Wolfram alpha calculates the integral
$$\int\limits\_0^\infty \frac{x^2\ln{x}}{e^x-1}dx=2\zeta^\prime(3)+3\zeta(3)-2\gamma\zeta(3).$$
However, I need to cite the source of this identity (the table of integrals, or the article where this integral was calculated). Could you indicate any?



The source of the Integral
One way to get the claimed value of the given integral, $J$ in notation below, is by starting from the standard relation
$$
\begin{aligned}
\zeta(s) 
&= \frac 1{\Gamma(s)}\int\_0^\infty \frac {x^{s-1}}{e^x-1}\; dx\ ,\qquad
\text{ so }\\
\zeta'(s)
&= 
\frac\partial{\partial s}
\left(\ 
\frac 1{\Gamma(s)}\int\_0^\infty \frac {x^{s-1}}{e^x-1}\; dx
\ \right)
\\
&= 
\frac 1{\Gamma(s)}\int\_0^\infty \frac {x^{s-1}\; \ln x}{e^x-1}\; dx
-
\underbrace{\frac{\Gamma'(s)}{\Gamma(s)}}\_{=\psi(s)}\zeta(s)\ ,\qquad\text{leading to }
\\
\zeta'(3)&=\frac 1{\Gamma(3)}\underbrace{\int\_0^\infty \frac {x^2\; \ln x}{e^x-1}\; dx}\_{\text{our integral }J}
-
\psi(3)\zeta(3)\ .
\\[2mm]
&\qquad\text{Extracting $J$ from above,}\\[2mm]
J &=\Gamma(3)\; \zeta'(3) \ +\ \Gamma(3)\; \psi(3)\; \zeta(3)
\\
&=2\zeta'(3)+(3-2\gamma)\zeta(3)\ ,
\end{aligned}
$$
where we have used the relations $\displaystyle\psi(x+1)=\frac 1x+\psi(x)$ and $\psi(1)=-\gamma$, e.g. from [here](https://www.nbi.dk/%7Epolesen/borel/node5.html). Explicitly, this gives the value for $\displaystyle\psi(3)=\frac 12+\psi(2)=\frac 12+1+\psi(1)=\frac 12(3-2\gamma)$.


$\square$




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The above shows how to get similar relations when there is an other power of $x$ instead of $x^2$ in the numerator of then integrand, and gives an interpretation of the involved coefficients.


