let-A-p-0-sin-px-e-x-1-dx-with-p-gt-0-1-give-A-p-at-form-of-serie-2-give-A-1-at-form-of-serie-

Question Number 42191 by maxmathsup by imad last updated on 19/Aug/18

l e t A_{p} = \int_{0}^{\infty} \frac{s i n (p x)}{e^{x} - 1} d x w i t h p > 0

1) g i v e A_{p} a t f o r m o f s e r i e

2) g i v e A_{1} a t f o r m o f s e r i e .

Commented by maxmathsup by imad last updated on 20/Aug/18

1) w e h a v e A_{p} = \int_{0}^{\infty} \frac{e^{- x} s i n (p x)}{1 - e^{- x}} d x = \int_{0}^{\infty} (\sum_{n = 0}^{\infty} e^{- n x}) e^{- x} s i n (p x)) d x

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

= \sum_{n = 0}^{\infty} \frac{1}{n + 1} \int_{0}^{\infty} e^{- t} s i n (\frac{p}{n + 1} t) d t l e t c a l c u l a t e

I_{λ} = \int_{0}^{\infty} e^{- t} s i n (λ t) d t \Rightarrow I_{λ} = I m (\int_{0}^{\infty} e^{- t + i λ t} d t) b u t

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

= \frac{1 + i λ}{1 + λ^{2}} \Rightarrow I_{λ} = \frac{λ}{1 + λ^{2}} \Rightarrow A_{p} = \sum_{n = 0}^{\infty} \frac{1}{n + 1} (\frac{\frac{p}{n + 1}}{1 + {(\frac{p}{n + 1})}^{2}})

= \sum_{n = 0}^{\infty} \frac{1}{n + 1} (\frac{p}{n + 1} \frac{{(n + 1)}^{2}}{{(n + 1)}^{2} + p^{2}}) = \sum_{n = 0}^{\infty} \frac{p}{{(n + 1)}^{2} + p^{2}} \Rightarrow

A_{p} = \sum_{n = 0}^{\infty} \frac{p}{{(n + 1)}^{2} + p^{2}}

2) A_{1} = \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2} + 1} = \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1} .