# Lebesgue Integration

#### Mimi89

Hi all!

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.

#### Opalg

MHF Hall of Honor
Hi all!

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.

• Mimi89