$\int_{-\infty}^{\infty} \frac{e^{mx}}{(ae^{nx}+b)}dx$

$\int_{-\infty}^{\infty} \frac{e^{mx}}{(ae^{nx}+b)}dx$

i tried substitution $e^x = y $ . But stuck at $\int_{0}^{\infty} y^{m-1}(a+by)^{-1} dy$.

Limits are giving me suspicion of beta function but I am failing to convert into in that form .

How I proceed ?

Solution 1:

Let's substitute $ \left\lbrace\begin{matrix}y=\frac{a}{b}\operatorname{e}^{nx}\\ \mathrm{d}x=\frac{\mathrm{d}y}{ny}\ \ \ \ \ \ \end{matrix}\right. $\begin{aligned}\int_{-\infty}^{+\infty}{\frac{\operatorname{e}^{mx}}{a\operatorname{e}^{nx}+b}\,\mathrm{d}x}=\frac{b^{\frac{m}{n}-1}}{na^{\frac{m}{n}}}\int_{0}^{+\infty}{\frac{y^{\frac{m}{n}-1}}{y+1}\,\mathrm{d}y}\end{aligned}

Now, substitute $ \left\lbrace\begin{matrix}u=\frac{1}{1+y}\\ \mathrm{d}y=-\frac{\mathrm{d}u}{u^{2}}\end{matrix}\right. $, we get : \begin{aligned}\int_{0}^{+\infty}{\frac{y^{\frac{m}{n}-1}}{1+y}\,\mathrm{d}y}=\int_{0}^{1}{u^{-\frac{m}{n}}\left(1-u\right)^{\frac{m}{n}-1}\,\mathrm{d}u}=\Gamma\left(1-\frac{m}{n}\right)\Gamma\left(\frac{m}{n}\right)=\frac{\pi}{\sin{\left(\frac{\pi m}{n}\right)}}\end{aligned}

Thus : $$ \int_{-\infty}^{+\infty}{\frac{\operatorname{e}^{mx}}{a\operatorname{e}^{nx}+b}\,\mathrm{d}x}=\frac{\pi b^{\frac{m}{n}-1}}{na^{\frac{m}{n}}\sin{\left(\frac{\pi m}{n}\right)}} $$