I need to evaluate the following integral $$\int_{0}^{\infty} \frac{\sinh bx}{\sinh ax} dx \space \space \space , \space \space 0<b<a$$
Here is my attempt -
I can write $\sinh ax $ as $\frac{e^{ax} - e^{-ax}}{2}. $ So , substituting this in the integral I get, $$\int_{0}^{\infty} \frac{\sinh bx }{1 - e^{-2ax}} { 2e^{ax}} dx $$ Now using the following expansion $$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$ I can rewrite the above integral as $$\int_{0}^{\infty} (2 \sinh bx ) \space e^{ax} \sum_{n=0}^{\infty} e^{-2nax} \space dx$$ Now since the integrand is a non-negative measureable function so I can interchange the integral and summation position, rewriting the same integral as, $$\sum_{n=0}^{\infty} \int_{0}^{\infty} (2 \sinh bx ) \space e^{ax} e^{-2nax} \space dx$$ I am stuck at this point. I tried using the ILATE rule for definite integrals to solve this but it turns out that it is not getting me any answer. Do I need to make any kind of substitution here or any other approach is required?