let-give-x-gt-0-and-S-x-0-sint-e-xt-1-dt-developp-S-at-form-of-series-

Question Number 28613 by abdo imad last updated on 27/Jan/18

l e t g i v e x > 0 a n d S (x) = \int_{0}^{\infty} \frac{s i n t}{e^{x t} - 1} d t .

d e v e l o p p S a t f o r m o f s e r i e s .

Commented by abdo imad last updated on 01/Feb/18

S (x) = \int_{0}^{\infty} \frac{e^{- x t} s i n t}{1 - e^{- x t}} d t = \int_{0}^{\infty} (\sum_{n = 0}^{\infty} e^{- n x t}) e^{- x t} s i n t d t

= \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- (n + 1) x t} s i n t d t = \sum_{n = 0}^{\infty} A_{n} (x)

A_{n} (x) = \int_{0}^{\infty} e^{- (n + 1) x t} s i n t d t = - I m (\int_{0}^{\infty} e^{- ((n + 1) x + i) t} d t) b u t

\int_{0}^{\infty} e^{- ((n + 1) x + i) t} d t = - \frac{1}{(n + 1) x + i} {[e^{- ((n + 1) x + i) t}]}_{0}^{+ \infty}

= \frac{1}{(n + 1) x + i} = \frac{(n + 1) x - i}{{(n + 1)}^{2} x^{2} + 1} = \frac{(n + 1) x}{1 + {(n + 1)}^{2} x^{2}} - \frac{i}{1 + {(n + 1)}^{2} x^{2}}

A_{n} (x) = \frac{1}{1 + {(n + 1)}^{2} x^{2}} \Rightarrow S (x) = \sum_{n = 0}^{\infty} \frac{1}{1 + {(n + 1)}^{2} x^{2}} w i t h x > 0