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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?

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    $\begingroup$ This is a relatively standard integral. I don't think it requires citation of any sources. $\endgroup$
    – Negan
    Dec 1 '21 at 14:16
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    $\begingroup$ Perhaps you can cite this questions and answers... :-) $\endgroup$
    – Pablo H
    Dec 2 '21 at 18:22
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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. Explicitly, this gives the value for $\displaystyle\psi(3)=\frac 12+\psi(2)=\frac 12+1+\psi(1)=\frac 12(3-2\gamma)$.

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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.

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    $\begingroup$ So, as a reference, it would be enough to mention that we differentiate in the standard relation for $\zeta(s)$, then set $s=3$. $\endgroup$
    – dan_fulea
    Dec 1 '21 at 15:38
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I don't have a published source for this integral, but if need be you could refer to the following derivation: $$\int_0^\infty \frac{x^2\ln{x}}{e^x-1}\,dx=\int_0^\infty x^2 e^{-x}\ln x\sum_{k=0}^\infty e^{-kx}$$ $$=\sum_{k=0}^\infty\frac{3-2\gamma-2 \ln (k+1)}{(k+1)^3}$$ $$=(3-2\gamma)\zeta(3)+2\zeta^\prime(3).$$ The integral over $x^2 e^{-(k+1)x}\ln x$ follows upon partial integration and in the final equation I used the identity $$\sum_{k=1}^\infty k^{-p}\ln k=-\zeta'(p).$$

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    $\begingroup$ Thanks. Both answers are nice. It's a pity that I can "accept" only one. $\endgroup$ Dec 2 '21 at 8:13

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