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I have the idea of this integral when I see $$\int_{0}^{\infty }\!{\frac {\arctan \left( x \right) }{{{\rm e}^{\pi\,x}}-1}}\,{\rm d}x$$ and so I know that the closed form is $${\frac{1}{2}}-{\frac {\ln \left( 2 \right) }{2}}.$$

But really, I don't know any paper where I can evaluate for example $$\int_{0}^{\infty }\!{\frac {\ln \left( x \right) \arctan \left( x \right) }{{{\rm e}^{\pi\,x}}-1}}\,{\rm d}x.$$

In the same time, I see that Wolfram can't give a closed form.

Does someone has an idea please? Thanks

Gary
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Dens
  • 353

2 Answers2

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I really tried a lot but couldn't find a closed form. Maybe the closed form for this doesn't exist but I found something else;

A general form for your former integral is

$\int_{0}^{\infty} (\frac{e^{\pi x}}{e^{2\pi x}-1}) \arctan({\frac{x}{q}}) \,dx =\frac{1}{2}[\frac{1}{2}\ln(\frac{q}{\pi})+(q-\frac{3}{2})\ln{(2)}+\ln{[B(\frac{q}{2},\frac{q}{2})]}]$

also a cool result I found was;

$\int_{0}^{\infty} (\frac{x}{e^{\pi x}+1}) \ln({1+\frac{1}{x^2}}) \,dx = 6\ln{(A)}+\frac{1}{4}-\frac{2}{3}\ln{(2)}-\ln{(\pi)}$ where A is the Glaisher–Kinkelin constant

Nucleo
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$$\int\limits_{0}^{\infty}\frac{\log\left(x\right)\arctan\left(x\right)}{e^{\pi x}-1}\,dx$$ $$\sum_{k=0}^{\infty}\frac{\left(-1\right)^k}{2k+1}\int\limits_{0}^{\infty}\frac{x^{2k+1}\log\left(x\right)}{e^{\pi x}-1}\,dx$$

From my notes on the Zeta function: $$ \int\limits_{0}^{\infty} \frac{x^{s-1}}{e^{ax^b}-1}\,dx=\frac{\zeta\left(\frac{s}{b}\right)\Gamma\left(\frac{s}{b}\right)}{ba^{\frac{s}{b}}}$$ $$s=2k+2,\quad b=1, \quad a=\pi$$ $$\frac{d}{dk}\left[\frac{\zeta\left(2k+2\right)\Gamma\left(2k+2\right)}{\pi^{2k+2}}\right]=2\int\limits_{0}^{\infty}\frac{x^{2k+2}\log\left(x\right)}{e^{\pi x}-1}\,dx$$ $$\int\limits_{0}^{\infty}\frac{x^{2k+1}\log\left(x\right)}{e^{\pi x}-1}\,dx =\frac{1}{2}\frac{d}{dk}\left[\frac{\zeta\left(2k+2\right)\Gamma\left(2k+2\right)}{\pi^{2k+2}}\right]=\zeta'\left(2k+2\right)\Gamma\left(2k+2\right)\pi^{-2k-2}+\zeta\left(2k+2\right)\Gamma\left(2k+2\right)\psi^{\left(0\right)}\left(2k+2\right)\pi^{-2k-2}-\log\left(\pi\right)\zeta\left(2k+2\right)\Gamma\left(2k+2\right)$$ $$\frac{1}{\pi^2}\left(\sum_{k=0}^{\infty}\frac{\left(-1\right)^k\zeta'\left(2k+2\right)\Gamma\left(2k+2\right)}{\pi^{2k}\left(2k+1\right)}+\sum_{k=0}^{\infty}\frac{\left(-1\right)^k\zeta\left(2k+2\right)\Gamma\left(2k+2\right)\psi^{\left(0\right)}\left(2k+2\right)}{\pi^{2k}\left(2k+1\right)} - \log\left(\pi\right)\sum_{k=0}^{\infty}\frac{\left(-1\right)^k\zeta\left(2k+2\right)\Gamma\left(2k+2\right)}{\pi^{2k}\left(2k+1\right)}\right)$$

$$\boxed{\int\limits_{0}^{\infty}\frac{\log\left(x\right)\arctan\left(x\right)}{e^{\pi x}-1}\,dx = \frac{1}{2\pi^2}\sum_{k=0}^{\infty}\frac{\left(-1\right)^k}{2k+1}\left(\frac{d}{dk}\left[\frac{\zeta\left(2k+2\right)\Gamma\left(2k+2\right)}{\pi^{2k}}\right]\right)}$$

$$\int\limits_{0}^{\infty}\frac{\log\left(x\right)\arctan\left(x\right)}{e^{\pi x}-1}\,dx = -\gamma\frac{1}{6}-\log\left(\pi\right)\frac{1}{6}+\frac{\zeta\left(2,1\right)}{\pi^2}-\left(-\gamma+1\right)\frac{1}{18}+\log\left(\pi\right)\frac{1}{18}+\dots$$

Daniel K
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