Show that $\int_0^\infty t e^{-xt}\log(1+x^2) \, dx=(\pi-2\mathrm{Si}(t))\sin t-2\mathrm{Ci}(t)\cos t$, $\forall t>0$. How to show that
$$\int_0^\infty t e^{-xt}\log(1+x^2) \, dx=(\pi-2\mathrm{Si}(t))\sin t-2\mathrm{Ci}(t)\cos t$$
for all $t>0$?
 A: If you needed it for a paper/proof, you might be able to show it with a reference to a table in this way:
Mellin transform with respect to $t$, of integrand gives
$$
\int_0^\infty t^{s-1}t e^{-xt}\log(1+x^2)\; dt = x^{-1-s}\Gamma(1+s)\log(1+x^2)
$$
integrate this over $x$ gives
$$
\int_0^\infty x^{-1-s}\Gamma(1+s)\log(1+x^2)\;dx = \pi\csc\left(\frac{\pi s}{2}\right)\Gamma(s)
$$
Find an identity in this link near the bottom or similar for the inverse Mellin transform of $\pi\csc\left(\frac{\pi s}{2}\right)\Gamma(s)$
$$
I=\frac{1}{2 \pi i}\int_{c - i \infty}^{c + i \infty} t^{-s}\pi\csc\left(\frac{\pi s}{2}\right)\Gamma(s) \; ds
$$
which looks like it should be in there as terms like $\pi \csc(\pi s)$ are very common in these inverses. Worst case, you'll need to convert the $\csc$ to a hypergeometric function and use the more general identities at the bottom, this might be how Mathematica solves the integral as it has a lot of behind the scenes lookups. The command

InverseMellinTransform[Pi Csc[Pi s /2]Gamma[s],s,t]

does return your result.
