Definite Integral problem: $\int_0^{\infty}\dfrac{e^{-sk}\sin (k x)}{k} \: dk$

Question

We're given : $\int_0^{\infty}e^{-sk}\sin (k x)\:dk$ = $\dfrac{x}{x^{2}+s^{2}}$

We need to evaluate : $\int_0^{\infty}\dfrac{e^{-sk}\sin (k x)}{k} \: dk$

I tried as follows : $\int_0^{\infty}\dfrac{e^{-sk}\sin (k x)}{k} \: dk$ = $\int_0^{\infty}\dfrac{1}{k} . \: (e^{-sk}\sin (k x))\: dk$

=> $\dfrac{1}{k} (\int e^{-sk}\sin (k x)\: dk)\: +\int[\dfrac{1}{k^2}(\int e^{-sk}\sin (k x)\: dk)]\: dk$

But that comes out to be zero. Can anyone tell what am I doing wrong ?

Answer 1

Assume $x>0,\,s>0$.

Then by differentiating the following identity with respect to $s$, $$ f(s)=\int_0^\infty {\frac{e^{−sk}} k}\sin(kx)\,dk $$ one obtains $$ f'(s)=-\int_0^\infty e^{−sk}\sin(kx)\,dk=-\frac{x}{x^2+ s^2} $$ which gives $$ f(s)=-\arctan \left( \frac{s}x\right)+C. $$ Observing that, as $s \to \infty$, $f(s) \to 0$, we then obtain $C=\dfrac\pi2$. Thus

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

The case $x<0$ is obtained with $\displaystyle \int_0^\infty {\frac{e^{−sk}} k}\sin(-kx)\,dk=-\int_0^\infty {\frac{e^{−sk}} k}\sin(kx)\,dk.$

Answer 2

One easy way to do it to notice that $$\frac{\partial}{\partial s}\int_0^{\infty}\dfrac{e^{-sk}sinkx}{k}=-\int_0^{\infty}e^{-sk}sinkx=-\dfrac{x}{x^{2}+s^{2}}$$ Then you just have to revert the derivative with respect to "s" $$-\int ds \dfrac{x}{x^{2}+s^{2}}=-\arctan(s/x)$$