TITLE: Integrating $\int\nolimits_0^{\infty}{\frac{e^{-ax}-e^{-bx}}{x}\sin{mx} \, dx} \quad (a > 0 \, , b >0)$
QUESTION [11 upvotes]: 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.

REPLY [3 votes]: The integral can also be evaluated as follows. First a change of variable $x \mapsto mx$ gives
$$
I = \int_0^\infty  {\frac{{e^{ - ax}  - e^{ - bx} }}{x}} \sin (mx)\,dx = \int_0^\infty  {\frac{{e^{ - (a/m)x}  - e^{ - (b/m)x} }}{x}\sin (x)\,dx} .
$$
Thus,
$$
I = \int_0^\infty  {e^{ - (a/m)x} \frac{{\sin (x)}}{x}dx}  - \int_0^\infty  {e^{ - (b/m)x} \frac{{\sin (x)}}{x}dx} .
$$
The Laplace transform of the sinc function ($\sin(x)/x$) is well known (*):
$$
\int_0^\infty  {e^{ - sx} \frac{{\sin (x)}}{x}} \,dx = \arctan \bigg(\frac{1}{s}\bigg), \;\; s > 0.
$$
Hence
$$
I = \arctan \bigg(\frac{m}{a}\bigg) - \arctan \bigg(\frac{m}{b}\bigg),
$$
or alternatively
$$
I = \bigg[\frac{\pi }{2} - \arctan \bigg(\frac{a}{m}\bigg)\bigg] - \bigg[\frac{\pi }{2} - \arctan \bigg(\frac{b}{m}\bigg)\bigg] = \arctan \bigg(\frac{b}{m}\bigg) - \arctan \bigg(\frac{a}{m}\bigg).
$$
(*): See here (starting from "Take an additional parameter $a$ to the defining integral...'') for an elementary proof.