# Math Help - Lebesgue Integration

1. ## Lebesgue Integration

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

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

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

2. Originally Posted by 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

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

Now, expanding the numerator as a Taylor series we get $\frac{\sum_{n=0}^{\infty} x^{2n+1}}{(2n+1)! sinh (bx)}$, and as each $f_n=\frac{x^{2n+1}}{(2n+1)! sinh (bx)}$ is non-negative, we will be able to apply the MCT... but how do you integrate the $f_n$'s? And where do you go from here? Any help is appreciated.
First, note that $\int_{-\infty}^{\infty} \frac{\sinh(ax)}{\sinh(bx)} dx = 2\int_0^{\infty} \frac{\sinh(ax)}{\sinh(bx)} dx$ (because the integrand is an even function). Then

\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 $(1-t)^{-1}$).

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