Proof for $\int_0^\infty dx\ \sin(y x) \left[ \frac{1}{x^2} - \mathrm{csch}^2(x) \right]$ For $y>0$ I have been able to numerically confirm that
$$
\int_0^\infty dx\ \sin(y x) \left[ \frac{1}{x^2} - \mathrm{csch}^2(x) \right] \ = \ \frac{y}{2} \; \psi^{(0)}\left( \frac{i y}{2} \right) + \frac{y}{2} \; \psi^{(0)}\left( - \frac{i y}{2} \right) - y\;  \log\left(\frac{y}{2}\right) .
$$
where $\psi^{(0)}$ is the digamma function.
Does anyone have an interesting proof for the above? I can't get anywhere.
 A: Starting with the first of these representations for $\psi:=\psi^{(0)}$, we have
\begin{align}
\psi(z)&=\int_0^\infty\left(\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^{-t}}\right)\,dt
\\&=\underbrace{\int_0^\infty\frac{e^{-t}-e^{-zt}}{t}\,dt}_{\text{Frullani integral}}+\underbrace{\int_0^\infty\left(\frac{1}{t}-\frac{1}{1-e^{-t}}\right)e^{-zt}\,dt}_{\text{integrate by parts}}
\\&=\ln z-\frac{1}{2z}-\frac{1}{z}\int_0^\infty\left(\frac{1}{t^2}-\frac{e^{-t}}{(1-e^{-t})^2}\right)e^{-zt}\,dt,
\end{align}
where $\Re z>0$ and the principal branch of $\ln z$ is chosen. This gives
$$I(z):=\int_0^\infty(t^{-2}-\operatorname{csch}^2 t)e^{-zt}\,dt=z\big(\psi(z/2)-\ln(z/2)\big)+1.$$
Now consider $I(x+iy)-I(x-iy)$ with $x\to 0$.
A: Let $y > 0$ and
\begin{align}
I(y) &\equiv\int \limits_0^\infty \sin(y x) \left[x^{-2} - \operatorname{csch}^2(x)\right] \mathrm{d} x = -\operatorname{Im} \left[\int \limits_0^\infty \mathrm{e}^{-\mathrm{i} y x} \left[x^{-2} - \operatorname{csch}^2(x)\right] \mathrm{d} x\right] \\
&= - \operatorname{Im} \left[\lim_{\varepsilon \rightarrow 0^+} \int \limits_0^\infty \mathrm{e}^{-(\mathrm{i} y + \varepsilon)x} \left[x^{-2} - \operatorname{csch}^2(x)\right] \mathrm{d} x\right] \equiv \operatorname{Im} \left[\lim_{\varepsilon \rightarrow 0^+} J_\varepsilon(y)\right] \, .
\end{align}
Taking the limit outside of the integral is justified by the dominated convergence theorem, since $|\mathrm{e}^{-(\mathrm{i}y + \varepsilon)x}| \leq 1$ and $\int_0^\infty \left[x^{-2} - \operatorname{csch}^2(x)\right] \mathrm{d} x = 1$ . Now integrate by parts to find
\begin{align}
J_\varepsilon (y) &= -(\mathrm{i}y + \varepsilon) \int \limits_0^\infty \mathrm{e}^{-(\mathrm{i}y + \varepsilon)x} \left[\coth(x) - 1 - \frac{1}{x}\right] \mathrm{d} x \stackrel{x \, = \,  t/2}{=} (\mathrm{i}y + \varepsilon) \int \limits_0^\infty \mathrm{e}^{-\frac{\mathrm{i}y + \varepsilon}{2} t} \left[\frac{1}{t} - \frac{1}{\mathrm{e}^t - 1}\right] \mathrm{d} t \\
&= (\mathrm{i}y + \varepsilon) \left[\psi^{(0)}\left(\frac{\mathrm{i}y + \varepsilon}{2} + 1\right) - \log \left(\frac{\mathrm{i}y + \varepsilon}{2}\right)\right] \, .
\end{align}
The last step follows from an integral representation for the digamma function derived from Binet's log-gamma formula. Therefore,
\begin{align}
I(y) &= \operatorname{Im} \left\{\mathrm{i} y\left[\psi^{(0)}\left(\frac{\mathrm{i}y}{2} + 1\right) - \log \left(\frac{\mathrm{i}y}{2}\right) \right]\right\} = y \operatorname{Re} \left[\psi^{(0)}\left(\frac{\mathrm{i}y}{2}\right) + \frac{2}{\mathrm{i} y} - \log \left(\frac{\mathrm{i}y}{2}\right)\right] \\
&= y \left\{\operatorname{Re} \left[\psi^{(0)}\left(\frac{\mathrm{i}y}{2}\right)\right] - \log \left(\frac{y}{2}\right)\right\} = y \left\{\frac{1}{2} \left[\psi^{(0)}\left(\frac{\mathrm{i}y}{2}\right) + \psi^{(0)}\left(-\frac{\mathrm{i}y}{2}\right)\right] - \log \left(\frac{y}{2}\right)\right\}\, . 
\end{align}
