<s>[INST] Infinite sums and integrals using residues I have no idea how to solve these two, any help?
$\mathtt{i)}$ $$\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\frac{e^{tz}}{\sqrt{z+1}}dz$$ $$ a,t\gt0$$
$\mathtt{ii)}$ 
$$ \sum_{n=1}^\infty \frac{\coth (n\pi)}{n^3}=\frac{7\pi^3}{180}$$
 [/INST] For the second problem, let $ \displaystyle f(z) = \frac{\pi\cot (\pi z) \coth(\pi z)}{z^{3}}$ and integrate around a square ($C_{N}$) with vertices at $\pm (N+\frac{1}{2}) \pm (N+\frac{1}{2})$ where $N$ is a positive integer.
Both $\cot (\pi z)$ and $\coth(\pi z)$ are uniformly bounded on the contour.
So $ \displaystyle \int_{C_{N}} f(z) \ dz$ vanishes as $N$ goes to infinity through the integers..
And we have
$$ 2 \sum_{n=1}^{\infty} \frac{\coth (\pi n)}{n^{3}} + \text{Res}[f(z),0] + \sum_{n=1}^{\infty} \text{Res} [f(z),in] +  \sum_{n=1}^{\infty} \text{Res} [f(z),-in]  =0 $$
where
$$ \text{Res} [f(z),in] = \lim_{z \to in}  \frac{\pi \cot (\pi z)}{\frac{d}{dz} (z^{3} \tanh (\pi z) )} = \lim_{z \to in} \frac{\pi \cot (\pi z)}{3z^{2} \tanh (\pi z) + \pi z^{3} \text{sech}^{2} (\pi z) } = \frac{\coth(\pi n)}{n^{3}}$$
and similarly
$$ \text{Res} [f(z),-in] = \frac{\coth (\pi n)}{n^{3}} \ .$$
And expanding $\cot (\pi z)$ and $\coth(\pi z)$ in Laurent expansions about the origin,
$$ \begin{align} \frac{\pi\cot (\pi z) \coth(\pi z)}{z^{3}} &= \frac{\pi}{z^{3}} \left( \frac{1}{\pi z} - \frac{\pi z}{3} - \frac{\pi^{3}z^{3}}{45} + \ldots \right) \left( \frac{1}{\pi z} + \frac{\pi z}{3} - \frac{\pi^{3}z^{3}}{45} + \ldots \right) \\ &= \frac{\pi}{z^{3}} \left(\frac{1}{\pi^{2}z^{2}} + \frac{1}{3} - \frac{\pi^{2}z^{2}}{45} - \frac{1}{3} - \frac{\pi^{2} z^{2}}{9} - \frac{\pi^{2}z^{2}}{45} + \ldots \right) \\ &= \frac{1}{\pi z^{5}} - \frac{7 \pi^{3}}{45 z} + \ldots \ \ .\end{align}$$
Therefore, $$\text{Res}[f(z),0] = - \frac{7 \pi^{3}}{45}$$
and $$\sum_{n=1}^{\infty} \frac{\coth (\pi n)}{n^{3}} = \frac{1}{4} \left( \frac{7 \pi^{3}}{45} \right) = \frac{7 \pi^{3}}{180} .$$
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