\( \int_{0}^{\infty}\frac{x}{\sqrt{e^x-1}}\,dx \)
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\( \int_{0}^{\infty}\frac{x}{\sqrt{e^x-1}}\,dx \)
Prove that: $$\int_{0}^{\infty}\frac{x}{\sqrt{e^x-1}}\,dx=2\pi \ln 2 $$ without complex analysis methods!
Imagination is much more important than knowledge.
Re: \( \int_{0}^{\infty}\frac{x}{\sqrt{e^x-1}}\,dx \)
Replied ex-member by aziiri:
Set \(e^x-1=t^2\) then : \
Now set \(t=\tan y\) to get : \ The latter integral is widely known to be \(\frac{-\pi\ln 2}{2}\), then the result is \(I=\pi \ln 4\).
Set \(e^x-1=t^2\) then : \
Now set \(t=\tan y\) to get : \ The latter integral is widely known to be \(\frac{-\pi\ln 2}{2}\), then the result is \(I=\pi \ln 4\).
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