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Question Number 213759 by Spillover last updated on 15/Nov/24

Answered by MrGaster last updated on 06/Feb/25

$$={\int }_0^{\mathrm{\infty }}\frac{1}{\frac{e^{nx}-1}{e^x-1}}dx={\int }_0^{\mathrm{\infty }}\frac{e^x-1}{e^{nx}-1}dx$$$${\psi }^{\left(m\right)}\left(z\right)=\frac{d^{m+1}}{{dz}^{m+1}}\ln \mathrm{\Gamma }\left(z\right)$$$${\psi }^{\left(0\right)}\left(z\right)=\frac{{\mathrm{\Gamma }}^{\prime }\left(z\right)}{\mathrm{\Gamma }\left(z\right)}$$$${\int }_0^{\mathrm{\infty }}\frac{e^x-1}{e^{nx}-1}dx=-\frac{1}{2}(\gamma +{\psi }^{\left(0\right)}(1-\frac{1}{2}))$$

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