How do I calculate $u(w)=\int_0^\infty \frac{1-\cos(wt)}{t}\,e^{-t}\,dt$? How do I calculate 
$$u(w)=\int_0^\infty \frac{1-\cos(wt)}{t}\,e^{-t}\,dt$$
I tried to do it, I use partial integration but I get lost. Is there any nice simple way to calculate it?

$\int \frac {(1−\cos\omega t)e^{−t}}{t}dt$ cannot be evaluated into elementary functions.  You need to get tricky.
$F(s) = \int_0^{\infty} \frac {(1−\cos\omega t)e^{−st}}{t}dt$
and if we can find $F(1)$ we are done. 
$\frac {dF}{ds} = $$\int_0^{\infty} -(1−\cos\omega t)e^{−st}dt\\
 \frac 1s e^{-st} + \frac {-s\cos\omega te^{−st} + \omega \sin\omega t e^{−st}}{s^2+\omega^2} |_0^\infty\\
 -\frac 1s + \frac {s}{s^2+\omega^2}$
$F(\infty) - F(1) = \int_1^{\infty}-\frac 1s + \frac {s}{s^2+\omega^2}\\
F(\infty) - F(1) = -\ln s + \frac 12 (s^2 + \omega^2)|_1^{\infty}$
I am going to leave it to you to prove to yourself that
$\lim_\limits{s\to\infty} \ln s + \frac 12 (s^2 + \omega^2) = 0$
$F(\infty) - F(1) = -\frac 12 \ln(1 + \omega^2)$
Going back to the definition of $F$, it should be clear that $F(\infty) = 0$
$F(1) = \frac 12 \ln(1 + \omega^2)$
