prove-that-0-sin-x-sinh-x-dx-pi-2-tanh-pi-2-

Question Number 173424 by mnjuly1970 last updated on 11/Jul/22

prove that

\int_{0}^{\infty} \frac{\sin (x)}{\sinh (x)} dx = \frac{π}{2} \tanh (\frac{π}{2})

Answered by Mathspace last updated on 11/Jul/22

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

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

= 2 \int_{0}^{\infty} e^{- x} s i n x \sum_{n = 0}^{\infty} e^{- 2 n x} d x

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

b u t \int_{0}^{\infty} e^{- (2 n + 1) x} s i n x d x

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

a n d \int_{0}^{\infty} e^{- (n + 1) + i) x} d x

{= \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

I m (\int_{0}^{\infty} \dots .) = \frac{1}{{(n + 1)}^{2} + 1} \Rightarrow

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

= 2 \sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1}

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

= \frac{Ψ (i) - Ψ (- i)}{2 i} a f t e r w e u s e

Ψ (z) - Ψ (1 - z) = π c o t a n (π z)

o r w e c a n d e v e l o p p c o s (α x)

a t f o u r i e r s e r i e t o f i n d t h e v a l u e \dots

Commented by mnjuly1970 last updated on 11/Jul/22

grateful sir

Commented by Tawa11 last updated on 13/Jul/22

Great sir

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