"Show that ∫0∞xcos⁡axsinhxdxπ24sech2(aπ2)" was answered by our community

Arthur Pratt

Answered question

2022-01-03

Show that

\int_{0}^{\infty} \frac{x \cos a x}{sinh x} d x \frac{π^{2}}{4} {sech}^{2} (\frac{a π}{2})

Answer & Explanation

puhnut1m

Beginner2022-01-04Added 33 answers

Heres

Chanell Sanborn

Beginner2022-01-05Added 41 answers

We have that,

\frac{1}{sinh x} = 2 \sum_{k = 2 n + 1, n \in N} \exp (- k x)

So,

\int_{0}^{\infty} \frac{\sin a x}{sinh x} d x = 2 \int_{0}^{\infty} \sin a x \sum_{k = 0}^{\infty} \exp (- (2 k + 1) x) d x

= 2 a \sum \frac{1}{a^{2} + {(2 k + 1)}^{2}} = π \coth π a - \frac{π}{2} \coth \frac{π a}{2}

Differentiating with respect to a we get the answer.

Vasquez

Expert2022-01-09Added 669 answers

$\begin{matrix} \int_{0}^{\infty} \frac{x \cos (a x)}{sinh x} d x = \frac{d}{d a} \int_{0}^{\infty} \frac{\sin (a x)}{sinh (x)} d x \\ 2 \frac{d}{d a} I \int_{0}^{\infty} \frac{e^{i a x} - 1}{e^{x} - e^{- x}} d x \\ = 2 \frac{d}{d a} I \int_{0}^{\infty} \frac{e^{- (1 - i a) x} - e^{- x}}{1 - e^{- 2 x}} d x \\ = \frac{d}{d a} I \int_{0}^{\infty} \frac{e^{- (1 / 2 - i a / 2) x} - e^{- x / 2}}{1 - e^{- 2 x}} d x \\ = \frac{d}{d a} I \int_{0}^{\infty} [\frac{e^{- x} - e^{- x / 2}}{1 - e^{- x}} d x - \int_{0}^{\infty} \frac{e^{- x} e^{- (1 / 2 - i a / 2) x}}{1 - e^{- x}} d x] \\ = \frac{d}{d a} I [ψ (\frac{1}{2}) - ψ (\frac{1}{2} - \frac{a}{2} i)] \\ = \frac{d}{d a} [\frac{- ψ (1 / 2) - a i / 2) + ψ (1 / 2 + a i / 2)}{2 i}] \\ = - \frac{i}{2} \frac{d}{d a} [π \cot (π [\frac{1}{2} - \frac{a}{2} i])] \\ = - \frac{π i i}{2} \frac{d}{d a} [\tan (\frac{π a}{2} i)] = \frac{π}{2} \frac{d}{d a} [\tanh (\frac{π a}{2})] \\ = \frac{π^{2}}{4} {sech}^{2} (\frac{π a}{2}) \end{matrix}$

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