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Abel-Plana Formula:
\begin{align}
\sum_{n = 0}^{\infty}\expo{-n\verts{a}} & =
\int_{0}^{\infty}\expo{-\verts{a}x}\,\dd x +
\bracks{{1 \over 2}\,\expo{-\verts{a}x}}_{\ x\ =\ 0} -
2\,\Im\int_{0}^{\infty}{\expo{-\verts{a}x\ic} \over \expo{2\pi x} - 1}\,\dd x
\end{align}
\begin{align}
{1 \over 1 - \expo{-\verts{a}}} & =
{1 \over \verts{a}} + {1 \over 2} +
2\,\mrm{sgn}\pars{a}\int_{0}^{\infty}
{\sin\pars{ax} \over \expo{2\pi x} - 1}\,\dd x
\end{align}
\begin{align}
&2\,\mrm{sgn}\pars{a}\int_{0}^{\infty}
{\sin\pars{ax} \over \expo{2\pi x} - 1}\,\dd x
=
{1 \over 1 - \expo{-\verts{a}}} - {1 \over 2} - {1 \over \verts{a}} =
{1 + \expo{-\verts{a}} \over 1 - \expo{-\verts{a}}}\,{1 \over 2} - {1 \over \verts{a}}
\end{align}
\begin{align}
&\int_{0}^{\infty}
{\sin\pars{ax} \over \expo{2\pi x} - 1}\,\dd x =
\bbx{\ds{{1 \over 4}\,\coth\pars{a \over 2} - {1 \over 2a}}}
\end{align}
To answer your third question, you can do whatever you want, which we will see later. Because we want to evaluate the integral over the real axis, we will include $\mathbb{R}$ (except $z=0$) in the contour. For this integral, this is possible, maybe not in other integrals. Because $\sinh(z)=0$ for all integer multiples of $i\pi$, this gives infinitely many singularities. We don't want this, so the idea is to only include $z=0$ and $z=\pi i$ in the contour.
Note that $$\sin(a(z+i\pi))=\sin(az+ai\pi)=\sin(az)\cos(ai\pi)+\cos(az)\sin(ai\pi)$$ and $$\sinh(z+i\pi)=-\sinh(z).$$ These facts suggest including $i\pi+\mathbb{R}$ in the integral. Note that $\frac{\sin(az)}{\sinh(z)}$ is an even function, so we can integrate over $\mathbb{R}$ instead of from $0$ to $\infty$. From this follows the rectangular contour ($R\rightarrow\infty$ and $\varepsilon\rightarrow0$):
- A line from $-R$ to $-\varepsilon$
- A semicircle of radius $\varepsilon$ around $z=0$, choose it to include $z=0$
- A line from $\varepsilon$ to $R$
- A line from $R$ to $R+i\pi$
- A line from $R+i\pi$ to $\varepsilon+i\pi$
- A semicircle of radius $\varepsilon$ around $z=i\pi$, choose it to include $z=i\pi$
- A line from $-\varepsilon+i\pi$ to $-R+i\pi$
- A line from $-R+i\pi$ to $-R$.
We know that $\int_1=\int_3$, because $\frac{\sin(az)}{\sinh(z)}$ is an even function. We also know that $\int_5=\int_7=\cos(ai\pi)\int_1$, because $$\int_{-\infty}^\infty\frac{\sin(az)\cos(ai\pi)+\cos(az)\sin(ai\pi)}{\sinh(z)}dz=\int_{-\infty}^\infty\frac{\sin(az)\cos(ai\pi)}{\sinh(z)}dz$$ and $\frac{\cos(az)}{\sinh(z)}$ is an odd function. Here the minus signs from $\sinh(z+i\pi)=-\sinh(z)$ and the opposite direction cancel each other. Note that we have to say something about convergence. We also have that $\int_4=\int_8=0$, where we again have to say something about convergence.
Now we calculate residues: around $z=0$ we have $\frac{\sin(az)}{\sinh(z)}\rightarrow0$ and around $z=i\pi$ we have $$\frac{\sin(az)}{\sinh(z)}=\frac{\sin(az)\cos(ai\pi)+\cos(az)\sin(ai\pi)}{-\sinh(z)}\overset{z\rightarrow0}{\rightarrow}-\frac{\sin(ai\pi)}{\sinh(z)},$$ which gives $-\sin(ai\pi)$ as residue.
Note that $\int_6=-2\pi i\frac{1}{2}\sin(ai\pi)$, (there is a theorem about poles of order 1 and half circles around it) and $\int_2=0$. From this follows that $$2(1+\cos(ai\pi))\int_0^\infty\frac{\sin(az)}{\sinh(z)}dz-\pi i\sin(ai\pi)=-2\pi i\sin(ai\pi),$$ which gives also the answer to your third question. If we don't include $z=i\pi$, this gives $$2(1+\cos(ai\pi))\int_0^\infty\frac{\sin(az)}{\sinh(z)}dz+\pi i \sin(ai\pi)=0,$$ because the halfcircle around $z=i\pi$ is traversed clockwise, instead of counterclockwise.
Finally, the answer is $$\int_0^\infty\frac{\sin(az)}{\sinh(z)}dz=-\pi i\frac{\sin(ai\pi)}{2+2\cos(ai\pi)}=\frac{\pi}{2}\tanh(\frac{\pi a}{2}).$$
Best Answer
After substituting $x\mapsto\log(x)$, we use the keyhole contour from this answer with $n=a-1$ and $m=1$: $$ \begin{align} \int_{-\infty}^\infty\frac{e^{ax}}{1+e^x}\,\mathrm{d}x &=\int_0^\infty\frac{x^{a-1}}{1+x}\,\mathrm{d}x\\ &=\pi\csc(\pi a) \end{align} $$