Convergence of Improper Integral with Hyperbolic Functions?

[DreamWeaver]

For \(\displaystyle a\, ,b\in\mathbb{R}\,\) and \(\displaystyle b>|a|\,\) show that:

\(\displaystyle \int_0^{\infty}\frac{\sinh ax}{\sinh bx}\, dx = \frac{\pi}{2b}\tan\frac{\pi a}{2b}\)

 

[polygamma]

Spoiler

$$ \begin{align} \int_{0}^{\infty} \frac{\sinh ax}{\sinh b x} \ dx &= 2 \int_{0}^{\infty} \frac{\sinh ax}{e^{bx}-e^{-b x}} \ dx \\ &= 2 \int_{0}^{\infty} \sinh ax \ \frac{e^{- b x}}{1-e^{-2 b x}} \ dx \\ &= 2 \int_{0}^{\infty} \sinh ax \sum_{n=0}^{\infty} e^{-(2n+1)b x} \ dx \\ &= 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} \sinh ax \ e^{-(2n+1) b x} \ dx \\ &= 2 \sum_{n=0}^{\infty} \frac{a}{(2n+1)^{2} b - t^{2}} \\ &= \frac{\pi}{2 b} \tan \left( \frac{\pi a}{2b}\right) \end{align} $$

where I used the partial fractions expansion of $\tan z$, that is,

$$ \tan z = \sum_{n=0}^{\infty} \frac{8z}{(2n+1)^{2}\pi^{2}-4z^{2}} $$

Partial fractions in complex analysis - Wikipedia, the free encyclopedia

 

[DreamWeaver]

Spoiler

$$ \begin{align} \int_{0}^{\infty} \frac{\sinh ax}{\sinh b x} \ dx &= 2 \int_{0}^{\infty} \frac{\sinh ax}{e^{bx}-e^{-b x}} \ dx \\ &= 2 \int_{0}^{\infty} \sinh ax \ \frac{e^{- b x}}{1-e^{-2 b x}} \ dx \\ &= 2 \int_{0}^{\infty} \sinh ax \sum_{n=0}^{\infty} e^{-(2n+1)b x} \ dx \\ &= 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} \sinh ax \ e^{-(2n+1) b x} \ dx \\ &= 2 \sum_{n=0}^{\infty} \frac{a}{(2n+1)^{2} b - t^{2}} \\ &= \frac{\pi}{2 b} \tan \left( \frac{\pi a}{2b}\right) \end{align} $$

where I used the partial fractions expansion of $\tan z$, that is,

$$ \tan z = \sum_{n=0}^{\infty} \frac{8z}{(2n+1)^{2}\pi^{2}-4z^{2}} $$

Partial fractions in complex analysis - Wikipedia, the free encyclopedia

Very nicely done, Sir! (Heidy)