Integral $\int_0^\infty\frac{x\cos(ax)}{x^2+1}\coth(sx)\mathrm{d}x$ I need help computing this integral:
\begin{equation}
\int_0^\infty\frac{x\cos(ax)}{x^2+1}\coth(sx)\mathrm{d} x \tag{1}
\end{equation}
I searched through many integral tables and I found this in Gradstein and Ryshik:
\begin{equation}
\int_0^\infty\frac{x\cos(ax)}{x^2+1}\coth(\pi x)\mathrm{d}x=-\frac{a}{2}e^{-a}-\frac{1}{2}-\cosh(a)\ln(1-e^{-a}) 
\end{equation}
Sadly the book does not have proof of this, so I might try the same method for evaluating $(1)$. Since these two look almost the same there has to be a way to solve $(1)$. If anyone else owns the book and would like to check it out there it's in chapter $4.11-4.12$.
 A: $$I = \int_{0}^{\infty} \frac{x \cos (a x)}{x^2+1} \coth (s x) \, dx$$
Recall the Mittag-Leffler pole expansion of $\cot (z)$:
$$\cot (z) = \frac{1}{z} + 2z \sum_{k=1}^{\infty} \frac{1}{z^2 - (k \pi)^2}$$
Letting $z \to i s x$ and multiplying by $i$ gives:
$$\coth (s x) = \frac{1}{s x} + 2s x \sum_{k=1}^{\infty} \frac{1}{(s x)^2 + (k \pi)^2}$$
Substituting this expression in, we have then:
$$I = \frac{1}{s} \int_{0}^{\infty} \frac{\cos (a x)}{x^2+1} \, dx + \sum_{k=1}^{\infty} \int_{0}^{\infty} \frac{2sx^2 \cos (a x)}{(x^2+1)((s x)^2 + (k \pi)^2)} \, dx$$
We can separate the integrals because each converges individually. The first integral has a well-known result:
$$\frac{1}{s} \int_{0}^{\infty} \frac{\cos (a x)}{x^2+1} \, dx = \frac{\pi e^{-a}}{2s}$$
The second integral is a little less trivial, but it can be done.
Mathematica gives:
$$\int_{0}^{\infty} \frac{2sx^2 \cos (a x)}{(x^2+1)((s x)^2 + (k \pi)^2)} \, dx=\frac{\pi^2 k e^{-\frac{\pi a k}{s}}-\pi e^{-a}s}{\pi^2k^2-s^2}$$
Furthermore, Mathematica gives:
$$\sum_{k=1}^{\infty} \frac{\pi^2 k e^{-\frac{\pi a k}{s}}-\pi e^{-a}s}{\pi^2k^2-s^2} = \frac{1}{2}e^{-\frac{\pi a}{s}}\left(\Phi\left(e^{-\frac{a\pi}{s}},1,1-\frac{s}{\pi}\right)+\Phi\left(e^{-\frac{a\pi}{s}},1,1+\frac{s}{\pi}\right)\right)-\frac{\pi e^{-a}}{2s}+\frac{\pi}{2} e^{-a}\cot(s)$$
Where $\Phi$ is the Lerch transcendent.
Thus summing the two results, we have:
$$I = \frac{1}{2}e^{-\frac{\pi a}{s}}\left(\Phi\left(e^{-\frac{a\pi}{s}},1,1-\frac{s}{\pi}\right)+\Phi\left(e^{-\frac{a\pi}{s}},1,1+\frac{s}{\pi}\right)\right)+\frac{\pi}{2} e^{-a}\cot(s)$$
However, this expression does not hold for $s$ being an integer multiple of $\pi$ due to $\cot (s)$ causing it to be an indeterminate when one takes the limit, so there needs to be some finesse for evaluating these cases. I'm not sure if there is simplification possible for the result I obtained for $I$ involving $\Phi$.
