Answer by dan_fulea for 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$$\begin{aligned}\zeta(s) &= \frac 1{\Gamma(s)}\int_0^\infty \frac...
View ArticleAnswer by Carlo Beenakker for The source of the Integral
I don't have a published source for this integral, but if need be you could refer to the following derivation:$$\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...
View ArticleThe source of the Integral
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...
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