I have this integral $$\int\nolimits_0^{\infty}{\frac{e^{-ax}-e^{-bx}}{x}\sin{mx} \, dx} \quad (a > 0 \, , b >0)$$
What I did was this
$$ \begin{align}
\int_0^{\infty}{\frac{e^{-ax}-e^{-bx}}{x}\sin{(mx)} \, dx} &= \int_0^{\infty}{\left[\int_a^b{e^{-xy} \, dy}\right]\sin{(mx)} \, dx}\\
&= \int_a^b{\left[\int_0^{\infty}{e^{-xy}\sin{(mx)} \, dx}\right] \, dy}\\
&= \int_a^b{\frac{m}{m^2+y^2} \, dy}\\
&= \tan^{-1}\left(\frac{b}{m}\right) - \tan^{-1}\left(\frac{a}{m}\right)
\end{align}$$
So my first question is. Is this procedure ok?
Also the book suggests to use parametric differentiation to solve this, but I don't see how to apply it in here, so my second question would be. How can I use parametric differentiation to solve this integral?
Any help is appreciated, thanks.