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I was self studying integral. I meet a difficult problem here: $$\int_{0}^{\infty }{{{x}^{n}}\frac{\sinh ax}{\cosh bx}}\text{d}x=\frac{\pi }{2b}\cdot \frac{{{\text{d}}^{n}}}{\text{d}{{a}^{n}}}\tan \left( \frac{a\pi }{2b} \right)\ \ \ \ \ \left| a \right|<b$$ Anyone know how to deal with it?

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up vote 7 down vote accepted

$$\dfrac{\sinh(ax)}{\cosh(bx)} = \dfrac{e^{ax} - e^{-ax}}{e^{bx} + e^{-bx}} = \dfrac{e^{(a-b)x} - e^{-(a+b)x}}{1 + e^{-2bx}} = \left(e^{(a-b)x} - e^{-(a+b)x}\right) \left(\sum_{k=0}^{\infty} (-1)^k e^{-2kbx} \right)$$ This simplifies to $$\sum_{k=0}^{\infty} (-1)^k \left(e^{(a-2kb-b)x} - e^{-(a+2kb+b)x}\right)$$ Hence, \begin{align} I & = \int_0^{\infty} x^n \dfrac{\sinh(ax)}{\cosh(bx)} dx = \sum_{k=0}^{\infty}(-1)^k \left(\int_0^{\infty} x^n e^{(a-2kb-b)x}dx - \int_0^{\infty} x^n e^{-(a+2kb+b)x} dx\right)\\ & = \sum_{k=0}^{\infty} (-1)^k \left(\dfrac{\Gamma(n+1)}{(b+2kb-a)^{n+1}} - \dfrac{\Gamma(n+1)}{(a+2kb+b)^{n+1}}\right)\\ & = n! \left(\sum_{k=0}^{\infty} (-1)^k \left(\dfrac1{(b+2kb-a)^{n+1}} - \dfrac1{(a+2kb+b)^{n+1}}\right)\right) \end{align} Now recall that $$\tan(\pi x/2) = \cot(\pi/2-\pi x/2) = \cot(\pi(1-x)/2)$$ Further, we have $$\cot(\pi y) = \dfrac1{\pi} \lim_{N \to \infty}\left(\sum_{k=-N}^N \dfrac1{y+k}\right)$$ Hence, $$\tan(\pi x/2) = \dfrac1{\pi} \lim_{N \to \infty}\left(\sum_{k=-N}^N \dfrac2{1-x+2k}\right)$$ Plugging in $x=a/b$ and taking derivative term by term you should be able to obtain your desired result.

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Wow! Thx so much! Beatiful proof : ) – Ryan Jan 27 '13 at 10:18

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