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Question Number 198695 by universe last updated on 23/Oct/23

\int_{0}^{\infty} \frac{e^{at} - e^{- at}}{e^{π t} - e^{- π t}} dt = ? ?

Answered by witcher3 last updated on 25/Oct/23

existe \Leftrightarrow a \in] - π, π [

f (a) = \int_{0}^{\infty} \frac{e^{at} - e^{- at}}{e^{π t} - e^{- π t}} dt, f (- a) = - f (a)

\forall t \in [0, a [

\int_{0}^{\infty} \frac{e^{at} - e^{- at}}{e^{π t} - e^{- π t}} dt = \int_{0}^{\infty} (e^{at} - e^{- at}) e^{- π t} Σ e^{- 2 n π t} dt

= \sum_{n ⩾ 0} \int_{0}^{\infty} e^{- (π - a + 2 n π) t} - e^{- (a + π + 2 n π) t} dt

= \sum_{n ⩾ 0} (\frac{1}{π - a + 2 n π} - \frac{1}{a + π + 2 n π})

= \frac{1}{2 π} \sum_{n ⩾ 0} (\frac{1}{n + \frac{1}{2} - \frac{a}{2 π}} - \frac{1}{n + \frac{1}{2} + \frac{a}{2 π}})

= \frac{1}{2 π} \sum_{n ⩾ 0} (\frac{1}{n + 1} - \frac{1}{n + \frac{1}{2} + \frac{a}{2 π}} - (\frac{1}{n + 1} - \frac{1}{n + \frac{1}{2} - \frac{a}{2 π}}))

= \frac{1}{2 π} \sum_{n ⩾ 0} \frac{1}{n + 1} - \frac{1}{n + \frac{1}{2} + \frac{a}{2 π}} - \frac{1}{2 π} \sum_{n ⩾ 0} \frac{1}{n + 1} - \frac{1}{n + \frac{1}{2} - \frac{a}{2 π}}

= \frac{1}{2 π} (Ψ (\frac{1}{2} + \frac{a}{2 π}) - Ψ (\frac{1}{2} - \frac{a}{2 π}))

= \frac{1}{2 π} (Ψ (1 - (\frac{1}{2} - \frac{a}{2 π})) - Ψ (\frac{1}{2} - \frac{a}{2 π}))

= \frac{1}{2 π} (. π \cot (π (\frac{1}{2} - \frac{a}{2 π})) = \frac{\tan (\frac{a}{2})}{2}

\int_{0}^{\infty} \frac{e^{ax} - e^{- ax}}{e^{π x} - e^{- π x}} dx = \frac{\tan (\frac{a}{2})}{2}

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