Exercises: Contour integration

The integrand $f\in { \mathsf{O}} \qty{1\over z^4} \subseteq { \mathsf{O}} \qty{1\over z^{1+{\varepsilon}}}$, so a semicircular contour works. Factoring the denominator: find a principal root $$\begin{align*} \omega^4 = -1 = e^{i\pi} \implies \omega = e{i\pi\over 4} ,\end{align*}$$ so $$\begin{align*} z^4 + 1 = (z-\omega)(z-\omega\zeta_4)(z-\omega\zeta_4^2)(z-\omega\zeta_4^3) \leadsto {\pi \over 4}, {3\pi \over 4}, {5\pi \over 4}, {7\pi \over 4} .\end{align*}$$ So there are two simple poles in ${\mathbb{H}}$:

Write them as $z_1 = e^{i\pi \over 4}$ and $z_2 = e^{3 i \pi \over 4}$. Computing the residues: $$\begin{align*} \mathop{\mathrm{Res}}_{z=z_k} f(z) &= \lim_{z\to z_k} {z-z_k \over z^4 + 1} \\ &{ \overset{\scriptscriptstyle\text{LH}}{=} }\lim_{z\to z_k} {1\over 4z^3} \\ &= {4z^{-3} }\Big|_{z=z_k} ,\end{align*}$$ so $$\begin{align*} 2\pi i \sum \mathop{\mathrm{Res}}_{z=z_k} &= {2\pi i \over 4}\qty{z_1^{-3} + z_2^{-3}} \\ &= {\pi i \over 2}\qty{e^{-3 i \pi \over 4} + e^{-9 i \pi \over 4}} \\ &= {\pi i \over 2}\qty{e^{-3 i \pi \over 4} + e^{-i \pi \over 4}} \\ &= {\pi i \over 2}\cdot e^{-2i\pi \over 4}\qty{e^{- i \pi \over 4} + e^{i \pi \over 4}} \\ &= \pi i\, {1\over i} \cos\qty{\pi \over 4} \\ &= {\pi \sqrt 2 \over 2} .\end{align*}$$

Note that the function is even, so $$\begin{align*} \int_{0^\infty} f(x) \,dx= {1\over 2} \int_{\mathbb{R}}f(x)\,dx= {1\over 2} {\pi \over \sqrt 2} = {\pi \over 2 \sqrt 2} ,\end{align*}$$ using the solution from a previous problem.

A sector will work, since there is a symmetry under $z\to \zeta_4 z$ and $f(z) \sim z^{-4}$, so the semicircular piece will vanish. Take the contour $\Gamma$ comprised of

• $\gamma_1: \left\{{t + 0i {~\mathrel{\Big\vert}~}t\in [0, R]}\right\}$,
• $C_R: \left\{{Re^{it}{~\mathrel{\Big\vert}~}t\in [0, \pi/2]}\right\}$,
• $\gamma_2: \left\{{0 + it {~\mathrel{\Big\vert}~}t\in [0, R]}\right\}$,

oriented counter-clockwise. Note that $z^4+1 = \prod_{k=0}^3 (z-\omega \zeta_4)$ where $\omega = e^{i\pi \over 4}$ and $\zeta_4 = e^{2\pi i\over 4} = i$, so there is only one pole at $z_0 \coloneqq e^{i\pi\over 4}$ within this contour.

Computing the symmetry: $$\begin{align*} \int_{\gamma_2} f(z) \,dz &= \int_R^0 {i \over (ti)^4 + 1}\,dt\qquad z=ti,\, \,dz= i\,dt\\ &= -i \int_0^R {1\over t^4 + 1}\,dt\\ &= -i \int_{\gamma_1} f(z) \,dz ,\end{align*}$$ so applying the residue theorem and noting that $\int_{C_R}f\to 0$, $$\begin{align*} 2\pi \mathop{\mathrm{Res}}_{z=z_0} f(z) = \int_\Gamma f(z) \,dz= \qty{\int_{\gamma_1} + \int_{C_R} + \int_{\gamma_2}}f \longrightarrow(1-i) I .\end{align*}$$ Computing the residue: $$\begin{align*} \mathop{\mathrm{Res}}_{z=z_0}f(z) &= {1\over 4z^3}\Big|_{z= e^{i\pi \over 4}} \\ &= {1\over 4}e^{-3i\pi \over 4} \\ &= {1\over 4}e^{5i\pi \over 4} .\end{align*}$$

Thus $$\begin{align*} I &= (1-i)^{-1}2\pi i \mathop{\mathrm{Res}}_{z=z_0} f(z) \\ &= \qty{\mkern 1.5mu\overline{\mkern-1.5mu1-i\mkern-1.5mu}\mkern 1.5mu \over {\left\lvert {1-i} \right\rvert}^2} 2\pi i \cdot {1\over 4}e^{5i\pi \over 4} \\ &= {1+i \over 2} {\pi i \over 2}e^{5i\pi \over 4} \\ &= {e^{i\pi \over 4} \over \sqrt{2} } {\pi i \over 2}e^{5i\pi \over 4} \\ &= {\pi i \over 2\sqrt{2}}e^{6i\pi \over 4} \\ &= {\pi i \over 2\sqrt{2}}(-i) \\ &= {\pi \over 2\sqrt{2}} .\end{align*}$$

Consider the auxiliary function $g(z) \coloneqq\log(z) f(z)$, and take a keyhole contour:

Let $\Gamma$ be the counterclockwise contour consisting of

• $C_{\varepsilon}= \left\{{{\varepsilon}e^{it}{~\mathrel{\Big\vert}~}t\in [0+{\varepsilon}, 2\pi - {\varepsilon}]}\right\}$
• $\gamma_+ = \left\{{x+i{\varepsilon}{~\mathrel{\Big\vert}~}x\in [{\varepsilon}, R]}\right\}$
• $C_R= \left\{{R e^{it}{~\mathrel{\Big\vert}~}t\in [0+{\varepsilon}, 2\pi - {\varepsilon}]}\right\}$
• $\gamma_- = \left\{{x-i{\varepsilon}{~\mathrel{\Big\vert}~}x\in [{\varepsilon}, R]}\right\}$

Computing the symmetry: $$\begin{align*} \int_{\gamma_-}{\log(z) \over z^{4} + 1} \,dz &= \int_{R}^{\varepsilon}{\log(x-i{\varepsilon}) \over (x-i{\varepsilon})^4 + 1} \,dx\qquad z=x-i{\varepsilon}, \,dz= \,dx\\ &\to - \int_{{\varepsilon}}^R {\log(x) + 2\pi i\over x^4 + 1}\,dz\\ &= -\int_{\gamma_+} {\log(z) \over z^4+1}\,dz- 2\pi i\int_{\gamma_+} {1\over z^4+1} \,dz ,\end{align*}$$ so $$\begin{align*} \int_{\gamma_+} f + \int_{\gamma_-}f \longrightarrow-2\pi i I .\end{align*}$$ By the ML estimate, since $\log(z)/z^4\to 0$ as ${\left\lvert {z} \right\rvert}\to \infty$, $\int_{C_R}g(z) \to 0$. Similarly, since $\log(z) / (z^4+1)\to 0$ as ${\left\lvert {z} \right\rvert}\to 0$, $\int_{C_{\varepsilon}}\to 0$. We’re then left with the sum of residues at $e^{k i \pi \over 4}$ for $k = 1,3,5,7$. We have $$\begin{align*} \mathop{\mathrm{Res}}_{z=z_k} f(z) &= {1\over 4z^3}\Big|_{z=z_k} \\ &= {z\over 4z^4}\Big|_{z=z_k} \\ &= - {z\over 4}\Big|_{z=z_k} \qquad \text{ since } z_k^4 = -1 \\ &= -{z_k \over 4} ,\end{align*}$$ so $$\begin{align*} \mathop{\mathrm{Res}}_{z=z_k}g(z) = -{z_k \over 4}\log(z_k) .\end{align*}$$

Now use that in the limit, $$\begin{align*} 2\pi i \sum_k \mathop{\mathrm{Res}}_{z=z_k}g(z) &= \int_\Gamma g(z) \,dz\\ &= \qty{\int_{\gamma_+} + \int_{\gamma_i}} f \\ &= -2\pi i I ,\end{align*}$$ so $I = -\sum_k \mathop{\mathrm{Res}}_{z=z_k} g(z)$.

Being very careful to note that we’ve chosen a branch of $\log$ where $\operatorname{Arg}(z) \in (0, 2\pi)$ in order to get the signs right, $$\begin{align*} -\sum_k \mathop{\mathrm{Res}}(z=z_k) g(z) &= {1\over 4}\qty{ z_1\log(z_1) + z_2\log(z_2)+z_3\log(z_3)+z_4\log(z_4) } \\ &= {1\over 4}\qty{ z_1 {i\pi \over 4} + z_2 {3i\pi \over 4} + z_3 {5i\pi \over 4}+ z_4 {7i\pi \over 4} } \\ &= {i\pi \over 16}\qty{ z_1 + 3z_2 + 5z_3+ 7z_4 } \\ &= {i\pi \over 16}\qty{ \omega \zeta_4^0 + 3\omega\zeta_4^1 + 5\omega\zeta_4^2+ 7\omega \zeta_4^3 } \\ &= {i\pi \omega \over 16}\qty{ 1 + 3i + -5 + -7i } \\ &= {i\pi \omega \over 16}(-4-4i) \\ &= -{i\pi \omega \over 4}(1+i) \\ &= -{i\pi \omega \over 2}{ \omega \over \sqrt{2} } \\ &= -i\omega^2 {\pi \over 2 \sqrt 2} \\ &= {\pi \over 2\sqrt{2} } ,\end{align*}$$ using that ${\sqrt 2\over 2}(1+i) = e^{i\pi \over 4} = \omega$ and $\omega^2 = i$.

Use that $f(z) \sim 1/z^4$:

Then $$\begin{align*} \int_{C_1 + C_R} f = \int_{C_1} f + \int_{C_R} f = 2\pi i \mathop{\mathrm{Res}}_{z=i} {1\over (1+z^2)^2} .\end{align*}$$

Note $f$ factors: $$\begin{align*} f(z) = {1 \over \qty{(z-i)(z+i)}^2 } ,\end{align*}$$ so $z=i$ is a pole of order 2.

Compute the residue within the contour: $$\begin{align*} \mathop{\mathrm{Res}}_{z=i} f(z) = \lim_{z\to i} {\frac{\partial }{\partial z}\,} {1\over (z+i)^2} = - {2 \over (z+i)^3 }\Big|_{z=i} = - {2\over (2i)^3 } = {1\over 4i} = -{i\over 4} .\end{align*}$$

Now solve: $$\begin{align*} 2\pi i \qty{- {i\over 4}} = \int_{C_1}f + \int_{C_R} f \coloneqq I_R + \int_{C_R}f \implies I_R = {\pi \over 4} - \int_{C_R} f .\end{align*}$$

Note $I_R \overset{R\to \infty}\longrightarrow I \coloneqq\int_{\mathbb{R}}f$, so it suffices to show the semicircular error term vanishes as $R\to \infty$. Parameterize $C_R = \left\{{z=R e^{it} {~\mathrel{\Big\vert}~}t\in [0, \pi]}\right\}$, so ${\left\lvert {z} \right\rvert} = R$ on $C_R$. Letting $z=e^{it}$, $\,dz=iR e^{it}\,dt$, $$\begin{align*} {\left\lvert {\int_{C_R} f(z) \,dz} \right\rvert} &= {\left\lvert {\int_{C_R} {1\over (1+z^2)^2} \,dz} \right\rvert} \\ &\leq \int_{C_R}{\left\lvert { {1\over (1+z^2)^2} } \right\rvert} \,dz\\ &= \int_{C_R} {1\over {\left\lvert {1+z^2} \right\rvert}^2 } \,dz\\ &\leq \int_{C_R} {1\over \qty{1 - {\left\lvert {z} \right\rvert}^2 }^2 } \,dz\\ &= \int_{C_R} {1\over \qty{1 - R^2 }^2 } \,dz\\ &= {1\over \qty{1 - R^2 }^2 } \int_{C_R} \,dz\\ &= {1\over \qty{1 - R^2 }^2 } \cdot \mathop{\mathrm{length}}(C_R) \\ &= {1\over \qty{1 - R^2 }^2 } \cdot \qty{1\over 2}2\pi R \\ &= {\pi R \over R^4 + O(R^3)} \\ &= \pi\qty{ 1 \over R^3 + O(R^2)} \\ &\overset{R\to\infty}\longrightarrow 0 ,\end{align*}$$ where we’ve used a variant of the triangle inequality: $$\begin{align*} {\left\lvert {a\pm b} \right\rvert} \geq {\left\lvert { {\left\lvert {a} \right\rvert} - {\left\lvert {b} \right\rvert} } \right\rvert} \implies {1\over {\left\lvert {a \pm b} \right\rvert}} \leq {1\over {\left\lvert {a} \right\rvert} - {\left\lvert {b} \right\rvert} } .\end{align*}$$

• Factor $(1+z^2)^2 = ((z-i)(z+i))^2$, so $f$ has poles at $\pm i$ of order 2.
• Take a semicircular contour $\gamma \coloneqq I_R \cup D_R$, then $f(z) \approx 1/z^4 \to 0$ for large $R$ and $\int_{D_R} f \to 0$.
• Note $\int_{I_R} f \to \int_{\mathbb{R}}f$, so $\int_\gamma f \to \int_{\mathbb{R}}f$.
• $\int_\gamma f = 2\pi i \sum_{z_0} \mathop{\mathrm{Res}}_{z=z_0} f$, and $z_0 = i$ is the only pole in this region.
• Compute $$\begin{align*} \mathop{\mathrm{Res}}_{z=i} f &= \lim_{z\to i} {1\over (2-1)!} {\frac{\partial }{\partial z}\,} (z-i)^2 f(z) \\ &= \lim_{z\to i} {\frac{\partial }{\partial z}\,} {1\over (z+i)^2 } \\ &= \lim_{z\to i} {-2 \over (z+i)^3 } \\ &= -{2 \over (2i)^3 } \\ &= {1\over 4i} \\ \\ \implies \int_\gamma f &= {2\pi i \over 4i} = \pi/2 ,\end{align*}$$

The integrand is $f\in { \mathsf{O}} \qty{1\over z^{2n+2}} \subseteq { \mathsf{O}} \qty{1\over z^{1+{\varepsilon}}}$, so a semicircular contour will work:

Thus $$\begin{align*} I &= 2\pi i \sum_{z_k \in {\mathbb{H}}} \mathop{\mathrm{Res}}_{z=z_k} {1\over (1+z^2)^{n+1}} \\ &= 2\pi i \mathop{\mathrm{Res}}_{z=i} {1\over (z+i)^{n+1}(z-i)^{n+1} } \\ &= 2\pi i \lim_{z\to i}{1\over n!}\qty{{\frac{\partial }{\partial z}\,}}^n (z+i)^{-(n+1)} \\ &= 2\pi i \lim_{z\to i}{1\over n!}\qty{{\frac{\partial }{\partial z}\,}}^{n-1} -(n+1)(z+i)^{-(n+2)} \\ &= 2\pi i \lim_{z\to i}{1\over n!}\qty{{\frac{\partial }{\partial z}\,}}^{n-2} -(n+1) \cdot -(n+2) (z+i)^{-(n+3)} \\ &= \qquad \vdots \\ &= 2\pi i \lim_{z\to i}{1\over n!}\qty{{\frac{\partial }{\partial z}\,}}^{n-k} (-1)^k (n+1)(n+2)\cdots(n+k) (z+i)^{-(n+k-1)} \\ &= \qquad \vdots \\ &= 2\pi i \lim_{z\to i} {1 \over n!}(-1)^n (n+1)(n+2)\cdots(2n)(z+i)^{-(2n-1)} \\ &= { \qty{2n}_{ (n) } \over n!} 2\pi i (-1)^n (2i)^{-(2n-1)} \\ &= { \qty{2n}_{ (n) } \over n!} 2\pi i (-1)^n {1\over 2^{2n+1} i^{2n+1}} \\ &= { \qty{2n}_{ (n) } \over n!} \pi (-1)^n {1\over 4^n i^{2n}} \\ &= { \qty{2n}_{ (n) } \over n!} \pi (-1)^n {1\over 4^n (-1)^n } \\ &= { \qty{2n}_{ (n) } \over n!} {\pi \over 4^n } .\end{align*}$$

Write the integrand as $f$.

Factor the denominator: $$\begin{align*} x^2 + 4x + 13 = 0 &\implies x^2 + 4x + \qty{4\over 2}^2 = -13 + \qty{4\over 2}^2 \\ &\implies (x+2)^2 = -9 \\ &\implies x = -2 \pm 3i ,\end{align*}$$ one of which is in ${\mathbb{H}}$. Write these as $$\begin{align*} z_1 \coloneqq-2+3i && z_2 \coloneqq-2 - 3i .\end{align*}$$

So let $\Gamma$ be comprised of

• $C_1 = [-R,R]$,
• $C_2 = \left\{{Re^{it} {~\mathrel{\Big\vert}~}t\in [0, \pi]}\right\}$,
• $\Gamma = C_1 + C_2$, then $$\begin{align*} 2\pi i \sum_{z_k\in {\mathbb{H}}} \mathop{\mathrm{Res}}_{z=z_k}f(z) = \int_\Gamma f = \qty{\int_{C_1} + \int_{C_2}}f ,\end{align*}$$ where $\int_{C_2} f\to 0$ and $\int_{C_1} f\to I$. So in the limit, $I = 2\pi i \mathop{\mathrm{Res}}_{z=z_1} f(z)$. Computing this residue: note $z_1$ is a pole of order 2, so $$\begin{align*} \mathop{\mathrm{Res}}_{z=z_1} &= \lim_{z\to z_1} {\frac{\partial }{\partial z}\,} (z-z_1)^2f(z) &= \lim_{z\to z_1} {\frac{\partial }{\partial z}\,} {z \over (z-z_2)^2} \\ &= \lim_{z\to z_1} { (z-z_2)^2 - 2z(z-z_2 ) \over (z-z_2)^4} \\ &= \lim_{z\to z_1} { (z-z_2) - 2z \over (z-z_2)^3} \\ &= \lim_{z\to z_1} -{ z+z_2 \over (z-z_2)^3} \\ &= - {z_1 + z_2 \over (z_1 - z_2)^2}\\ &= - {z_1 + \mkern 1.5mu\overline{\mkern-1.5muz_1\mkern-1.5mu}\mkern 1.5mu \over (z_1 - \mkern 1.5mu\overline{\mkern-1.5muz_1\mkern-1.5mu}\mkern 1.5mu )^2}\\ &= - {2\Re(z_1) \over (2i\Im(z_1))^3} \\ &= - {2\cdot (-2) \over (2i\cdot 3) ^3} \\ &= {4\over 2^3 \cdot 3^3 \cdot i^3} \\ &= {-i \over 2 \cdot 3^3 i^2} \\ &= {i\over 54} ,\end{align*}$$ so $$\begin{align*} I = 2\pi i \cdot {i\over 54} = -{\pi \over 27} .\end{align*}$$

The integrand is even, so $$\begin{align*} I = \Re{1\over 2} \tilde I \coloneqq{1\over 2} \int_{\mathbb{R}}{e^{iz} \over (z+ib)(z-ib)} \end{align*}$$

Since $f \sim 1/x^2$, the ML estimate on a semicircular contour works:

Then $\int_{C_R} f\to 0$ and $\int_{C_1} f\to \tilde I$. Thus $$\begin{align*} \tilde I &= 2\pi i \mathop{\mathrm{Res}}_{z=ib} \\ &= 2\pi i \lim_{z\to ib}{e^{iz}\over (z+ib)} \\ &= 2\pi i {e^{-b} \over 2i b} \\ &= {\pi e^{-b} \over b} \end{align*}$$ and $$\begin{align*} I = \Re {1\over 2} \tilde I = {\pi e^{-b} \over 2b} .\end{align*}$$

Write $f(z) = {ze^{iz} \over 1+z^2}$, and note that $f\in { \mathsf{O}} \qty{1\over z}$, so the usual semicircular contour with the ML estimate won’t work. Claim: a semicircular contour with a better estimate will work:

Writing $f(z) = e^{iz}g(z)$ where $g(z) \coloneqq{z\over 1 + z^2}$, we have $g\in { \mathsf{O}} \qty{1\over z} \to 0$ as ${\left\lvert {z} \right\rvert}\to \infty$, so Jordan’s lemma applies. Write $C_1 = [-R, R]$ and $C_R = \left\{{Re^{it} {~\mathrel{\Big\vert}~}t\in [0, \pi]}\right\}$, then $$\begin{align*} {\left\lvert {\int_{C_R} e^{iz} g(z)\,dz} \right\rvert}\leq \pi M_R,\, \qquad M_R \coloneqq\sup_{z\in C_R}{\left\lvert {z\over 1+z^2} \right\rvert} .\end{align*}$$ Now use that ${z+1\over z^2}\leq M{\left\lvert {z} \right\rvert}$ for ${\left\lvert {z} \right\rvert}$ large enough to conclude this integral goes to zero. By the residue theorem, $$\begin{align*} 2\pi i \sum_{z_k\in {\mathbb{H}}}\mathop{\mathrm{Res}}_{z=z_k}f(z) = \int_{C_1 + C_R} f = \qty{\int_{C_1} + \int_{C_R}}f \overset{R\to\infty} \int_{C_1} f = I ,\end{align*}$$ so it suffices to compute the residues of $f$. Check that $1+z^2 = (1+i)(1-i)$, so $z_1 = i \in {\mathbb{H}}$ is a simple pole and $$\begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=i} f(z) &= 2\pi i \lim_{z\to i} {e^{iz} \over z+i} \\ &= 2\pi i {i\over 2ei} = {\pi \over e} ,\end{align*}$$ so $$\begin{align*} I = {\pi \over e} .\end{align*}$$

Note that the usual thing won’t work, since ${\cos(z) \over z+i}\neq \Re\qty{e^{iz}\over z+i};$ the complex constant in the denominator throws this off! Instead, use $\cos(z) = {1\over 2}(e^{iz} + e^{-iz})$ to decompose into two integrals: $$\begin{align*} I \coloneqq\int_{\mathbb{R}}{\cos(z) \over z+i} = {1\over 2} \int_{\mathbb{R}}{e^{iz} \over z+i} + {1\over 2}\int_{\mathbb{R}}{e^{-iz} \over z+i} \coloneqq\int_{\mathbb{R}}f_1 + \int_{\mathbb{R}}f_2 ,\end{align*}$$ These both have $\deg(f_i) = -1$, so Jordan’s lemma on semicircular contours will work. For $e^{i\alpha z}$, one needs to take the upper half-plane for $\alpha>0$ (so $f_1$) and the lower for $\alpha<0$ (for $f_2$). For $f_1$, use the upper contour:

Then by Jordan’s lemma, since $f(z) = e^{iz}g(z)$ with $g(z) \to 0$ as ${\left\lvert {z} \right\rvert}\to \infty$, $\int_{\gamma_R} f \to 0$ and we’re left with the residues in ${\mathbb{H}}$. Here, the only residue is at $z=-i$, so this integral is zero. For $f_2$, use the lower contour:

This is parameterized counterclockwise, and so the piece along ${\mathbb{R}}$ converges to $-I$. By Jordan’s lemma $$\begin{align*} -I &= 2\pi i \mathop{\mathrm{Res}}_{z=-i} {e^{iz}\over 2(z+i)} \\ &= 2\pi i \lim_{z\to -i} {e^{iz}\over 2} \\ &= \pi i e^{-1} ,\end{align*}$$ so $$\begin{align*} I = -{i\pi \over e} .\end{align*}$$

Replication: find $b$ such that $f(z) = f(z+ib)$ and use a rectangle. $$\begin{align*} f(z+ib) = {e^{z\over 2}e^{ib\over 2} \over 1 + e^{ib}e^z} = e^{ib\over 2} {e^{z\over 2} \over 1 + e^{ib}e^z } = f(z) \impliedby e^{ib} = 1 \impliedby b=2\pi ,\end{align*}$$ in which case $$\begin{align*} f(z + 2\pi i) = e^{ib\over 2}f(z) = e^{2\pi i \over 2}f(z) = -f(z) .\end{align*}$$

So take the following rectangle where $H_-$ is at $2\pi i$ and $H_+$ is at 0, with sides at $\pm R$:

Write $\Gamma$ for the entire contour. Note that integrating left-to-right on $H_-$ yields $-I$, since $w = z+2\pi i$ for $w\in H_-$ and $f(w) = -f(z)$. Then reversing the orientation, going right-to-left yields $\int_{H_i} f = I$.

Claim: the integrals over the sides $V_{\pm}$ vanish. For the right, $$\begin{align*} {\left\lvert {\int_{V_+} f(z) \,dz} \right\rvert} &\leq 2\pi \sup_{t\in [0, 2\pi]} {\left\lvert {e^{R + it\over 2} \over 1 + e^{R + it} } \right\rvert} \\ &\leq 2\pi \sup_t {{\left\lvert { e^{R\over 2} e^{it \over 2} } \right\rvert} \over {\left\lvert {e^{R}e^{it}} \right\rvert} - 1} \\ &\leq 2\pi \sup_t {{\left\lvert { e^{R\over 2} } \right\rvert} \over {\left\lvert {e^{R}} \right\rvert} - 1} \\ &\in { \mathsf{O}} (e^{- {R\over 2} }) \to 0 .\end{align*}$$

For the left: $$\begin{align*} {\left\lvert {\int_{V_-} f(z) \,dz} \right\rvert} &\leq \sup_{t\in [0, 2\pi] } {\left\lvert { e^{-R-it\over 2} \over 1 + e^{-R- it}} \right\rvert} \\ &\leq \sup_{t\in [0, 2\pi]}{\left\lvert {e^{-R\over 2} e^{-it\over 2}} \right\rvert} \\ &\leq \sup_{t\in [0, 2\pi]}{\left\lvert {e^{-R\over 2}} \right\rvert} \\ &\in { \mathsf{O}} (e^{- {R\over 2} }) \to 0 ,\end{align*}$$ where we’ve thrown away positive terms in the denominator, which only makes this quantity larger.

Finding the poles within $\Gamma$: by inspection, there are poles when $e^z=-1$, so at $z=(2k+1)\pi i$ for $k\in {\mathbb{Z}}$. Exactly one falls into this contour, $z_k = i\pi$. By the residue theorem, $$\begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=i\pi} f(z) = \int_\Gamma f(z) \,dz= \qty{\int_{H_-} + \int_{H_+}} f = 2I \implies I = \pi i \mathop{\mathrm{Res}}_{z=i\pi } f .\end{align*}$$ Computing this residue: $$\begin{align*} \pi i \mathop{\mathrm{Res}}_{z=i\pi }f(z) &= \pi i \lim_{z\to i\pi} { (z-i\pi) e^{z\over 2}\over e^z + 1} \\ &{ \overset{\scriptscriptstyle\text{LH}}{=} }\pi i \lim_{z\to i\pi} {e^{z\over 2} \over e^z} \\ &= \pi i e^{-i\pi \over 2} \\ &= \pi .\end{align*}$$

DZG: This is much easier than trying to find the Laurent expansion about $z=i\pi$ – trust me!

Heuristically, $\int e^{ax} \operatorname{sech}(x)$ should converge since $\operatorname{sech}(x) \sim e^{-x}$, so ${\left\lvert {f} \right\rvert} \sim {\left\lvert {e^{(a-1)x}} \right\rvert}\sim e^{\Re(a-1)x} \sim e^{-x}$ when $\Re(a-1)$ is negative, so $\Re(a) < 1$.

We’ll need a contour along ${\mathbb{R}}$, so immediate options are a semicircular contour or a rectangle. A semicircular contour is not a good idea here, since there are infinitely many poles of this function: $$\begin{align*} \cosh(z) = 0 \implies e^z + e^{-z} = 0 \implies e^{2z} = -1 \implies z = {i\pi \cdot k \over 2} .\end{align*}$$ It turns out the residues at these poles are all 1, the residue theorem would yield a divergent series. So take the rectangle contour with one side long ${\mathbb{R}}$ and one along $\left\{{t+ib {~\mathrel{\Big\vert}~}t\in [-R, R]}\right\}$ respectively, where we’ll choose $ib$ so that the two integrals differ by a scalar. Computing the symmetry by looking at $f(z+ib)$: $$\begin{align*} {e^{a(z+ib)} \over \cosh(z+ib)} = e^{aib} {e^{z} \over e^z e^{ib} + e^{-z} e^{-ib} } ,\end{align*}$$ and we now need $e^{ib} = e^{-ib}$ in order to scale it out. Noting that if $z\coloneqq e^{ib}$ then $\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = e^{-ib}$, this forces $z\in {\mathbb{R}}$, so $z=\pm 1$. Taking $z=+1$ forces $b=0$, which is the original contour, so taking $z=-1$ yields $b=\pi i$. So we take the following contour:

Computing the integral on the upper contour: $$\begin{align*} \int_{\gamma_1} f(z) \,dz &= \int_R^{-R} f(t+ib) \,dt\qquad z=t+ib, \,dz= \,dt\\ &= - \int_{-R}^R f(t+ib) \,dt\\ &= - \int_{-R}^R {e^{a(t+i\pi)} \over \cosh(t+i\pi ) } \,dt\\ &= - \int_{-R}^R e^{ai\pi} {e^{t} \over e^t e^{i\pi} + e^{-t} e^{-i\pi} } \,dt\\ &= - \int_{-R}^R e^{ai\pi} {e^{t} \over -e^t - e^{-t} } \,dt\\ &= \int_{-R}^R e^{ai\pi} {e^{t} \over e^t + e^{-t} } \,dt\\ &= e^{ai\pi} \int_{-R}^R {e^{t} \over e^t + e^{-t} } \,dt\\ &= e^{ai\pi} \int_{-R}^R {e^{t} \over \cosh(t) } \,dt\\ &= e^{ai\pi} \int_{\gamma_0} f(z) \,dz ,\end{align*}$$ so $$\begin{align*} \qty{\int_{\gamma_0} + \int_{\gamma_1}} f = (1+e^{ai\pi}) I .\end{align*}$$

Given thus, noting that only the pole $z_0 = {i\pi \over 2}$ is enclosed, the residue theorem yields $$\begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=z_0}f(z) = \int_\Gamma f = (1+e^{ai\pi})I \implies I = {2\pi i \mathop{\mathrm{Res}}_{z=z_0} f(z) \over 1 + e^{ai\pi}} .\end{align*}$$ Computing the residue: $$\begin{align*} \mathop{\mathrm{Res}}_{z=z_0}f(z) &= \lim_{z\to z_0} {(z-z_0)e^{az} \over \cosh(z)}\\ &{ \overset{\scriptscriptstyle\text{LH}}{=} }\lim_{z\to z_0} {a(z-z_0) e^{az} \over \sinh(z)} + {e^{az}\over \sinh(z)} \\ &= {a \cdot 0 \cdot e^{ai\pi \over 2} \over i \sin\qty{\pi \over 2} } + {e^{a i \pi \over 2} \over i\sin\qty{\pi \over 2}} \\ &= {e^{ai\pi\over 2} \over i} \\ &= -i e^{ai\pi \over 2} ,\end{align*}$$ where we’ve used that $\sinh(iz) = i\sin(z)$. Putting it all together: $$\begin{align*} I &= {2\pi i \cdot -i e^{ai\pi \over 2} \over 1 + e^{a i \pi}} \\ &= {2\pi e^{ai\pi \over 2} \over 1 + e^{a i \pi}} \\ &= {2\pi \over e^{-ai\pi \over 2}\qty{ 1 + e^{a i \pi}} } \\ &= {2\pi \over e^{-ai\pi \over 2}+ e^{a i \pi\over 2} } \\ &= {\pi \over \cos\qty{a\pi \over 2}}\\ &= \pi \csc\qty{a\pi \over 2} .\end{align*}$$

Take $\Gamma$ an upper-half-plane semicircular contour indented at the origin; considering $f(z) = e^{iz}$, by Jordan’s lemma $\int_{C_R}f \to 0$ and the pieces along ${\mathbb{R}}$ converge to ${ \operatorname{PV} }\int f$. The singularity at $z_0 = 0$ contributes a fractional residue: $$\begin{align*} \int_{C_{\varepsilon}}f(z)\,dz\to i\pi \mathop{\mathrm{Res}}_{z=z_0} f(z) = i\pi \lim_{z\to 0} {(z-0) e^{iz}\over z} = i\pi \cdot 1 .\end{align*}$$ Thus $$\begin{align*} I \coloneqq{ \operatorname{PV} }\int_{\mathbb{R}}f(z)\,dz= 0 + i\pi \implies I_1 = \Im(I) = \pi, \quad I_2 = \Re(I) = 0 .\end{align*}$$

Factor the denominator as $(z-1)(z+1)$, then there are two poles of order 1 on ${\mathbb{R}}$. Define a contour

• $C_1: [-R, 1-{\varepsilon}_1]$
• $-C_2: -1 + Re^{it}, t \in [0, \pi]$
• $C_3: [1-{\varepsilon}, 1+{\varepsilon}]$
• $C_4: 1 + Re^{it}, t \in [0, \pi]$
• $C_R: Re^{it}, t\in [0, \pi]$
• $\Gamma = C_1 + \cdots + C_4 + C_R$

So this, but with a semicircular contour instead of a rectangle:

By Jordan’s lemma, $\int_{C_R}\to 0$, and $\qty{\int_{C_2} + \int_{C_3} + \int_{C_4}}\to 0$, while $\int_\Gamma = 0$ since it encloses no singularities. Thus $$\begin{align*} 0 = \int_{C_2} f + \int_{C_4} f ,\end{align*}$$ which converges to the fractional residues at $z=\pm 1$.

For $z=-1$, $$\begin{align*} \int_{C_2} &\to \pi i \mathop{\mathrm{Res}}_{z=-1} f(z) \\\ &= \pi i \lim_{z\to -1} {e^{2iz} \over z-1} \\ &= \pi i {e^{-2i} \over 2} .\end{align*}$$

For $z=-1$: $$\begin{align*} \int_{C_4} &\to \pi i \mathop{\mathrm{Res}}_{z=1} f(z) \\\ &= \pi i \lim_{z\to 1} {e^{2iz} \over z+1} \\ &= \pi i {e^{2i} \over 2} .\end{align*}$$

Putting it all together: $$\begin{align*} I &= \pi i {e^{-2i} \over 2} + \pi i {e^{2i}\over 2} \\ &=\pi i \cos(2) .\end{align*}$$

Set $z=e^{i\theta}$, so $\sin(\theta) = {z+z^{-1}\over 2i}$ and $\sin^2(\theta) = -{1\over 4}(z^{-2}-2+z^2)$. Then $$\begin{align*} I\coloneqq\int_{[-\pi, \pi]} {1\over 1 + \sin^2(\theta)} \,d\theta &= \int_{S^1} {1\over 1 - {1\over 4}(z^{-2} -2 + z^2) } {1\over iz}\,dz\\ &= \int_{S^1} {-4i \over z(4-(z^{-2} -2 + z^2))}\,dz\\ &= \int_{S^1} {-4iz \over z^2(6-z^{-2} - z^2)}\,dz\\ &= \int_{S^1} {4iz \over z^4-6z^2+1}\,dz\\ &= 2\pi i \sum_{z_k\in {\mathbb{D}}} \mathop{\mathrm{Res}}_{z=z_k} {4iz\over z^4-6z^2+1} .\end{align*}$$

Factoring the denominator: $$\begin{align*} w^2 - 6w - 1 =0 &\implies w^2 - 6w + \qty{-6\over 2}^2 - \qty{-6\over 2}^2 + 1 =0 \\ &\implies (w - 3)^2 = -1 + 9 = 8 \\ &\implies w-3 = \zeta_2^k \sqrt{8},\, k=0, 1 \\ &\implies w = 3 \pm \sqrt{8} \\ &\implies z= \pm\sqrt{3\pm \sqrt 8} .\end{align*}$$ Write these roots as

• $z_1 \coloneqq\sqrt{3-\sqrt 8}$
• $z_2 \coloneqq-\sqrt{3-\sqrt 8}$
• $z_3 \coloneqq\sqrt{3+\sqrt 8}$
• $z_4 \coloneqq-\sqrt{3 + \sqrt 8}$

Some numerology to figure out the modulus of these roots:

• $3 + \sqrt{8} = 3+2\sqrt{2} 3+2\cdot(1.4) \approx 5.8$, so ${\left\lvert { \pm \sqrt{3+\sqrt 8}} \right\rvert}>\sqrt{4}>2>1$.
• $3-\sqrt{8} \approx 3-2.8 \approx 0.2$ so ${\left\lvert { \pm \sqrt{3-\sqrt 8} } \right\rvert} < 1$.

So it suffices to compute the residues at $z_1, z_2 = \pm \sqrt{3-\sqrt 8}$: At $z_1$: $$\begin{align*} \mathop{\mathrm{Res}}_{z = z_1} &= \lim_{z\to z_1} {(z-z_1) 4i z \over z^4-6z^2+1 } \\ &\overset{\text{LH}}{=} \lim_{z\to z_1} { 4iz + 4i(z-z_1) \over 4z^3 -12z} \\ &= {4iz_1 \over 4z_1^3 - 12z_1} \\ &= {i\over z_1^2 - 3} \\ &= {i\over (3-\sqrt 8) - 3} \\ &= -{i\over \sqrt 8} .\end{align*}$$ At $z_2$: $$\begin{align*} \mathop{\mathrm{Res}}_{z=z_2} &= \lim_{z\to z_2} {(z-z_2) 4i z \over z^4-6z^2+1 } \\ &= {4iz_2\over 4z_2^2 - 12z_2}\\ &= {i \over z_2^2 - 3} \\ &= -{i\over \sqrt 8} .\end{align*}$$

Thus $$\begin{align*} I = 2\pi i \cdot -{2i\over \sqrt 8} = {4\pi \over \sqrt{8}} .\end{align*}$$

The usual substitution: $z=e^{i\theta}, \,dz= (iz)\,d\theta$. $$\begin{align*} I\coloneqq\int_{[0, 2\pi]} \qty{a^2 - 2a\cos(\theta) + 1}^{-1}\,d\theta &= \oint \qty{a^2-2(z+z^{-1}) + 1}^{-1}(iz)^{-1}\,dz\\ &= -i\oint \qty{za^2 - a(z^2+1) +z}^{-1}\,dz\\ &= -i \oint\qty{-az^2 + (a^2+1)z - a}^{-1}\,dz\\ &= {i\over a}\oint \qty{z^2 - \qty{a^2+a\over a}z + 1}^{-1}\,dz\\ &= {i\over a}\oint (z-a)^{-1}(z-a^{-1})^{-1}\,dz ,\end{align*}$$ noting that ${a^2+a\over a} = a+a^{-1}$. Now there are two cases:

• ${\left\lvert {a} \right\rvert} < 1$: then $a\in {\mathbb{D}},a^{-1}\in {\mathbb{D}}^c$, so there is a simple pole at $a$. Then $$\begin{align*} I &= {i\over a}\, 2\pi i \mathop{\mathrm{Res}}_{z=a} (z-a)^{-1}(z-a^{-1})^{-1}\\ &= -{2\pi \over a} (z-a^{-1})^{-1}\Big|_{z=a} \\ &= -{2\pi \over a(a-a^{-1})} \\ &= {2\pi \over 1 - a^2} .\end{align*}$$

• ${\left\lvert {a} \right\rvert}> 1$: then $a^{-1}\in {\mathbb{D}}, a\in {\mathbb{D}}^c$ so there is a simple pole at $a^{-1}$. Then $$\begin{align*} I &= {i\over a}\, 2\pi i \mathop{\mathrm{Res}}_{z=a^{-1}} (z-a)^{-1}(z-a^{-1})^{-1}\\ &= -{2\pi \over a} (z-a)^{-1}\Big|_{z=a^{-1}} \\ &= -{2\pi \over a(a^{-1}- a)} \\ &= {2\pi \over a^2 - 1} .\end{align*}$$

The usual substitution: $z=e^{i\theta}, \,d\theta= (iz)^{-1}\,dz$. $$\begin{align*} \int_{[0, 2\pi]} (a +b\cos(\theta))^{-1}\,d\theta &= \oint \qty{ a + {b\over 2}(z+z^{-1})}^{-1}(iz)^{-1}\,dz\\ &= -i\oint \qty{ za + {b\over 2}(z^2 + 1)} ^{-1}\,dz\\ &= -i \oint \qty{{b\over 2}z^2 + az + {b\over 2} }^{-1}\,dz\\ &= -{2i\over b} \oint \qty{z^2 + {2a\over b}z + 1}^{-1}\,dz\\ &= -{2i\over b}\oint (z-r_1)^{-1}(z-r_2)^{-1}\,dz ,\end{align*}$$ where the roots can just be found using the quadratic formula $$\begin{align*} z_k &= {1\over 2} \qty{-{2a\over b} \pm \sqrt{\qty{2a\over b}^2 - 4}} \\ &= -{a\over b}\pm {1\over 2}\sqrt{{4a^2 \over b^2} - 4} \\ &= -{a\over b}\pm \sqrt{{a^2 \over b^2} - 1} \\ &= -{a\over b}\pm \sqrt{{a^2 - b^2 \over b^2}} \\ &= -{a\over b}\pm {1\over b } \sqrt{{a^2 - b^2}} .\end{align*}$$ Thus $$\begin{align*} r_1 &\coloneqq b^{-1}\qty{-a + \sqrt{a^2-b^2}} \\ r_2 &\coloneqq b^{-1}\qty{-a - \sqrt{a^2-b^2}} .\end{align*}$$

Since $r_1 r_2 = 1$ and thus ${\left\lvert {r_1 r_2} \right\rvert} = 1$, only one root is in ${\mathbb{D}}$ and this yields one simple pole. Assume $a>b$. Note that for $r_2$, ${\left\lvert {a/b} \right\rvert} > 1$ and ${\left\lvert {a^2-b^2} \right\rvert}>0$, so $r_2 \approx -1 - {\varepsilon}< -1$, so $r_1\in {\mathbb{D}}$. Computing the residue here: $$\begin{align*} \mathop{\mathrm{Res}}_{z=r_1} (z-r_1)^{-1}(z-r_2)^{-1} &= (z-r_2)^{-1}\Big|_{z=r_1} \\ &= (r_1 - r_2)^{-1}\\ &= \qty{2b^{-1}\sqrt{a^2-b^2} }^{-1} ,\end{align*}$$ so $$\begin{align*} I &= 2\pi i \cdot -{2 i \over b}{b\over 2\sqrt{a^2-b^2}} \\ &= {2\pi \over \sqrt{a^2-b^2} } .\end{align*}$$

Noting the partial $\zeta_2 = -1$ symmetry, take a branch cut for $\log$ along $\theta = -\pi/2$ and the following semicircular contour:

Since $f(z) \coloneqq{\log(z) \over z^2 + 1}$ goes to zero as ${\left\lvert {R} \right\rvert}\to \infty$ and ${\varepsilon}\to 0$, only the horizontal contours will contribute. Parameterize, oriented counterclockwise:

• $\gamma_1 \coloneqq\left\{{t+0i {~\mathrel{\Big\vert}~}t\in [{\varepsilon}, R]}\right\}$
• $\gamma_2 \coloneqq\left\{{t+0i {~\mathrel{\Big\vert}~}t\in [-{\varepsilon}, -R]}\right\}$

Then $\int_{\gamma_1} f(z)\,dz\to I$. Computing the contribution from $\gamma_2$: $$\begin{align*} \int_{\gamma_2} f(z) \,dz &= \int_{-R}^{-{\varepsilon}} f(t) \,dt\qquad z=t+0i, \,dz=\,dt\\ &= \int_{-R}^{-{\varepsilon}} {\log(t) \over t^2 + 1}\,dt\\ &= -\int_{R}^{{\varepsilon}} {\log(-x) \over (-x)^2 + 1 }\,dx\qquad t=-x, \,dt= -\,dx\\ &= \int_{{\varepsilon}}^{R} {\log(x) + i\pi \over x^2 + 1 }\,dx\\ &= I + i\pi \cdot {\pi \over 2} \\ &= I + {i\pi^2\over 2} ,\end{align*}$$ using the known antiderivative $\arctan$.

Note that there are two simple poles at $\pm i$, so only the residue at $z_0=i$ contributes: $$\begin{align*} \mathop{\mathrm{Res}}_{z=i} = \lim_{z\to i} {\log(z) \over (z+i)} = {\log(i) \over 2i} = {i\pi/2 \over 2i} = {\pi \over 4} ,\end{align*}$$ so by the residue theorem, $$\begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=i}f(z) = \int_{\Gamma}f(z) \,dz= \qty{ \int_{\gamma_1} + \int{\gamma_2}}f = 2I + {i\pi^2 \over 2} \\ \implies 2\pi i \cdot {\pi \over 4} = I + {i\pi^2\over 2} \\ \implies {i\pi^2\over 4} = I + {i\pi^2\over 4} \\ \implies I = 0 .\end{align*}$$

Let $f$ be the integrand, then $f\sim \log(z)/(z^4+1)$, so an indented semicircular contour will work since ${\left\lvert {f} \right\rvert}\to 0$ as ${\left\lvert {z\to\infty} \right\rvert}$, and the inner integral will be dominated by a term of the form ${\varepsilon}\log({\varepsilon})/{\varepsilon}^4\to 0$ as ${\varepsilon}\to 0$. So take such a contour, branch cutting along $\theta = -\pi/2$:

Now consider the contribution from $\gamma_2$: $$\begin{align*} \int_{\gamma_2} f(z) \,dz &= \int_{-R}^{-{\varepsilon}} f(t+0i) \,dt\\ &= \int_{-R}^{-{\varepsilon}} {\log(t) \over (t^2 + 1)^2 }\,dt\\ &= -\int_{R}^{{\varepsilon}} {\log(-x) \over ((-x)^2 + 1)^2 }\,dx\\ &= \int_{{\varepsilon}}^{R} {\ln{\left\lvert {x} \right\rvert} + i\pi \over (x^2 + 1)^2 }\,dx\\ &\to I + i\pi \int_{0}^\infty {1\over (x^2 + 1)^2}\,dx .\end{align*}$$ This auxiliary integral can be handled easily with a usual semicircular contour, since the integrand is ${ \mathsf{O}} (x^4)$: $$\begin{align*} \int_{0}^\infty {1\over (z^2 + 1)}\,dx &= 2\pi i \sum_{z_k\in {\mathbb{H}}}\mathop{\mathrm{Res}}_{z=z_k} {1\over (z^2 + 1)} \\ &= 2\pi i \mathop{\mathrm{Res}}_{z=i} {1\over (z^2 + 1)} \\ &= 2\pi i \lim_{z\to i} {\frac{\partial }{\partial z}\,} {1\over (z+i)^2} \\ &= 2\pi i \cdot {-2\over (2i)^3 } \\ &= 2\pi i \cdot {-i\over 4} \\ &= {\pi \over 2} .\end{align*}$$ Computing the residue of the main integral: $$\begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=i} f(z) &= 2\pi i \lim_{z\to i} {\frac{\partial }{\partial z}\,} { \log(z) \over (z+i)^2} \\ &= 2\pi i \lim_{z\to i} { (z+i)^2 z^{-1}- 2(z+i)\log(z) \over (z+i)^4 } \\ &= 2\pi i { -i(2i)^2 - 2(2i)\qty{i\pi \over 2} \over (2i)^4 } \\ &= \pi i { 2i + \pi \over 2^2 }\\ &= -{\pi \over 2} + {i\pi^2 \over 4} .\end{align*}$$ Combining all of this: $$\begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=i} f(z) = \int_\Gamma f(z) \,dz= \qty{\int_{\gamma_1} + \int_{ \gamma_2} } f = I + \qty{I + {i\pi^2\over 4}} \\ \implies -{\pi \over 2} + {i\pi^2 \over 4} = 2I + {i\pi^2\over 4} \\ \implies -{\pi\over 2} = 2I \\ implies I = -{\pi \over 4} .\end{align*}$$

Factor $(1+z^2)^2 = (z+i)^2(z-i)^2$. Apropos of nothing, considering the auxiliary function $$\begin{align*} g(z) \coloneqq\qty{\log(z) \over 1+x^2}^2 = {\log^2(z) \over (1+x^2)^2} .\end{align*}$$ Use a keyhole contour for a branch cut along $\theta = -\pi$, so $\operatorname{Arg}(z) \in (-\pi, \pi)$:

• $\gamma_+: \left\{{t + i{\varepsilon}{~\mathrel{\Big\vert}~}t \in [-{\varepsilon}, -R]}\right\}$,
• $\gamma_{{\varepsilon}}: \left\{{{\varepsilon}e^{it} {~\mathrel{\Big\vert}~}t\in [-\pi + {\varepsilon}, \pi - {\varepsilon}] }\right\}$,
• $\gamma_+: \left\{{t - i{\varepsilon}{~\mathrel{\Big\vert}~}t \in [-{\varepsilon}, -R]}\right\}$,
• $\gamma_{R}: \left\{{R e^{it} {~\mathrel{\Big\vert}~}t\in [-\pi + {\varepsilon}, \pi - {\varepsilon}] }\right\}$,

and $\Gamma$ the combined contour oriented counterclockwise. Note the symmetry: $$\begin{align*} \int_{\gamma_+}g(z) &= \int_{-R}^{-{\varepsilon}} { \log^2(t) \over (t^2 +1)^2 } \,dt\\ &= - \int_{-eps}^{-R} { \log^2(t) \over (t^2 +1)^2 } \,dt\\ &= \int_{{\varepsilon}}^{R} { \log^2(e^{i\pi} s) \over ( ( e^{i\pi} s)^2 +1)^2 } \,ds&& t = e^{i\pi}s,\, \,ds= -\,dt\\ &= \int_{{\varepsilon}}^{R} { \qty{ \log(s) + i\pi}^2 \over (s^2 +1)^2 } \,ds\\ ,\end{align*}$$ and similarly $$\begin{align*} \int_{\gamma_-}g(z) &= \int_{-{\varepsilon}}^{-R} { \log^2(t) \over (t^2 +1)^2 } \,dt\\ &= -\int_{{\varepsilon}}^{R} { \log^2(e^{-i\pi} s) \over ((e^{-i\pi} s)^2 +1)^2 } \,ds&& t=e^{-i\pi }s,\, \,dt= -\,ds\\ &= -\int_{{\varepsilon}}^{R} { \qty{ \log(s) - i\pi}^2 \over (s^2 +1)^2 } \,dt\\ .\end{align*}$$ Now note that $$\begin{align*} (\log(s) + i\pi)^2 - (\log(s) - i\pi)^2 &= \cdots \\ &= \qty{\log^2(s) + 2i\pi \log(s) - \pi^2 } - \qty{\log^2(s) -2i\pi\log(s) - \pi^2} \\ &= 4i\pi\log(s) ,\end{align*}$$ and so miraculously $$\begin{align*} \int_{\gamma_+}f(z)\,dz+ \int_{\gamma_-}f(z)\,dz &= \int_{{\varepsilon}}^{R} { \qty{ \log(s) + i\pi}^2 \over (s^2 +1)^2 } \,ds- \int_{{\varepsilon}}^{R} { \qty{ \log(s) - i\pi}^2 \over (s^2 +1)^2 } \,dt\\ &= \int_{\varepsilon}^R { \qty{ \log(s) + i\pi}^2 - \qty{ \log(s) - i\pi}^2 \over (s^2+1)^2 }\,ds\\ &= 4i\pi \int_{{\varepsilon}}^R {\log(s) \over (s^2+1)^2}\,ds\\ &\longrightarrow 4i \pi I .\end{align*}$$ The contributions from $\gamma_R$ vanish since ${\log(z) \over z^4}\to 0$ as ${\left\lvert {z} \right\rvert}\to \infty$, and the contribution from $\gamma_{\varepsilon}$ vanish since ${{\varepsilon}\log({\varepsilon}) \over {\varepsilon}^4+c}\to 0$ as ${\varepsilon}\to 0$ (and applying the ML estimate). Thus $$\begin{align*} 2\pi i \sum_{z_k\in {\mathbb{C}}}\mathop{\mathrm{Res}}_{z=z_k}g(z) = \int_\Gamma g(z) \,dz= 4i\pi I .\end{align*}$$ Factoring the denominator as $(1+z^2)^2 = (z-i)^2(z+i)^2$, there are two order 2 poles at $\pm i$. At $z=i$: $$\begin{align*} \mathop{\mathrm{Res}}_{z=i}g(z) &= \lim_{z\to i} {\frac{\partial }{\partial z}\,} {\log^2(z) \over (z+i)^2} \\ &= \lim_{z\to i} {(z+i)^2 2\log(z)z^{-1}- \log^2(z) 2(z+i) \over (z+i)^4} \\ &= 2^{-4}\qty{(2i)^2 \cdot 2 \cdot {i\pi \over 2} {1\over i} - \qty{i\pi \over 2}^2 \cdot 2 \cdot 2i} \\ &= 2^{-4}\qty{2^3 i^2 {\pi \over 2} - 2^{-2}i^2 \pi^2 2^2 i} \\ &= 2^{-4}\qty{-2^2\pi + i\pi ^2} \\ &= r_1 \coloneqq- {\pi \over 4} + i {\pi^2\over 16} .\end{align*}$$ Similarly, $$\begin{align*} \mathop{\mathrm{Res}}_{z=-i}g(z) &= \lim_{z\to -i} {\frac{\partial }{\partial z}\,} {\log^2(z) \over (z+i)^2} \\ &= \lim_{z\to -i} {(z-i)^2 2\log(z) z^{-1}- \log^2(z) 2(z-i) \over (z-i)^4} \\ &= 2^{-4} \qty{ (-2i)^2 \cdot 2 \qty{-i\pi \over 2}{1\over -i} - \qty{-i\pi\over 2}^2 \cdot 2(-2i) } \\ &= 2^{-4} \qty{ 2^2 i^2 \pi - 2^{-2} \pi^2 (-4i) }\\ &= 2^{-4}\qty{-2^2\pi - \pi^2} \\ &= r_2 \coloneqq-{\pi \over 4} - i {\pi^2\over 16} .\end{align*}$$ Solving for $I$ above, we have $$\begin{align*} I &= {2\pi i \over 4\pi i}(r_1 + r_2) \\ &= {1\over 2} \qty{- {\pi \over 4} - {\pi \over 4}} \\ &= -{\pi \over 4} .\end{align*}$$

Note that the poles are at $z=\pm ia$, and since $\lim_{{\left\lvert {z} \right\rvert}\to\infty}f(z) = 0$ and $\lim{R\to 0} {R\log(R)\over R^2 + a^2} = 0$, an indented semicircular contour will work.

Computing the contribution from the residues: $$\begin{align*} \mathop{\mathrm{Res}}_{z=ia}f(z) &= \lim_{z\to ia }{\log(z) \over (z+ia)} \\ &= {\log(ia) \over 2ia} \\ &= {\log(a) + i\pi/2 \over 2ia} \\ &= {\pi \over 4a} + {\log(a) \over 2ia} \\ \\ \implies 2\pi i \mathop{\mathrm{Res}}_{z=ia}f(z) &= {i\pi^2 \over 2a} + {\pi\log(a) \over a} .\end{align*}$$ The contribution from the integrals will come from $\qty{\int_{\gamma_1} + \int_{\gamma_2}}f$ where

• $\gamma_1 \coloneqq\left\{{tR + (1-t){\varepsilon}{~\mathrel{\Big\vert}~}t\in [{\varepsilon}, R] }\right\}$
• $\gamma_2 \coloneqq\left\{{t(-{\varepsilon}) + (1-t)(-R) {~\mathrel{\Big\vert}~}t\in [{\varepsilon}, R] }\right\}$,

so that the overall contour is oriented counterclockwise. Noting that $\int_{\gamma_1}f(z)\,dz\to I$ the desired integral, the other contribution is $$\begin{align*} \int_{\gamma_2}f(z)\,dz &= \int_{-R}^{-{\varepsilon}} {\log(t)\over t^2 + a^2}\,dt\\ &= - \int_{-{\varepsilon}}^{-R} {\log(t) \over t^2+a^2} \,dt\\ &= \int_{{\varepsilon}}^{R} {\log(-x) \over x^2+a^2} \,dt\qquad x=-t,\, \,dx= -\,dt\\ &= \int_{\varepsilon}^R {\log(x) \over x^2 + a^2}\,dx+ i\pi \int_{\varepsilon}^R {1\over x^2 +a^2}\,dx\\ &= I + i\pi\qty{\pi \over 2a} .\end{align*}$$ In the limit, by the residue theorem we have $$\begin{align*} {i\pi^2 \over 2a} + {\pi\log(a) \over a} &= 2I + i\pi\qty{\pi \over 2a} \\ \implies I &= {\pi \log(a) \over 2a} .\end{align*}$$

For the usual reasons, integrals along semicircles of radius $R$ and ${\varepsilon}$ go to zero, so noting the poles at $\omega_a \coloneqq e^{i\pi\over a}$, take an indented sector:

Set $\zeta_a \coloneqq e^{2\pi i \over a}$. Contributions from the contours: let $\gamma_1$ be the contour along ${\mathbb{R}}$ and $\gamma_2$ along $\zeta_a {\mathbb{R}}$, oriented so the overall contour is counterclockwise. Then $\int_{\gamma_1}f(z)\,dz\to I$ for $f(z) \coloneqq{\log(z) \over 1+z^a}$, so compute the monodromy term: parameterize $\gamma_2 \coloneqq\left\{{\zeta_a t {~\mathrel{\Big\vert}~}t\in [{\varepsilon}, R]}\right\}$, so $$\begin{align*} \int_{\gamma_2}f(z) \,dz &=\int_R^{\varepsilon}f(\zeta_a t) \zeta_a \,dt\\ &= -\zeta_a \int_{\varepsilon}^R {\log(\zeta_a t) \over (\zeta_a t)^a + 1}\,dt\\ &= -\zeta_a \int_{{\varepsilon}}^R {\log(z) + {2\pi i \over a} \over t^a+1}\,dt\\ &\to -\zeta_a I - \zeta_a {2\pi i \over a} \int_0^\infty {1\over t^a + 1 }\,dt\\ &\coloneqq-\zeta_a I - {2\pi i\over a}\zeta_a I' .\end{align*}$$

Given this, the RHS of the residue theorem limits to $$\begin{align*} (1-\zeta_a) I - {2\pi i\over a}\zeta_a \qty{ {\pi \over a}\csc\qty{\pi\over a} } = (1-\zeta_a) I - {2\pi^2\over a^2}i\zeta_a \csc\qty{\pi\over a} .\end{align*}$$

For the LHS, we compute the residue at $\omega_a$: $$\begin{align*} \mathop{\mathrm{Res}}_{z=\omega_a} f(z) &= \lim_{z\to \omega_a} {(z-\omega_a) \log(z) \over z^a + 1} \\ &{ \overset{\scriptscriptstyle\text{LH}}{=} }\lim_{z\to \omega_k} {\log(z) \over az^{a-1}}\\ &= {\log\qty{e^{i\pi \over a}} \over ae^{i\pi \qty{a-1\over a}} } \\ &= -{i\pi /a \over ae^{- i\pi \over a} } \\ &= -{i\pi \over a^2} e^{i\pi\over a} ,\end{align*}$$ so $$\begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=\omega_a} f(z) = -2\pi i \qty{i\pi\over a^2}e^{i\pi \over a} = {2\pi^2\over a^2}e^{i\pi\over a} .\end{align*}$$

After some truly arduous arithmetic, this assembles to: $$\begin{align*} {2\pi^2\over a^2}e^{i\pi\over a} &= (1-\zeta_a) I - {2\pi^2\over a^2}i\zeta_a \csc\qty{\pi\over a} \\ \\ \implies {2\pi^2\over a^2}e^{i\pi\over a} &= (1-e^{2\pi i\over a}) I - {2\pi^2\over a^2}ie^{2\pi i \over a} \csc\qty{\pi\over a} \\ \\ \implies {2\pi^2\over a^2} &= (e^{-{i\pi\over a}}-e^{\pi i\over a}) I - {2\pi^2\over a^2}ie^{\pi i \over a} \csc\qty{\pi\over a} \\ \\ &= -2i\sin\qty{\pi\over a}I - {2\pi^2\over a^2}ie^{\pi i \over a} \csc\qty{\pi\over a} \\ \\ \implies I &= { {2\pi^2\over a^2} + {2\pi^2\over a^2}ie^{\pi i \over a} \csc\qty{\pi\over a} \over -2i\sin\qty{\pi \over a} } \\\\ &= {2\pi^2\over a^2} \qty{ 1 + ie^{\pi i \over a} \csc\qty{\pi\over a} \over -2i\sin\qty{\pi \over a} } \\ \\ &= -{\pi^2\over a^2}\csc\qty{\pi\over a} \cdot (-i)\qty{1 + ie^{i\pi\over a}\csc\qty{\pi\over a}} \\ \\ &= -{\pi^2\over a^2}\csc\qty{\pi\over a} \qty{ -i + \qty{\cos\qty{\pi\over a} +i\sin\qty{\pi\over a}}\csc\qty{\pi\over a}} \\ \\ &= -{\pi^2\over a^2}\csc\qty{\pi\over a} \qty{-i + \cot\qty{\pi\over a} + i } \\ \\ &= -{\pi^2\over a^2}\csc\qty{\pi\over a} \cot\qty{\pi\over a} \\ &= -{\pi^2\over a^2}\cos\qty{\pi\over a} \csc^2\qty{\pi\over a} \\ .\end{align*}$$

Note the single pole of order 2 at $z=-1$, and also the branch singularity. Choose a branch cut of $\log$ by deleting $\theta=0$, and take a keyhole contour.

Write the contours as

• $\gamma_{\varepsilon}= \left\{{{\varepsilon}e^{it} {~\mathrel{\Big\vert}~}t\in[0+{\varepsilon}, 2\pi - {\varepsilon}]}\right\}$
• $\gamma_+ = \left\{{x+i{\varepsilon}{~\mathrel{\Big\vert}~}x\in [{\varepsilon}, R]}\right\}$
• $\gamma_R = \left\{{Re^{it} {~\mathrel{\Big\vert}~}t\in [0+{\varepsilon}, 2\pi - {\varepsilon}]}\right\}$
• $\gamma_- = \left\{{x-i{\varepsilon}{~\mathrel{\Big\vert}~}t\in [{\varepsilon}, R]}\right\}$,

all oriented so that the total curve $\Gamma$ is traversed counter-clockwise.

The claim is that $\int_{\gamma_{\varepsilon}} f, \int_{\gamma_R} f\to 0$, and $\int_{\gamma_+} f$ is a multiple of $\int{\gamma_-} f$. For $z=x-i{\varepsilon}$ on $\gamma_-$, we have $$\begin{align*} \log(z) = \log(x-i{\varepsilon}) = \ln{\left\lvert {x-i{\varepsilon}} \right\rvert} + i\operatorname{Arg}(x-i{\varepsilon})\overset{{\varepsilon}\to 0}\longrightarrow \ln{\left\lvert {x} \right\rvert} + 2\pi i = e^{2\pi i}z ,\end{align*}$$ and $$\begin{align*} f(e^{2\pi i}z) = {(e^{2\pi i}z)^\alpha \over ((e^{2\pi i}z)^2+1)^2 } = e^{2\pi i\alpha } {z \over z^2+1} = e^{2\pi i\alpha}f(z) .\end{align*}$$ Thus $$\begin{align*} \int_{\gamma_-} f(z)\,dz &\longrightarrow\int_R^{\varepsilon}f(e^{2\pi i }z)\,dz\\ &= \int_R^{\varepsilon}e^{2\pi i \alpha}f(z)\,dz\\ &= -e^{2\pi i \alpha}\int_{\varepsilon}^R f(z)\,dz\\ &= -e^{2\pi i\alpha}\int_{\gamma_+}f(z)\,dz .\end{align*}$$

Thus in the limit, $$\begin{align*} 2\pi i \sum_{z_k\in {\mathbb{C}}\setminus{\mathbb{R}}_{\geq 0}} \mathop{\mathrm{Res}}_{z=z_k}f(z) &= \int_\Gamma f(z)\,dz\\ &= \int_{\gamma_+}f(z)\,dz+ \int_{\gamma_-}f(z)\,dz\\ &= (1-e^{2\pi i\alpha})\int_{\gamma_+}f(z)\,dz\\ &= (1-e^{2\pi i\alpha})\int_{{\mathbb{R}}}f(z)\,dz\\ .\end{align*}$$

Computing the residue at $z_0 = -1$: $$\begin{align*} \mathop{\mathrm{Res}}_{z=-1}f(z) &= \lim_{z\to -1} {\frac{\partial }{\partial z}\,} (z+1)^2 f(z) \\ &= \lim_{z\to -1} {\frac{\partial }{\partial z}\,} z^\alpha \\ &= \alpha (-1)^{\alpha - 1} \\ &= \alpha e^{i\pi(\alpha - 1)} \\ &= -\alpha e^{i\pi \alpha} .\end{align*}$$ Thus $$\begin{align*} \int_{\mathbb{R}}f(z) \,dz &= 2\pi i \cdot {-\alpha e^{i\pi \alpha}\over 1 - e^{2\pi i \alpha}} \\ &=-2\pi i \alpha {1\over e^{-i\pi\alpha} (1- e^{2\pi i \alpha})} \\ &=-2\pi i \alpha {1\over e^{-i\pi\alpha} - e^{i\pi\alpha}} \\ &=2\pi i \alpha {1\over e^{-i\pi\alpha} - e^{-i\pi\alpha}} \\ &= \pi \alpha \csc(\pi\alpha) .\end{align*}$$

Write $f(z) \coloneqq{z^{1\over 3}\over z^2+1}$, the claim is that an indented semicircular contour will work:

Why: after parameterizing $C_R$, the integrand is approximately $R\cdot R^{1\over 3}/ R^2 \sim R^{{4\over 3} - 2} = R{-{2\over 3}}$, which goes to zero as $R\to \infty$. Similarly, on $C_{\varepsilon}$, the integrand is approximately ${\varepsilon}^{4\over 3}/({\varepsilon}^2+1)$, which goes to zero as ${\varepsilon}\to 0$.

Note the poles at $z=\pm i$. Computing the residue contribution at $z=i$: $$\begin{align*} \mathop{\mathrm{Res}}_{z=i} f(z) &= {i^{1\over 3} \over 2i} = {1\over 2e^{i\pi \over 3}} .\end{align*}$$

Computing the contribution from the integrals: let $\gamma_1$ be the contour along ${\mathbb{R}}_{\geq {\varepsilon}}$, and $\gamma_2$ along ${\mathbb{R}}_{\leq {\varepsilon}}$. Noting that $I = \int_{\gamma_1}f(z)\,dz$, $$\begin{align*} \int_{\gamma_2}f(z)\,dz &= \int_{-R}^{-{\varepsilon}} { t^{1\over 3} \over t^2 +1 } \,dt\\ &= -\int_R^{\varepsilon}{(-x)^{1\over 3} \over x^2+1}\,dx\qquad x=-t,\, \,dx= -\,dt\\ &= \int_{\varepsilon}^R {(\zeta_2 x)^{1\over 3} \over x^2+1}\,dx\\ &= \zeta_2^{1\over 3} I \\ &= e^{i\pi\over 3}I ,\end{align*}$$ so $\qty{\int_{\gamma_1} + \int_{\gamma_2}} f$ contributes $(1+e^{i\pi \over 3})I$. By the residue theorem, $$\begin{align*} 2\pi i \cdot {1\over 2e^{i\pi \over 3} } &= (1+e^{i\pi \over 3})I \\ \\ \implies I &= {i\pi \over 2e^{i\pi \over 3} (1+e^{i\pi \over 3}) } \\ &= {i\pi \over 2} \qty{ e^{i\pi \over 3} + e^{2i\pi \over 3} }^{-1}\\ &= {i\pi \over 2} \qty{ e^{i\omega} + e^{2i\omega} }^{-1},\qquad \omega={\pi\over 3} \\ &= {i\pi \over 2} \qty{ e^{3i\omega\over 2} \qty{ e^{-i\omega\over 2} + e^{i\omega\over 2}} }^{-1}\\ &= i\pi e^{-3i\omega\over 2}{1\over \cos\qty{\omega\over 2}} \\ &= i\pi e^{-i\pi\over 2 }{1\over \cos\qty{\pi \over 6}} \\ &= {\pi \over \sqrt{3}} ,\end{align*}$$ where we’ve used the “exponential balancing trick” (see complex arithmetic section).

For the same reasons as in the semicircular solution, a keyhole will work:

The contributions from $C_2$: $$\begin{align*} \int_{C_2}f(z) \,dz &= \int_R^{\varepsilon}{ (t-i{\varepsilon})^{1\over 3} \over (t-i{\varepsilon})^2 + 1 }\,dt\\ &= - \int^R_{\varepsilon}{ e^{{1\over 3}\qty{\ln{\left\lvert {t-i{\varepsilon}} \right\rvert} + i\operatorname{Arg}(t-i{\varepsilon}) } } \over (t-i{\varepsilon})^2 + 1 }\,dt\\ &\to - \int^R_{\varepsilon}{ e^{{1\over 3}\qty{\ln{\left\lvert {t} \right\rvert} + 2\pi i } } \over t^2 + 1 }\,dt\\ &= -\int_{\varepsilon}^R { e^{2\pi i \over 3} t^{1\over 3} \over t^2 + 1}\,dt\\ &= - \zeta_3 I ,\end{align*}$$ so the contributions from the contours sums to $(1-\zeta_3 I)$.

The contributions from residues: $$\begin{align*} \mathop{\mathrm{Res}}_{z=\pm i} f(z) &= { (\pm i)^{1\over 3} \over \pm 2i} \cdot \\ &= { (e^{k\pi \over 2})^{1\over 3} \over \pm 2i},\qquad k=1,3\\ &= { e^{k\pi \over 6} \over \pm 2i} \\ &= \begin{cases} { e^{\pi\over 6} \over 2i} & z=i \\ {e^{3\pi \over 6} \over -2i} = -{1\over 2} & z=-i. \end{cases} .\end{align*}$$ So the contribution from the residue theorem is $$\begin{align*} 2\pi i\qty{ {e^{\pi\over 6} \over 2i} - {i\over 2i} } = \pi\qty{e^{\pi\over 6} - 1} .\end{align*}$$

Solving for the integral: $$\begin{align*} I &= {\pi (e^{\pi \over 6} - i) \over 1 - e^{2i\pi \over 3}} \\ &=\pi { e^{i\omega} - e^{3i\omega} \over e^{0i\omega} - e^{4i\omega}},\qquad \omega = {\pi\over 6} \\ &= \pi {e^{2i\omega} \qty{e^{-i\omega} - e^{i\omega} } \over e^{2i\omega}\qty{e^{-2i\omega} - e^{2i\omega} } } \\ &= \pi {-2i\sin(\omega) \over -2i\sin(2\omega)} \\ &= \pi {\sin\qty{\pi\over 6} \over \sin\qty{\pi\over 3}} \\ &= \pi{ 1/2\over \sqrt{3}/2}\\ &= {\pi \over \sqrt 3} .\end{align*}$$

Take a branch cut $[-1, 1]$ and $\Gamma$ the standard dogbone contour:

Orient $\Gamma$ positively about infinity, i.e. counterclockwise.

Contribution from $\gamma_1 \coloneqq\left\{{t+i{\varepsilon}{~\mathrel{\Big\vert}~}t\in [-1, 1]}\right\}$, the upper horizontal piece: $$\begin{align*} \int_{\gamma_1}f(z)\,dz\to \int_{-1}^1 \sqrt{1-t^2} \,dt= I .\end{align*}$$

Contribution from $\gamma_2\coloneqq\left\{{t-i{\varepsilon}{~\mathrel{\Big\vert}~}t\in [-1, 1]}\right\}$, the lower horizontal piece: note that following $e^{2\pi i t}z$ to $t=1$ sends $\sqrt{z}$ to $-\sqrt{z}$, so $$\begin{align*} \int_{\gamma_2}f(z)\,dz\to \int_{1}^{-1} - \sqrt{1-t^2} \,dt= I .\end{align*}$$

Contributions from the circles: use that $f(z) \coloneqq\sqrt{z^2-1}$ is a continuous function and these arcs are compact, so they are uniformly bounded. Thus $\int f\to 0$ by the ML estimate on these arcs.

The total contribution: $$\begin{align*} \qty{\int_{\gamma_1} + \int_{\gamma_2}} f = 2I .\end{align*}$$

The residue at infinity: $$\begin{align*} \mathop{\mathrm{Res}}_{z=\infty}f(z) &= \mathop{\mathrm{Res}}_{z=0} - {1\over z^2}\sqrt{1 - {1\over z^2}} \\ &= \mathop{\mathrm{Res}}_{z=0} - {1\over z^2}\sqrt{z^2-1 \over z^2} \\ &= \mathop{\mathrm{Res}}_{z=0} - {i\over z^3}\sqrt{1-z^2} \\ &= \mathop{\mathrm{Res}}_{z=0} - {i\over z^3}\sum_{k\geq 0} {1/2\choose k}z^{2k} \\ &= \mathop{\mathrm{Res}}_{z=0} - {i\over z^3} \qty{1 - {1\over 2}z^2 - { \mathsf{O}} (z^3) } \\ &= {i\over 2} ,\end{align*}$$ thus $$\begin{align*} 2\pi i \cdot -{i\over 2} = 2I \implies I = {\pi \over 2} .\end{align*}$$

Write $f(z) = \sqrt{z^2-1} = \sqrt{(z+1)(z-1)}$. First note $f$ is even, so $$\begin{align*} I = {1\over 2}I',\qquad I' \coloneqq\int_{-1}^1 {1\over \sqrt{z^2-1}} \,dz .\end{align*}$$

Each branch point $\pm 1$ introduces a monodromy factor of $\sqrt{e^{2i\pi}} = e^{i\pi} = -1$, which cancel provided loops are not able to encircle a single branch point. So take the branch cut to be the slit $[-1, 1]$, forcing any loop to encircle neither or both of $\pm 1$ – now use a dogbone contour $\Gamma$ around the slit and apply the residue theorem to the exterior region:

The contribution from the top segment $\gamma_1$: $$\begin{align*} \int_{\gamma_1}f(z)\,dz\to \int_1^{-1} {1\over \sqrt{x^2-1}}\,dx= -I' .\end{align*}$$ The contribution from the bottom segment $\gamma_2$: a monodromy factor of $-1$ is introduced to $g(z)\coloneqq\sqrt{z}$ over a path that traces an angle of $2\pi$, so $$\begin{align*} \int_{\gamma_2}f(z)\,dz= \int_{-1}^1 {1\over -\sqrt{x^2-1}} = -I' .\end{align*}$$

These combine to contribute $$\begin{align*} \qty{\int_{\gamma_1} + \int_{\gamma_2}}f = -2I' .\end{align*}$$

Note – we’ll want the contour to actually be positively oriented with respect to $z=\infty$, so we should reverse the orientation of $\Gamma$ to get a total contribution to $2I$ instead.

The contribution from the small circles: parameterize the first as $-1 + R e^{2\pi i t}$, then $$\begin{align*} {\left\lvert { \int_{C_{\varepsilon}^1}f(z)\,dz} \right\rvert} = {\left\lvert {\int_0^1 {2\pi i R e^{2\pi i t}\over \sqrt{ (-1 + R e^{2\pi i t} )^2 - 1 } } \,dt} \right\rvert} \sim \int_0^1 {R\over \sqrt{R^2-1}} \,dt\overset{R\to 0}\longrightarrow 0 .\end{align*}$$ A similar bound works for the second circle using the parameterization $1+ Re^{2\pi i t}$.

Contributions from residues: take the residue at infinity, $$\begin{align*} \mathop{\mathrm{Res}}_{z=\infty}f(z) &= \mathop{\mathrm{Res}}_{z=0}-{1\over z^2}f\qty{1\over z} \\ &= \mathop{\mathrm{Res}}_{z=0} {-1\over z^2\sqrt{z^{-2} - 1 }} \\ &= \mathop{\mathrm{Res}}_{z=0} {-1\over z\sqrt{1-z^2}} \\ &= \lim_{z\to 0} {-1\over \sqrt{1-z^2}}\\ &= -1 .\end{align*}$$

Putting this together $$\begin{align*} -2\pi i \mathop{\mathrm{Res}}_{z=\infty}f(z) &= \oint_\Gamma f(z)\,dz= 2I' \\ \implies 2\pi i &= 2I' \\ \implies \pi i &= I' = 2I \\ \implies I &= {i\pi \over 2} .\end{align*}$$

Note the simple pole at $x=a$ and the branch points $x=\pm 1$, coming from factoring $\sqrt{1-z^2} = \sqrt{(1-z)(1+z)}$. The standard dog bone contour will work, but will involve a residue at $z=a$ and at $z=\infty$. Instead of taking the usual branch cut $[-1, 1]$, take instead $(-\infty, -1] \cup[1, \infty)$ and the following T-bone contour (noting that it is oriented negatively):

Note that in the limit, the two vertical pieces cancel since there is no phase introduced in $f$ when taking a path from $Q_1$ to $Q_2$, and the two upper horizontal segment limit to $\gamma_1\coloneqq\left\{{t+i{\varepsilon}{~\mathrel{\Big\vert}~}t\in[-1, 1]}\right\}$. Also note that $\int_{\gamma_1}f(z)\,dz\to I$.

Let $\gamma_2$ be the bottom horizontal segment. As in the arguments for the standard bone contour, $$\begin{align*} \int_{\gamma_2}f(z)\,dz= -\int_{\gamma_1} -f(z)\,dz= I ,\end{align*}$$ so $$\begin{align*} \qty{\int_{\gamma_1} + \int_{\gamma_2} }f \to 2I .\end{align*}$$

The contribution from the residue: $$\begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=a} {1\over (z-a) \sqrt{1-z^2} } = {2\pi i \over \sqrt{1-a^2}} = {2\pi \over \sqrt{a^2-1}} ,\end{align*}$$ so $$\begin{align*} {2\pi \over \sqrt{a^2-1}} = 2I \implies I = {\pi \over \sqrt{a^2-1}} .\end{align*}$$

Write $f(z) \coloneqq(z^2-1)^{-{1\over 2}}/z$. In order for $(z^2-1)^{-{1\over 2}}$ to be well-defined, one needs to introduce a branch cut. Note that $f$ has a simple pole at $z=0$ and is holomorphic away from $z=0$ if $z^2-1$ is not on the positive real axis, where we’ve chosen the branch cut $\theta = 0$ for $\operatorname{Log}(z)$ and define $z^{1\over 2} = e^{{1\over 2}\operatorname{Log}(z)}$. But $z^2-1 \in {\mathbb{R}}_{\geq 0} \iff z\in [-1, 1]^c$, which is what we’ve cut.

So take the branch cut $(-\infty, 1] \cup[1, \infty)$ and use the following indented double-keyhole contour:

Contributions along $C_8$ and $C_6$: note that $\int_{C_8}f \to I$, the desired integral. For reference, note that $z^2-1 = (z+1)(z-1)$, and we can parameterize $$\begin{align*} C_8 = \left\{{t+1+ i{\varepsilon}{~\mathrel{\Big\vert}~}t\in [{\varepsilon}, R] }\right\} \implies \\ \int_{C_8}f(z)\,dz\to \int_0^\infty t^{-1}(t+2)^{-{1\over 2}} t^{-{1\over 2}} \,dt ,\end{align*}$$ where on $C_8$ we choose a branch of the square root so that $\arg(z+1) \in [-\pi, \pi)$ and $\arg(z-1)\in [-\pi, \pi)$. Now consider $C_6$. Write $\zeta_0 \coloneqq e^{2\pi i}$, then $$\begin{align*} C_6 = \left\{{\zeta_0 t + 1 - i{\varepsilon}{~\mathrel{\Big\vert}~}t\in [{\varepsilon}, R]}\right\} \implies \\ \int_{C_6} f(z)\,dz &\to \int_R^{{\varepsilon}} (\zeta_0 t)^{-1}(\zeta_0 t + 2)^{-{1\over 2}}(\zeta_0 t)^{-{1\over 2}}\,dt\\ &\to -\zeta_0^{-{3 \over 2}} \int_0^\infty t^{-1}(t+2)^{-{1\over 2}} t^{-{1\over 2}}\,dt\\ &= I ,\end{align*}$$ since $-\zeta_0^{-{3\over 2}} \coloneqq-e^{-3\pi i} = 1$. So in the limit ${\varepsilon}\to 0, R\to\infty$, $$\begin{align*} \qty{ \int_{C_8} + \int_{C_6}}f \longrightarrow 2I .\end{align*}$$

The contribution from $C_2$: parameterize $$\begin{align*} C_2 = \left\{{s + i{\varepsilon}{~\mathrel{\Big\vert}~}x\in [-R, -1-{\varepsilon}]}\right\} ,\end{align*}$$ which implies $$\begin{align*} \int_{C_2}f(z)\,dz &= \int_{-\infty}^{-1} {1\over s\sqrt{s^2-1}}\,ds\\ &= - \int_{\infty}^{1} {1\over (-x) \sqrt{(-x)^2-1}}\,dx,\qquad x=-s,\, \,dx= -\,ds\\ &= \int_\infty^1 {1\over (-x) \sqrt{ (-x)^2 - 1} }\,dx\\ &= - \int_\infty^1 {1\over x \sqrt{ x^2 - 1} }\,dx\\ &= \int_1^\infty {1\over x \sqrt{ x^2 - 1} }\,dx\\ &= I ,\end{align*}$$ so the same argument as above shows $$\begin{align*} \qty{ \int_{C_2} + \int_{C_4}}f \longrightarrow 2I .\end{align*}$$

Computing the residues: the full contour encloses a simple pole at $z=0$, so $$\begin{align*} 2\pi i \mathop{\mathrm{Res}}_{z=0} f(z) = 2\pi i \lim_{z\to 0} {1\over \sqrt{z^2-1}} = 2\pi \cdot (-i) = 2\pi .\end{align*}$$

So by the residue theorem, $$\begin{align*} 2\pi = 2I + 2I \implies I = {\pi \over 2} .\end{align*}$$

Note that $f(x) = (x+2)(x+1)$, so the singularities on on ${\mathbb{R}}_{< 0}$. The function isn’t even, so a semicircular contour won’t work. Attempting to find a ray-like symmetry only yields one option: $$\begin{align*} f(z) = f(e^{i\theta } z) \implies \theta = 2k\pi ,\end{align*}$$ which suggests a keyhole. But since there’s no $\log$ in $f$, there’s no monodromy, and the contributions cancel out. So introduce a log with a branch cut along $\theta = 0$, and consider $$\begin{align*} \int f(z) \log(z) .\end{align*}$$

Let $\gamma_+ = \left\{{t + i{\varepsilon}{~\mathrel{\Big\vert}~}t\geq 0}\right\}$ (right-to left) and $\gamma_0 = \left\{{t-i{\varepsilon}{~\mathrel{\Big\vert}~}t\geq 0}\right\}$ (left-to-right). Now use the general fact $$\begin{align*} \int_{\gamma_+}f(z)\log(z) \,dz&= \int_{\varepsilon}^R f(t+i{\varepsilon})\log(t+i{\varepsilon}) \,dt\longrightarrow\int_{\mathbb{R}}f(t)\log(t)\,dt\\ \int_{\gamma_-}f(z)\log(z) \,dz&= \int_R^{\varepsilon}f(t-i{\varepsilon})\log(t-i{\varepsilon}) \,dt\longrightarrow-\int_{\mathbb{R}}f(t)\qty{ \log(t) + 2\pi i} \,dt\\ ,\end{align*}$$ thus $$\begin{align*} \int_{\gamma_+}f(z)\log(z)\,dz+ \int_{\gamma_-}f(z)\log(z) &\longrightarrow\int_0^\infty f(t)\log(t)\,dt- \int_0^\infty f(t)(\log(t) + 2\pi i) \,dt\\ &= -2\pi i \int_0^\infty f(t) \,dt .\end{align*}$$

By the ML estimate, the semicircular piece vanishes. Miraculously, since $\lim_{x\to 0}{ x\log(x) \over x^n+c} = 0$ for any $c>0$ and $n\geq 1$, the inner indented pieces goes to zero. Parameterize by $z= {\varepsilon}e^{it}$ $$\begin{align*} \int_{C_{\varepsilon}} f(z)\log(z)\,dz &\approx \int_{{\varepsilon}}^{2\pi - {\varepsilon}} {\log({\varepsilon}e^{it}) \over {\varepsilon}^2 + 3{\varepsilon}+ 2} {\varepsilon}e^{it}\,dt\\ &= \int_{{\varepsilon}}^{2\pi - {\varepsilon}} {\log({\varepsilon}) + it \over {\varepsilon}^2 + 3{\varepsilon}+ 2} {\varepsilon}e^{it}\,dt\\ &\approx \int_{\varepsilon}^{2\pi - {\varepsilon}} {{\varepsilon}\log({\varepsilon}) + c_1 \over {\varepsilon}^2 + c_2}\,dt\\ &\overset{{\varepsilon}\to 0}\longrightarrow 0 ,\end{align*}$$ where I’ve been extremely sloppy and left out many negligible $e^{it}$ terms. By the residue theorem, $$\begin{align*} 2\pi i \sum_{z_k\in {\mathbb{C}}} \mathop{\mathrm{Res}}_{z=z_k} f(z)\log(z) = \int_\Gamma f(z)\log(z)\,dz= -2\pi i\int_{\mathbb{R}}f(z)\,dz .\end{align*}$$ There are just two simple poles at $z_1 = -1, z_2 = -2$, so For $z_1$: $$\begin{align*} \mathop{\mathrm{Res}}_{z=-1}{\log(z) \over (z+1)(z+2)} = \lim_{z\to -1} {\log(z) \over z+2} = \log(-1) ,\end{align*}$$ and for $z_2$, $$\begin{align*} \mathop{\mathrm{Res}}_{z=-2}{\log(z) \over (z+1)(z+2)} = \lim_{z\to -2} {\log(z) \over z+1} = -\log(-2) .\end{align*}$$

Thus $$\begin{align*} I &= -(\log(-1) - \log(-2)) \\ &= - (\ln(1) + i\pi) + (\ln(2) + i\pi) \\ &= \ln(2) ,\end{align*}$$ noting that we’ve chosen a branch of $\log(z) \coloneqq\ln\qty{{\left\lvert {z} \right\rvert}} + i\operatorname{Arg}(z)$ where $\operatorname{Arg}(z) \in (0, 2\pi)$.