find the integral $\int ^\infty_0e^{-xu}\frac{\sin u}{u}du$ Find the integral $$\int ^\infty_0e^{-xu}\frac{\sin u}{u}du$$
my idea
$L\left(\frac{1}{t}f(t)\right)=\int_{s}^{\infty}L(f(t))dt$
$L(\sin bt)=t=\frac{1}{1+s^2}$
$\int ^\infty_0e^{-xu}\frac{\sin u}{u}du$ 
??  how  we processed
 A: 
In honour of Cleo:

$$
\tan^{-1}\Big(\frac{1}{x}\Big)
$$
A: In honor of Feynman:
Let us define (with $x>0$)
$$I(x) = \int_0^\infty e^{-x u} \frac{\sin u}{u} du\;.$$
Differentiating (under the integral sign), we obtain
$$I'(x) = -\int_0^\infty e^{-x u} \sin u\;du = -\operatorname{Im}\int_0^\infty e^{-xu +i u}\;du =\operatorname{Im} \frac1{i-x} = - \frac{1}{1+x^2}\;. $$
We thus obtain
$$I(x) = -\arctan(x) + c \;.$$
It is easy to see that $I(u) \to 0$ for $x \to \infty$, this sets $c=\pi/2$ and thus
$$ I(x) = \frac{\pi}{2} -\arctan(x) = \arctan(1/x)\;.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
& \left.\int_{0}^{\infty}\expo{-xu}\,{\sin\pars{u} \over u}\,\dd u
\,\right\vert_{\ x\ >\ 0} =
\int_{0}^{\infty}\expo{-xu}\bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic ku}
\dd k}\dd u
\\[5mm] = & \
{1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}
\expo{-\pars{x - \ic k}u}\,\,\,\dd u\,\dd k
 =
{1 \over 2}\int_{-1}^{1}{\dd k \over x - \ic k}
\\[5mm] = & \
x\int_{0}^{1}{\dd k \over k^{2} + x^{2}} =
\int_{0}^{1/x}{\dd k \over k^{2} + 1} =
\bbx{\arctan\pars{1 \over x}}
\end{align}
A: Sure your idea works: Let $f(t)=\sin(t)$, then
\begin{align}
\int_0^{\infty}e^{-st} \frac{\sin(t)}{t}
&= \mathcal{L}\left\{\tfrac{1}{t} f(t)\right\}(s)
= \int_s^{\infty} \mathcal{L}\left\{f(t)\right\}(s')\;ds'\\
&= \int_s^{\infty} \frac{ds'}{1+s'^2}
= \left[\tan^{-1}(s')\right]_s^{\infty}\\
&= \tfrac{\pi}{2} - \tan^{-1}(s)
\end{align}
