Question-213759

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

Answered by MrGaster last updated on 06/Feb/25

= \int_{0}^{\infty} \frac{1}{\frac{e^{n x} - 1}{e^{x} - 1}} d x = \int_{0}^{\infty} \frac{e^{x} - 1}{e^{n x} - 1} d x

ψ^{(m)} (z) = \frac{d^{m + 1}}{{d z}^{m + 1}} \ln Γ (z)

ψ^{(0)} (z) = \frac{Γ^{'} (z)}{Γ (z)}

\int_{0}^{\infty} \frac{e^{x} - 1}{e^{n x} - 1} d x = - \frac{1}{2} (γ + ψ^{(0)} (1 - \frac{1}{2}))

Facebook Tweet Pin