Question Number 213759 by Spillover last updated on 15/Nov/24
Answered by MrGaster last updated on 06/Feb/25
$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\frac{{e}^{{nx}} −\mathrm{1}}{{e}^{{x}} −\mathrm{1}}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{{x}} −\mathrm{1}}{{e}^{{nx}} −\mathrm{1}}{dx} \\ $$$$\psi^{\left({m}\right)} \left({z}\right)=\frac{{d}^{{m}+\mathrm{1}} }{{dz}^{{m}+\mathrm{1}} }\mathrm{ln}\Gamma\left({z}\right) \\ $$$$\psi^{\left(\mathrm{0}\right)} \left({z}\right)=\frac{\Gamma'\left({z}\right)}{\Gamma\left({z}\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{{x}} −\mathrm{1}}{{e}^{{nx}} −\mathrm{1}}{dx}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\gamma+\psi^{\left(\mathrm{0}\right)} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$