac{1}{(z-2)(1+2z)^4(1-3z)^7}\,dz\) using residues at infinity.

<1>1. The integrand \(f(z) = \frac{1}{(z-2)(1+2z)^4(1-3z)^7}\) has poles inside \(|z| = 1\) at \(z = -1/2\) (order \(4\)) and \(z = 1/3\) (order \(7\)), and a simple pole at \(z = 2\) outside.

<1>2. The sum of all residues over the extended plane is \(0\).

<1>3. Compute \(\operatorname{Res}(f, \infty)\). <2>1. \(\operatorname{Res}(f, \infty) = -\operatorname{Res}\qty(\frac{1}{z^2} f(1/z), 0)\).

<2>2. \(\frac{1}{z^2} f(1/z) = \frac{1}{z^2} \cdot \frac{1}{(1/z - 2)(1 + 2/z)^4(1 - 3/z)^7} = \frac{z^{10}}{(1-2z)(z+2)^4(z-3)^7}\).

<2>3. This is analytic at \(z = 0\) (numerator has \(z^{10}\)), so \(\operatorname{Res}\qty(\frac{1}{z^2}f(1/z), 0) = 0\).

<2>4. Hence \(\operatorname{Res}(f, \infty) = 0\).

<1>4. Compute \(\operatorname{Res}(f, 2)\). <2>1. \(z = 2\) is a simple pole, so \(\operatorname{Res}(f, 2) = \lim_{z \to 2} (z-2) f(z) = \frac{1}{(1+2\cdot 2)^4 (1 - 3\cdot 2)^7} = \frac{1}{5^4 \cdot (-5)^7} = -\frac{1}{5^{11}}\).

<1>5. Sum of residues inside \(|z| = 1\). <2>1. \(\sum_{\text{inside}} \operatorname{Res} = -(\operatorname{Res}(f, 2) + \operatorname{Res}(f, \infty))\).

<2>2. \(= -\qty(-\frac{1}{5^{11}} + 0) = \frac{1}{5^{11}}\).

<1>6. \(\oint_{|z|=1} f(z)\,dz = 2\pi i \cdot \frac{1}{5^{11}}\).

<1>7. Q.E.D.