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**Mimi89** Hi all!

Could you please help me with the following (we're allowed to use the MCT and DCT):

Justifying your calculations, show that if 0<a<b then

$\displaystyle \int_{-\infty}^{\infty} \frac{sinh(ax)}{sinh(bx)} dx = 4a ( \frac{1}{b^2-a^2} + \frac{1}{9b^2 - a^2} +\frac{1}{25b^2 - a^2} + ... )$

Now, expanding the numerator as a Taylor series we get $\displaystyle \frac{\sum_{n=0}^{\infty} x^{2n+1}}{(2n+1)! sinh (bx)} $, and as each $\displaystyle f_n=\frac{x^{2n+1}}{(2n+1)! sinh (bx)}$ is non-negative, we will be able to apply the MCT... but how do you integrate the $\displaystyle f_n $'s? And where do you go from here? Any help is appreciated.

First, note that$\displaystyle \int_{-\infty}^{\infty} \frac{\sinh(ax)}{\sinh(bx)} dx = 2\int_0^{\infty} \frac{\sinh(ax)}{\sinh(bx)} dx$ (because the integrand is an even function). Then

$\displaystyle \begin{aligned}\int_0^{\infty} \frac{\sinh(ax)}{\sinh(bx)} dx = \int_0^{\infty} \frac{e^{ax}-e^{-ax}}{e^{bx}-e^{-bx}} dx &= \int_0^{\infty} \bigl(e^{(a-b)x}-e^{(-a-b)x}\bigr)\bigl(1-e^{-2bx}\bigr)^{-1}\, dx \\ &= \int_0^{\infty} \sum_{k=0}^\infty\bigl(e^{(a-(2k+1)b)x}-e^{(-a-(2k+1)b)x}\bigr) dx,\end{aligned}$

(using the binomial expansion for $\displaystyle (1-t)^{-1}$).

Now over to you. Use your Lebesgue convergence theorems to justify integrating the series term by term.