Computing Integrals

By the ML estimate, \(\int_{C_R} f \to 0\).

The residue contribution: note the simple pole at \(\omega_n \coloneqq e^{i\pi \over n}\), \begin{align*} \mathop{\mathrm{Res}}_{z=\omega_n} f(z) = {1\over n\omega_n^{n-1}} = {\omega \over n\omega^n} = -{\omega_n \over n} .\end{align*}

The segment contributions: \(\int_{\gamma_1}f\to I\), and \begin{align*} \int_{\gamma_2}f(z) \,dz= \int_\infty^0 {1\over 1 + (\zeta_nt)^n} \zeta_n \,dt= -\zeta_n I ,\end{align*} so the contour contributions sum to \((1-\zeta_n)I\).

Solving: \begin{align*} I = -{2\pi i \omega_n\over n(1-\zeta_n)} = -2\pi i {1\over \omega_n^{-1}- \omega_n} = -2\pi i {1\over 2i \sin\qty{\pi \over n}} = {\pi \over n} \csc\qty{\pi \over n} .\end{align*}

Write \(\omega_{n, k} = \exp\qty{(2k+1)i\pi \over n}\) and factor \(z^n+1\) as \begin{align*} z^n+1 = \prod_{1\leq k \leq n}(z-\omega_{n, k}) = (z-e^{i\pi \over n})(z-e^{3i\pi \over n}) \cdots (z-e^{(2n-1)i\pi \over n}) .\end{align*} Note that only the root \(e^{i\pi\over n}\) lies in the \(2\pi/n\) wedge, so it is the only (simple) pole of \(f(z) \coloneqq{1\over 1+z^n}\) in this region. Since the pole is simple, we can compute the residue easily. Write \(r_0 \coloneqq e^{e\pi\over n}\), then By L’Hopital, \begin{align*} \mathop{\mathrm{Res}}_{z = r_0} {1\over 1+z^n} &= \lim_{z\to r_0} {z-r_0 \over 1 + z^n} \\ &= \lim_{z\to r_0} {1\over nz^{n-1}} \\ &= {1\over nr_0^{n-1}} \\ &= {1\over n e^{i\pi (n-1) \over n}} \\ &= n^{-1}{\exp\qty{-i\pi (n-1)\over n }} .\end{align*}

Take a contour \(\Gamma\) comprised of

• \(\gamma_1 = [0, R] \subseteq {\mathbf{R}}\)
• \(\gamma_2 = \left\{{Re^{it} {~\mathrel{\Big\vert}~}t\in [0, 2\pi/n]}\right\}\)
• \(\gamma_3 = \zeta_n [0, R]\)

By the residue theorem \begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=r_0} f(z) = I \coloneqq\int_\Gamma f = \qty{\int_{\gamma_1} + \int_{\gamma_2} + \int_{\gamma_3}}f .\end{align*}

so in the limit we have \begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=r_0}f(z) &= \qty{1 - \zeta_n}\int_{\gamma_1}f \\ \implies \int_{\gamma_1} f &= {2\pi i \mathop{\mathrm{Res}}_{z=r_0}f(z) \over 1 - \zeta_n}\\ &= {2\pi i e^{-\pi (n-1) \over n} \over n\qty{1-e^{2\pi i \over n}}} \\ &= {2\pi i \over n} \left[ e^{i\pi} e^{-i\pi \over n}\qty{1 - e^{2\pi i \over n}} \right]^{-1}\\ &= {2\pi i \over n} \left[ -1\qty{e^{-i\pi \over n} - e^{\pi i \over n}} \right]^{-1}\\ &= {2\pi i \over n} \left[ 2i \sin\qty{\pi\over n} \right]^{-1}\\ &= {\pi \over n\sin\qty{\pi \over n}} .\end{align*}

Sketch: see here.

Write \(I\) for the integral, \(\zeta_3\coloneqq e^{2\pi i\over 3}, \omega_3 \coloneqq e^{i\pi\pver 3}\). Take a indented semicircular wedge \(\Gamma\) at an angle of \(2\pi/3\), noting the pole at \(\omega_3 \coloneqq e^{i \pi \over 3}\):

Choosing a branch cut of \(\log\) along \(\theta = -\pi/2\), so \(\arg(z) \in (-\pi/2, 3\pi/2)\), this makes \(f(z) \coloneqq z^{\alpha-1}/(1+z^3)\) meromorphic on \(\Gamma\).

By the ML estimate, the integrals along \(C_{\varepsilon}, C_R\) will vanish in the limit.

The contribution from the contour: parameterize \(\gamma_2\) as \(\left\{{\zeta_3 t{~\mathrel{\Big\vert}~}t\in [{\varepsilon}, R]}\right\}\), then \begin{align*} \int_{\gamma_2}f(z) \,dz &= \int_R^{\varepsilon}{(\zeta_3 t)^{\alpha - 1} \over 1 + (\zeta_3 t)^3} \zeta_3\,dt\\ &= \zeta_3 \zeta_3^{\alpha-1}\int_R^{\varepsilon}{t^{\alpha-1} \over 1 + t^3}\,dt\\ &\to -\zeta_3^\alpha I ,\end{align*} so the total contribution is \begin{align*} \oint_\Gamma f \to \qty{\int_{\gamma_1} + \int_{\gamma_2}}f = (1-\zeta_3^{\alpha}) I .\end{align*}

Computing the contributions from residues: \begin{align*} \mathop{\mathrm{Res}}_{z=\omega_3} f(z) &= \lim_{z\to \omega_3} {(z-\omega_3) z^{\alpha-1} \over 1+z^3} \\ &{ \overset{\scriptscriptstyle\text{LH}}{=} }\lim_{z\to\omega_3} {z^{\alpha-1} \over 3z^2} \\ &= {1\over 3} \omega_3^{\alpha-3} \\ &= -{1\over 3}\omega_3^\alpha .\end{align*}

Combining it all using the residue theorem: \begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=\omega_3} f(z) &= \oint_\gamma f \\ \implies I &= {-2\pi i\over 3} {\omega_3^\alpha \over 1-\zeta_3^\alpha } \\ &= {-2\pi i\over 3} {1 \over \omega_3^{-\alpha} - \omega_3^\alpha } \\ &= {-2\pi i\over 3} {1 \over -2i\sin\qty{\pi\alpha\over 3} } \\ &= {\pi\over 3}\csc\qty{\alpha\pi\over 3} .\end{align*}

Sketch:

• Set \(z=e^{i\theta}\) to get \begin{align*} \frac{2}{i} \int_{|z|=1} \frac{\mathrm{d} z}{z^{2}+2 a z+1} .\end{align*}

• Factor into two roots \(r_1, r_2\). Use that without loss of generality, \(r_1\in {\mathbb{D}}\) and \(r_2\in {\mathbb{D}}^c\), with neither on \(S^1\) to compute the residue \(4\pi/(r_1 -r_2)\)

Note \(\cosh(z) \coloneqq{1\over 2}(e^z + e^{-z})\), and \begin{align*} \cosh(z) &= 0 \\ \iff e^z + e^{-z} &= 0 \\ \iff e^{-z}(e^{2z} + 1) &= 0 \\ \iff e^{2z} &= -1 \quad \text{since }{\left\lvert {e^{-z}} \right\rvert} = e^{\Re(z)} > 0 \\ \iff 2z &= (2k+1)i\pi \\ \iff z &\in \left\{{\cdots, {-3i\pi \over 2}, {-i\pi \over 2}, {i\pi \over 2}, {3i\pi \over 2}, \cdots}\right\} .\end{align*} So take the following rectangular contour enclosing the singularity \(z= i\pi/2\):

Then letting \(\Gamma\) be the entire contour and \(I\) be the desired integral, we can solve for \(I\): \begin{align*} \int_\Gamma f &= \int_{-R}^R f + \int_{\gamma_{R_1}} f + \int_{\gamma_2}f + \int_{\gamma_{R_2}} f \\ \int_\Gamma f &= I + \int_{\gamma_{R_1}} f + \int_{\gamma_2}f + \int_{\gamma_{R_2}} f \\ \\ I &= \int_{\gamma_{R_1}} f + \int_{\gamma_2}f + \int_{\gamma_{R_2}} f - \int_\Gamma f ,\end{align*} being very sloppy about the fact that we’re going to take \(R\to \infty\).

Computing the residue term \(\int_\Gamma f = 2\pi i \mathop{\mathrm{Res}}_{z=i\pi/2} f(z)\): \begin{align*} {1\over 2\ pi i } \int_\Gamma f &= \mathop{\mathrm{Res}}_{z = i\pi/2} f(z)\\ &= \mathop{\mathrm{Res}}_{z = i\pi/2} {e^{i\xi z} \over \cosh(z) } \\ &= {e^{i\xi z} \over {\frac{\partial }{\partial t}\,} \cosh(z) } \Big|_{i\pi/2}\\ &= {e^{i\xi \cdot i\pi/2} \over \sinh(i\pi/2)} \\ &= {e^{-\xi \pi/ 2} \over i} \\ \implies 2\pi i \mathop{\mathrm{Res}}_{z=i\pi/2} f(z) &= 2\pi e^{-\xi \pi/ 2} .\end{align*} using that \(2\sinh(i\pi/2) = e^{i\pi/2} - e^{-i\pi/2} = i-(-i) = 2i\).

The \(\gamma_2\) term: parameterize \(\gamma_2 = \left\{{t + i\pi {~\mathrel{\Big\vert}~}t\in [-R, R]}\right\}\), then \begin{align*} \int_{\gamma_2} f &= -\int_{-\gamma_2} f \\ &= -\int_{-R}^{R} { e^{iz} \over \cosh(t + i\pi) } \,dz, \quad z = t+i\pi, \,dz= \,dt\\ &= -\int_{-R}^{R} { e^{i\xi(t+i\pi)} \over \cosh(t + i\pi) }\,dt\\ &= -e^{-\xi \pi} \int_{-R}^R {e^{i\xi t} \over \cosh(t+i\pi)} \,dt\\ &= e^{-\xi \pi} \int_{-R}^R {e^{i\xi t} \over \cosh(t)} \,dt\\ &\coloneqq e^{-\xi \pi} I ,\end{align*} using that \(\cosh(z + i\pi) = -\cosh(z)\).

The two \(\gamma_{R_i}\) terms: the claim is that these vanish in the limit \(R\to \infty\). Parameterize \(\gamma_{R_2} = \left\{{R = i \pi t {~\mathrel{\Big\vert}~}t\in [0, 1]}\right\}\), then \begin{align*} {\left\lvert {\int_{\gamma_{R_2}} f} \right\rvert} &= {\left\lvert { \int_0^1 {e^{i\xi(R+i\pi t)} \over \cosh(R + i\pi t)} \,dt} \right\rvert} \\ &= {\left\lvert { 2e^{i\xi R} \int_0^1 {e^{-\xi \pi t } \over {e^{R+ i\pi t} + e^{-R - i\pi t}} } \,dt} \right\rvert} \\ &\leq 2 \int_0^1 {\left\lvert { {e^{-\xi \pi t } \over {e^{R+ i\pi t} + e^{-R - i\pi t}} } } \right\rvert} \,dt\\ &\leq 2 \int_0^1 {e^{-\xi \pi t } \over { {\left\lvert { e^{R+ i\pi t} } \right\rvert} - {\left\lvert { e^{-R - i\pi t} } \right\rvert} } }\,dt\\ &= 2 \int_0^1 {e^{-\xi\pi t} \over e^R - e^{-R} } \\ &= { 2\over e^R - e^{-R}} \int_0^1 e^{-\xi \pi t} \,dt\\ &= {2\over e^R - e^{-R}} \qty{-1\over \xi \pi} e^{\xi \pi t}\Big|_{t=0}^{t=1} \\ &= {2\over e^R - e^{-R}} \qty{1 - e^{-\pi \xi } \over \xi \pi} \\ &\overset{R\to\infty}\longrightarrow 0 .\end{align*}

Putting it all together: \begin{align*} (1 + e^{-\xi \pi}) I &= 2\pi e^{-\xi \pi / 2} \\ \implies I &= {2\pi e^{-\xi \pi / 2} \over 1 + e^{-\xi \pi }} \\ &= {2\pi \over e^{\xi\pi/2} (1 + e^{-\xi \pi })} \\ &= {2\pi \over e^{\xi\pi/2} + e^{-\xi\pi/2} } \\ &= {2\pi \over 2\cosh(\xi\pi/2)} \\ &= {\pi \over \cosh\qty{\xi\pi\over 2}} \\ \\ &= \pi { \mathrm{sech} }\qty{\xi\pi\over 2 } .\end{align*}

• Let \(I\) be the integral over \({\mathbf{R}}\). Since \(f(x)\) is even, the original integral is \({1\over 2}I\).

• Write \(f(z) = e^{iz} / (z^2 + b^2)\). Take a semicircular contour \(\Gamma \coloneqq\gamma_1 + \gamma_2\) where \(\gamma_1\) is \([-R, R]\) on \({\mathbf{R}}\) and \(\gamma_2\) is the usual half-circle of radius \(R\).

• Claim: \(\int_{\gamma_2} f \overset{R\to\infty}\longrightarrow 0\), so \(\int_\Gamma \to \int_{\mathbf{R}}f(z)\).

  • Easy estimate, just be careful with the \(i\) in the exponent: \begin{align*} {\left\lvert {f} \right\rvert} = { {\left\lvert {e^{iz} } \right\rvert} \over {\left\lvert {z^2 + b^2} \right\rvert} } = {e^{-\Re z} \over {\left\lvert {z^2 + b^2} \right\rvert} } \leq {1\over {\left\lvert {z^2 + b^2} \right\rvert}} \overset{R\to\infty}\longrightarrow 0 .\end{align*}

• Compute \(\int_\Gamma f\) by residues: factor \(z^2 + b^2 = (z+ib)(z-ib)\), so the contour only contains the order 1 pole \(z_0 = ib\).

• Compute the residue: \begin{align*} \mathop{\mathrm{Res}}_{z=ib}f = \lim_{z\to ib} (z-ib) {e^{iz} \over (z+ib)(z-ib) } = { e^{iz} \over z+ib} \Big|_{z=ib} = {e^{i(ib)} \over 2ib} = {e^{-b} \over 2ib} .\end{align*}

• So the intermediate integral is \(I\) is \(2\pi i\) times this, i.e. \(I = \pi e^{-b} / b\).

• And the original integral is \({1\over 2}I = \pi e^{-b} \over 2b\).

Part 1: S&S claim this can be done without a calculation – maybe \({\left\lvert {\psi_a'(z)} \right\rvert}^2\) is constant on a circle of radius \(r\)…?

Part 2: \begin{align*} {1\over \pi} \iint_{\mathbb{D}}{\left\lvert {\psi_a'(z)} \right\rvert} \,dz &= {1\over \pi} \iint_{\mathbb{D}}{\left\lvert {1- {\left\lvert {a} \right\rvert}^2 \over (1-\overline{a} z)^2 } \right\rvert} \,dz\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 \int_0^{2\pi} {1 \over {\left\lvert { 1-\overline{a} re^{it} } \right\rvert}^2 } r\,dt\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 \int_0^{2\pi} {1 \over (1-\overline{a} re^{it} )(1 - a re^{-it}) } r\,dt\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 \oint_{{{\partial}}{\mathbb{D}}} {1 \over (1-\overline{a} rz )(1 - a r \overline{z}) } {r\over iz} \,dz\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 \oint_{{{\partial}}{\mathbb{D}}} {1 \over (1-\overline{a} rz )(1 - a r z^{-1}) } {r\over iz} \,dz\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 {r\over i} \cdot \oint_{{{\partial}}{\mathbb{D}}} {1 \over (1-\overline{a} rz )(z - a r) } \,dz\,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 {r\over i} \cdot 2\pi i \mathop{\mathrm{Res}}_{z=ar} {1 \over (1-\overline{a} rz )(z - a r) } \,dr\\ &= {1- {\left\lvert {a} \right\rvert}^2 \over \pi} \int_0^1 {r\over i} \cdot 2\pi i {1 \over (1-\overline{a} rz ) }\Big|_{z=ar} \,dr\\ &= {2\pi \qty{ 1- {\left\lvert {a} \right\rvert}^2} \over \pi} \int_0^1 {r \over 1- {\left\lvert {a} \right\rvert}^2r^2 } \,dr\\ &= {1-{\left\lvert {a} \right\rvert}^2 \over {\left\lvert {a} \right\rvert}^2} \log\qty{1 \over 1 - {\left\lvert {a} \right\rvert}^2 } .\end{align*}