The forum discussion focuses on the convergence of the improper integral involving hyperbolic functions, specifically the integral $$\int_0^{\infty}\frac{\sinh ax}{\sinh bx}\, dx$$ for real numbers $$a$$ and $$b$$ where $$b > |a|$$. The conclusion reached is that this integral evaluates to $$\frac{\pi}{2b}\tan\frac{\pi a}{2b}$$. The discussion highlights the mathematical rigor involved in proving this result, showcasing the relationship between hyperbolic functions and their integrals.
PREREQUISITESMathematicians, calculus students, and anyone interested in advanced integral calculus and the properties of hyperbolic functions.
Very nicely done, Sir! (Heidy)Random Variable said:$$ \begin{align} \int_{0}^{\infty} \frac{\sinh ax}{\sinh b x} \ dx &= 2 \int_{0}^{\infty} \frac{\sinh ax}{e^{bx}-e^{-b x}} \ dx \\ &= 2 \int_{0}^{\infty} \sinh ax \ \frac{e^{- b x}}{1-e^{-2 b x}} \ dx \\ &= 2 \int_{0}^{\infty} \sinh ax \sum_{n=0}^{\infty} e^{-(2n+1)b x} \ dx \\ &= 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} \sinh ax \ e^{-(2n+1) b x} \ dx \\ &= 2 \sum_{n=0}^{\infty} \frac{a}{(2n+1)^{2} b - t^{2}} \\ &= \frac{\pi}{2 b} \tan \left( \frac{\pi a}{2b}\right) \end{align} $$
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