prove-that-o-1-sint-e-t-1-n-o-1-n-2-1-

Question Number 184992 by SANOGO last updated on 15/Jan/23

p r o v e t h a t :

\int_{o}^{1} \frac{s i n t}{e^{t} - 1} = \sum_{n = o} \frac{1}{n^{2} + 1}

Answered by ARUNG_Brandon_MBU last updated on 15/Jan/23

\int_{0}^{\infty} \frac{\sin t}{e^{t} - 1} d t = \int_{0}^{\infty} \frac{e^{- t} \sin t}{1 - e^{- t}} d t

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

= \underset{n = 0}{\sum^{\infty}} {[\frac{e^{- (n + 1) t}}{{(n + 1)}^{2} + 1} (- (n + 1) \sin t - \cos t)]}_{0}^{\infty}

= \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{2} + 1} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + 1} = \frac{π}{2} \coth π - \frac{1}{2}

Commented by SANOGO last updated on 15/Jan/23

m e r c i b i e n

Facebook Tweet Pin