nice-calculus-find-0-xe-x-coth-x-2-dx-

Question Number 136225 by mnjuly1970 last updated on 19/Mar/21

\dots . . n i c e c a l c u l u s

f i n d :: Ω = \int_{0}^{\infty} {x e}^{- x} c o t h (\frac{x}{2}) d x = ?

Answered by Dwaipayan Shikari last updated on 19/Mar/21

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

= \underset{n = 1}{\sum^{\infty}} \int_{0}^{\infty} {x e}^{- n x} (1 + e^{- x}) d x

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2}} \int_{0}^{\infty} {u e}^{- u} + \underset{n = 1}{\sum^{\infty}} \frac{1}{{(n + 1)}^{2}} \int_{0}^{\infty} {u e}^{- u} d u

= \frac{π^{2}}{6} + \frac{π^{2}}{6} - 1 = \frac{π^{2}}{3} - 1

Commented by mnjuly1970 last updated on 19/Mar/21

g r a t e f u l a n d t h a n k y o u \dots