nice-calculus-evaluate-0-sin-x-e-x-1-dx-

Question Number 130693 by mnjuly1970 last updated on 28/Jan/21

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

e v a l u a t e ::

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

\dots \dots \dots \dots \dots \dots . .

Answered by mnjuly1970 last updated on 28/Jan/21

\int_{0}^{\infty} \frac{s i n (x)}{e^{x} - 1} d x = \frac{1}{2 i} \int_{0}^{\infty} \frac{e^{i x} - e^{- i x}}{e^{x} - 1} d x

= \frac{1}{2 i} \int_{0}^{\infty} \frac{e^{i x - x} - e^{- i x - x}}{1 - e^{- x}} d x

= \frac{1}{2 i} \int_{0}^{\infty} {\underset{n = 0}{\sum^{\infty}} e^{i x - x - n x} - e^{- i x - x - n x}} d x

= \frac{1}{2 i} \underset{n = 0}{\sum^{\infty}} {\frac{e^{x (i - 1 - n)}}{i - n - 1} + \frac{e^{- x (i + 1 + n)}}{i + 1 + n}}_{0}^{\infty}

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

= \frac{- 1}{2 i} \underset{n = 1}{\sum^{\infty}} \frac{1}{i - n} + \frac{1}{i + n} = \frac{- 1}{2 i} \underset{n = 1}{\sum^{\infty}} \frac{2 i}{i^{2} - n^{2}}

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + 1} \overset{⟨ u p s i l o n f u n c t i o n ⟩}{=} \frac{π c o t h (π) - 1}{3}

Answered by Dwaipayan Shikari last updated on 28/Jan/21

\frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} \int_{0}^{\infty} e^{- n x + i x} - e^{- n x - i x} d x

= \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} + 1} = \frac{1}{2} (π c o t h (π) - 1)

Answered by mathmax by abdo last updated on 28/Jan/21

Φ = \int_{0}^{\infty} \frac{sinx}{e^{x} - 1} dx \Rightarrow Φ = \int_{0}^{\infty} \frac{e^{- x} sinx}{1 - e^{- x}} dx = \int_{0}^{\infty} e^{- x} sinx \sum_{n = 0}^{\infty} e^{- nx} dx

= \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- (n + 1) x} sinx dx = \sum_{n = 0}^{\infty} u_{n}

u_{n} = Im (\int_{0}^{\infty} e^{- (n + 1) x} e^{ix} dx) but

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

= \frac{1}{n + 1 - i} = \frac{n + 1 + i}{{(n + 1)}^{2} + 1} \Rightarrow u_{n} = \frac{1}{{(n + 1)}^{2} + 1} \Rightarrow Φ = \sum_{n = 0}^{\infty} \frac{1}{{(n + 1)}^{2} + 1}

= \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1}, the value of this serie is known see the platform