Definite Integral problem: $\int_0^{\infty}\dfrac{e^{-sk}\sin (k x)}{k} \: dk$ 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 ?
 A: 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 $\displaystyle \int_0^\infty {\frac{e^{−sk}} k}\sin(-kx)\,dk=-\int_0^\infty {\frac{e^{−sk}} k}\sin(kx)\,dk.$
A: 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"
$$-\int ds \dfrac{x}{x^{2}+s^{2}}=-\arctan(s/x)$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\int_{0}^{\infty}{\expo{-sk}\sin\pars{kx} \over k}\,\dd k} & =
x\int_{0}^{\infty}\expo{-sk}\,{\sin\pars{kx} \over kx}\,\dd k =
x\int_{0}^{\infty}\expo{-sk}\,\half\int_{-1}^{1}\expo{\ic kxt}\,\dd t\,\dd k
\\[4mm] & =
\half\,x\int_{-1}^{1}\int_{0}^{\infty}\expo{\pars{-s + \ic xt}k}
\,\,\,\dd k\,\dd t =
\half\,x\int_{-1}^{1}{\dd t \over s - \ic xt}
\\[4mm] & =
\half\,x\int_{-1}^{1}{s + \ic xt \over \pars{xt}^{2} + s^{2}}\,\dd t =
x\int_{0}^{1}{s \over \pars{xt}^{2} + s^{2}}\,\dd t =
\int_{0}^{1}{1 \over \pars{xt/s}^{2} + 1}\,{x\,\dd t \over s}
\\[4mm] & =
\int_{0}^{x/s}{\dd t \over t^{2} + 1} =
\color{#f00}{\arctan\pars{x \over s}}
\end{align}
