<s>[INST] Integration of $\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$ Integration of $$\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$$
I am thinking about this integral from last half and hour but still can't figure it out how to solve this. what i was think is to make this in form
$$\int_{0}^{\infty}\frac{\sinh(\lambda+is)t}{\sinh (\pi t)} dt$$ but still this will get me nowhere. please just give me a hint how to proceed this.
Thank you.
 [/INST] Considering $\displaystyle \mathcal{I}(\alpha, \beta) = \int_{0}^{\infty} \frac{\sinh\alpha z}{\sinh \beta z}\,\mathrm{d}z$, we show that $\displaystyle \mathcal{I}(\alpha, \beta) = \frac{\pi}{2\beta} \tan\left(\frac{\alpha \pi}{2\beta}\right)$.
Then your integral is $\displaystyle \mathcal I(\lambda+is, \pi)$. Starting with   $z \mapsto \frac{1}{\beta} \log(z)$, we have:
\begin{aligned} \displaystyle \mathcal{I}(\alpha, \beta) & = \frac{1}{\beta} \int_0^1 \frac{z^{\frac{\alpha}{\beta}} -z^{-\frac{\alpha}{\beta}}}{z^2-1}\;{dz} \\& = \frac{1}{\beta} \int_0^{1} \left( z^{-\frac{\alpha}{\beta}}-z^{\frac{\alpha}{\beta}}\right) \sum_{k \ge 0}z^{2k} \;{dz}\\& = \frac{1}{\beta} \sum_{k \ge 0}\int_0^1 z^{-\frac{\alpha}{\beta}+2k} -z^{\frac{\alpha}{\beta}+2k}\;{dz} \\& = \frac{1}{\beta} \sum_{k \ge 0} \frac{1}{1-\frac{\alpha}{\beta}+2k}-\frac{1}{1+\frac{\alpha}{\beta}+2k} \\& = \frac{1}{\beta} \sum_{k \ge 0} \frac{2\alpha /\beta }{(1+2 k)^2-(\alpha/\beta)^2}  \\& = \frac{1}{2\beta} \sum_{k ~ \text{odd} } \frac{4\alpha/\beta}{k^2-(\alpha/\beta)^2} \end{aligned}
Consider the series $\displaystyle \displaystyle \pi\cot(\pi z) = \frac{1}{z}+ \sum_{n \ge 1} \frac{2z}{z^2-n^2} $. Map $\displaystyle z \mapsto \frac{z}{2} $ and divide by two then we have:
$\displaystyle \displaystyle \frac{\pi}{2}\cot\left(\frac{\pi}{2} z\right) = \frac{1}{z}+\frac{1}{2}\sum_{n \ge 1} \frac{z}{(z/2)^2-n^2} $$\displaystyle \displaystyle =  \frac{1}{z}+ \frac{1}{2}\sum_{n \ge 1} \frac{4z}{z^2-(2n)^2} = \frac{1}{z}+\frac{1}{2}\sum_{n ~ \text{even}} \frac{4z}{z^2-n^2}$
Thus $\displaystyle \displaystyle \frac{1}{z}+\sum_{n \ge 1} \frac{2z}{z^2-n^2} -\frac{1}{z}-\frac{1}{2} \sum_{n ~ \text{even}} \frac{4z}{z^2-n^2}$ $\displaystyle \displaystyle  =\pi\cot(\pi z)-\frac{\pi}{2}\cot\left(\frac{\pi}{2} z\right) = -\frac{\pi}{2} \tan\left(\frac{z\pi}{2}\right)$
Hence  $\displaystyle \displaystyle \frac{1}{2} \sum_{n \textrm{ odd}} \frac{4z}{z^2-n^2} = -\frac{\pi}{2}\tan\left(\frac{\pi z}{2}\right)$ and so $\displaystyle  \displaystyle  \mathcal I(\alpha, \beta)=\frac{1}{2\beta} \sum_{k ~ \text{odd} } \frac{4\alpha/\beta}{(\alpha/\beta)^2-k^2} = \frac{\pi}{2\beta} \tan\left(\frac{\alpha \pi}{2\beta}\right).$
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