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


In honour of Cleo:

$$
\tan^{-1}\Big(\frac{1}{x}\Big)
$$
