Evaluate the improper integral $\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk$.

Question

$$\int_0^\infty {\exp(−sk)\over k}\sin(kx)\,dk$$

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!

Answer 1

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

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

Answer 2

@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.