Integrating $\int\nolimits_0^{\infty}{\frac{e^{-ax}-e^{-bx}}{x}\sin{mx} \, dx} \quad (a > 0 \, , b >0)$ 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.
 A: I don't have the theorems to justify all the tricks I am manipulating (mainly because it's late and I just thought I'd give you an idea) but here's how this goes : let 
$$
f(t) = \int_0^{\infty} \frac{e^{-atx} - e^{-btx}}x \sin(mx) \, dx
$$
Now clearly $f(0) = 0$ because when $t=0$ the integrand is $0$. The trick here is to evaluate
$$
f(1) = \int_0^1 f'(t) \, dt
$$
and to differentiate under the integral sign, as you suggested. So here it goes :
$$
\begin{gather*}
\begin{aligned}
f'(t) & = \frac{d}{d t} \int_0^{\infty} \frac{e^{-atx} - e^{-btx}}x \sin (mx) \, dx \\\
& = \int_0^{\infty} \frac{\partial}{\partial t} \left( \frac{e^{-atx} - e^{-btx}}x \sin (mx) \right) \, dx \\\
& = \int_0^{\infty} \frac{-ax e^{-atx} + bx e^{-btx}}x \sin(mx) \, dx \\\
& = \int_0^{\infty} (b e^{-btx} - a e^{-atx}) \sin(mx) \, dx. \\\
& = b \left( \frac{m}{m^2+(bt)^2} \right) - a \left( \frac{m}{m^2+(at)^2} \right). \\\
\end{aligned}
\end{gather*}
$$
Integrating this between $0$ and $1$ gives you the right answer. If you need clarification on some aspects I may edit the answer for you.
Hope that helps,
P.S. : I almost forgot you had a first question :P right now, your procedure is fine! Modulo some justification on your arguments (mine lacks arguments too, but the sketch is all there!) Good question, seriously. Most undergraduates I know don't even know about parametric differentiation to evaluate complicated integrals. Is there any course in which we encounter this? I'm a first year, so I just did basic calculus, analysis, algebra... and read some books in my leisure time. =) 
A: Concerning the first question ``Is this procedure ok?'', the key step, that is the equality 
$$
\int_0^\infty  {\bigg[\int_a^b {e^{ - xy} \sin (mx)\,dy\bigg]dx} }  = \int_a^b {\bigg[\int_0^\infty  {e^{ - xy} \sin (mx)} \,dx\bigg]dy} ,
$$
is justified by Fubini's theorem (see here).
Specifically, letting
$$
f(x,y) = e^{ - xy} \sin (mx),
$$
it suffices to show that
$$
\int_0^\infty  {\bigg[\int_a^b {|f(x,y)|\,dy\bigg]dx} }  < \infty \;\;{\rm or}\;\; \int_a^b {\bigg[\int_0^\infty  {|f(x,y)|} \,dx\bigg]dy} < \infty.
$$
Since $|f(x,y)| \leq e^{-xy}$, it thus suffices to show that 
$$
\int_0^\infty  {\bigg[\int_a^b {e^{-xy}\,dy\bigg]dx} }  < \infty \;\;{\rm or}\;\; \int_a^b {\bigg[\int_0^\infty  {e^{-xy}} \,dx\bigg]dy} < \infty.
$$
Both are very easy to show, the latter in particular:
$$
\int_a^b {\bigg[\int_0^\infty  {e^{-xy}} \,dx\bigg]dy} = \int_a^b {\frac{1}{y}\,dy}  < \infty .
$$
A: 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.
