INTEGRAL RELATED TO DERIVATIVE OF BETA FUNCTION

Today we will evaluate the following integral found in this post


\int_0^\infty\frac{x}{\left(e^{x}-1 \right)^{\frac{2}{3}}}dx=\sqrt{3} \pi \ln 3-\frac{\pi^2}{3}

Lets consider the more general case


\begin{aligned} I&=\int_0^\infty\frac{x}{\left(e^{x}-1 \right)^{s}}dx \qquad e^x \mapsto x\\ &=\int_1^\infty\frac{\ln x}{\left(x-1 \right)^{s}} \frac{dx}{x} \qquad x \mapsto \frac{1}{x}\\ &=-\int_0^1\frac{x^{s-1}\ln x}{\left(1-x \right)^{s}} \end{aligned}


Now observe the following: First recall the Beta function


\int_0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma\left( a\right)\Gamma\left( b\right)}{\Gamma\left(a+b \right)}


Differentiating the above expression w.r. to a we obtain


\int_0^1 x^{a-1}(1-x)^{b-1} \ln x \,dx=\frac{\Gamma\left( a\right)\Gamma\left( b\right)}{\Gamma\left(a+b \right)}\left(\psi(a)-\psi(a+b) \right)

Letting a \rightarrow s \, \, \text{and} \,\, b \rightarrow 1-s\, we obtain

\int_0^\infty\frac{x}{\left(e^{x}-1 \right)^{s}}dx&=-\frac{\Gamma\left( s\right)\Gamma\left( 1-s\right)}{\Gamma\left(1 \right)}\left(\psi(s)-\psi(1) \right)

Now let  s \rightarrow \frac{2}{3}

\begin{aligned} \int_0^\infty\frac{x}{\left(e^{x}-1 \right)^{\frac{2}{3}}}dx&=-\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)\left(\psi\left(\frac{2}{3}\right)-\psi(1) \right) \right)\\ &=-\frac{\pi}{\sin \frac{\pi}{3}}\left(\psi\left(\frac{2}{3}\right)-\psi(1) \right)\\ &=-\frac{2\pi}{\sqrt{3}}\left(\frac{\pi}{2 \sqrt{3}}-\frac{3 \ln 3}{2}-\gamma-(-\gamma) \right) \\ &=\sqrt{3} \pi \ln 3-\frac{\pi^2}{3} \qquad \blacksquare \end{aligned}

We used the reflection formula of the Gamma function and the special values of the Digamma function at  1 \,\,\text{and}\,\, \, \frac{2}{3}


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