Evaluate the improper integral $\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk$. $$\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk$$
I've tried hard for this but of no use.I've applied integration by parts by which I get $$\int_0^\infty \exp(-sk)\sin(kx)\,dk=\frac{x}{x^2+ s^2}.$$ But, I'm not getting how to adjust $\displaystyle\frac1k$.
Please Help!
 A: We assume $x>0$ and $s>0$. 
Then by differentiating the following identity with respect to $s$,
$$
f(s)=\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk
$$ one may write
$$
f'(s)=-\int_0^\infty \exp(−sk)\sin(kx)\,dk=-\frac{x}{x^2+ s^2}
$$ giving
$$
f(s)=-\arctan \left( \frac{s}x\right)+C.
$$ Observing that, as $s \to \infty$, $f(s) \to 0$, we then obtain $C=\dfrac\pi2$. Thus

$$
\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk=\frac\pi2-\arctan \left( \frac{s}x\right), \qquad s>0,\,x>0.
$$

A: @ChristopherCarlHeckman : it is a Laplace Transform (see remark at the end) 
Precisely, it is the Laplace Transform (LT) of a very important function, the cardinal sine (sinc).
This LT can be found in most LT tables as
$$\int_0^\infty {\sin(k)\over k}\exp(−sk)\,dk=\frac\pi2-\arctan(s)$$
from which it is easy to deduce by an elementary change of variables:
$$\int_0^\infty {\sin(kx)\over k}\exp(−sk)\,dk=\frac\pi2-\arctan \left( \frac{s}x\right)$$
Remark 1: This result can also be written $arccot\left( \frac{s}x\right)$.
Remark 2: In the reference dlmf.nist.gov/1.14#vii  given by ChristopherCarlHeckman the result they give is not the same... I don't understand.
