Can anyone help me prove this with the help of the Laplace transformation?
$\int_0^\infty \frac{w}{1+w^2} \sin wx dw=\frac{\pi}{2}e^{-x}$
where $x>0$
EDIT: So I was wondering if you could split up $\sin wx$ into $\frac{1}{2i}\big[e^{iwx}-e^{-iwx}\big]$
Then say
$\int_0^\infty e^{iwx}-e^{-iwx}=\int_{-\infty}^0 e^{-iwx}-\int_0^\infty e^{-iwx}=\int_{-\infty}^\infty -\operatorname{sgn}(x) e^{-iwx}$
And use this to calculate the above problem? (by using a Fourier transformation)
EDIT#2: ok, so if you define $f(t)=\operatorname{sgn}(x)e^{-|x|} \Rightarrow \hat{f}(w)=-2i\frac{w}{1+w^2}$
So if
$\int_0^\infty \frac{w}{1+w^2} \sin wx dw=\frac{1}{2i}\int_{-\infty}^\infty \frac{w}{1+w^2}e^{iwx}dw=\frac{\pi}{2} \frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(w)e^{iwx}dw=\frac{\pi}{2}f(x)$
Then for $x>0$ this gives
$\int_0^\infty \frac{w}{1+w^2} \sin wx dw=\frac{\pi}{2}e^{-x}$
:)