Integral $ \int_0^\infty \frac{x^n\ln x}{(x^2+\alpha^2)^2(e^x-1)}dx$ Hey I am trying to integrate
$$
\int_0^\infty \frac{x^n\ln x}{(x^2+\alpha^2)^2(e^x-1)}dx,\quad \alpha,n \in \mathbb{R}^{0+}.
$$
This integral is old.  I am also looking for literature on these integrals as I have seen many for small values of n, and variations of this.  Thanks.  Maybe we can use residues however the log function in the denominator is what is concerning me, without that I can see what to do
 A: This is inspired by the game that @BennettGardiner played in this question with Binet's second Log-Gamma integral formula:
$$
 \ln \Gamma(z)=\left(z-\frac{1}{2}\right)\ln z-z+\frac{1}{2}\ln(2\pi)+2\int_0^{\infty} \frac{\tan^{-1}(t/z)}{e^{2\pi t}-1} \ \mathrm{d}t 
$$
In the above, shift the integration variable $t$ to the coordinate $x=2\pi t$, and then define the constant $\alpha:= 2 \pi z$. Rearranging the terms, we are left with the following integral:
$$
\int_{0}^{\infty} \frac{\tan^{-1}\left( \tfrac{x}{\alpha} \right)}{e^{x}-1} dx \ = \ \pi \ln\left[ \Gamma\left(\frac{\alpha}{2\pi}\right) \right] - \frac{\alpha - \pi}{2} \ln\left( \frac{\alpha}{2\pi} \right) + \frac{\alpha}{2} - \frac{\pi}{2} \ln(2\pi)
$$
Now noting that $\frac{d}{d\alpha}\left\{ \tan^{-1}\left( \tfrac{x}{\alpha} \right) \right\} = - \frac{x}{x^2 + \alpha^2}$, we hit both sides of the above relation with $-\frac{d}{d\alpha}$ and get the following:
$$
\int_{0}^{\infty} \frac{x}{(e^{x}-1)(x^2+\alpha^2)} dx \ = \ \frac{1}{2} \ln\left( \frac{\alpha}{2 \pi} \right) - \frac{\pi}{2 \alpha} - \frac{1}{2} \psi^{(0)}\left( \frac{\alpha}{2\pi} \right)
$$
where $\psi^{(0)}$ is the digamma (aka polygamma-0) function. This is the generalization of the integral from that same question I mentioned above.
Next we take note of the following derivative:
$$
\frac{d}{d\alpha} \left\{ \frac{x}{x^2 + \alpha^2} \right\} = - \frac{2 \alpha x}{(x^2 + a^2)^2}
$$
So hitting both sides of the above integral with the operator $- \frac{1}{2\alpha} \frac{d}{d\alpha}$ yields the following integral:
$$
\int_{0}^{\infty} \frac{x}{(e^{x}-1)(x^2+\alpha^2)^2} dx \ = \ - \frac{1}{4\alpha^2} - \frac{\pi}{4 \alpha^3} + \frac{1}{8 \pi \alpha} \psi^{(1)}\left( \frac{\alpha}{2\pi} \right)
$$
where $\psi^{(1)}$ is the polygamma-$1$ function.
Next, we note the following:
$$
\frac{d}{d\alpha} \left\{ \frac{\alpha^2 x}{x^2+\alpha^2} \right\} = \frac{2 \alpha x^3}{(\alpha^2 + x^2)^2}
$$
So we take our second integral identity, multiply both sides by $\alpha^2$ and then hit both sides with $\frac{d}{d\alpha}$:
$$
\frac{d}{d\alpha} \left\{ \int_{0}^{\infty} \frac{x\alpha^2}{(e^{x}-1)(x^2+\alpha^2)} dx \right\} \ = \ \frac{d}{d\alpha} \left\{ \frac{\alpha^2}{2} \ln\left( \frac{\alpha}{2 \pi} \right) - \frac{\pi \alpha}{2} - \frac{\alpha^2}{2} \psi^{(0)}\left( \frac{\alpha}{2\pi} \right) \right\}
$$
Taking the derivative in the above, and then dividing everything by $2\alpha$ we are left with the formula:
$$
\int_{0}^{\infty} \frac{x^3}{(e^{x}-1)(x^2+\alpha^2)^2} dx \ = \ \frac{1}{4} - \frac{\pi}{4 \alpha} + \frac{1}{2} \log\left( \frac{\alpha}{2\pi} \right) - \frac{1}{2} \psi^{(0)}\left( \frac{\alpha}{2\pi} \right) -  \frac{\alpha}{8 \pi} \psi^{(1)}\left( \frac{\alpha}{2\pi} \right)
$$
In summary, I have found two integrals related to your integral (without the log term though) for $n=1,3$. I would like to find a general formula for any $n \geq 0$ so we could apply an operator $\frac{d}{dn}$ to get down a $\ln(x)$ as @BennettGardiner suggested above. I don't know where to go from here though.
