How do I can evaluate this integral: $\int_{1}^{\infty}\frac{y\cosh(yx)}{\sinh(y\pi)}dy$? I'm interested to know how do I evaluate this integral:
$$\int_{1}^{\infty}\frac{y\cosh(yx)}{\sinh(y\pi)}dy.$$
Wolfram alpha gives the output
$$\frac1{(\pi-x)^2}+\text{Li}(-2e^{-2\pi})-\frac{\log(1-e^{-2\pi})}{\pi}-\frac12+O(\pi-x).$$
Note: I have tried to put $x=y$ to show if the above integral easy for
evaluation where I took $t=\tan\frac{x}{2}$ but I got a complicated form which I can't use some standard and simple trigonometric transformations for,
Thank you for any help.

I presume you are actually interested in this integral for some application, and then you might want a convenient expression --- Mathematica does evaluate it in closed form [*], as indicated in the comments, but this involves special functions and might not be of much practical use. Here is what I propose to do. Consider the integral with a variable lower bound,
$$I_u(x)=\int_{u}^{\infty}\frac{y\cosh(yx)}{\sinh(y\pi)}dy,\;\;u\geq 0,\;\;|x|<\pi.$$
The function for $u=0$ has a simple form,
$$I_0(x)=\frac{1}{2+2\cos x},$$
which is already quite close to the desired $I_1(x)$. We can improve by adding an offset of $-1/4$, as is evident from the plot: blue = $I_1$, orange = $I_0$, green = $I_0-\frac{1}{4}=\frac{1}{4}\tan^2(x/2)$


[*] For the record, after some massaging this is the Mathematica output for $I_1$, with $\Phi$ the Lerch transcendent:
$$I_1(x)= \frac{1}{2\cos^2(x/2)}\\
\quad+\sum_{+x,-x}\left[\frac{e^{\pi+x}}{\pi+x}  \, _2F_1\left(1,\tfrac{x }{2 \pi }+\tfrac{1}{2};\tfrac{x}{2\pi} +\tfrac{3}{2};e^{2 \pi }\right)-\frac{e^{\pi+x}}{4 \pi ^2 } \Phi \left(e^{2 \pi },2,\tfrac{x }{2 \pi }+\tfrac{1}{2}\right)\right]$$
(the sum over $+x$ and $-x$ ensures that the result is an even function of $x$, as it should be)
