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?
 A: For any $w\in\mathbb{R}$, the identity
$$ \int_{0}^{+\infty}\frac{1-\cos(wt)}{t}e^{-t}\,dt = \frac{1}{2}\log(1+w^2) \tag{1} $$
follows from the complex version of Frullani's theorem, hence
$$ \frac{d}{dw}\int_{0}^{+\infty}\frac{1-\cos(wt)}{t}e^{-t}\,dt = \frac{w}{1+w^2}.\tag{2} $$
You may prove the same by checking first that, by the dominated convergence theorem, we may apply differentiation under the integral sign, leaving us with
$$ \frac{d}{dw}\int_{0}^{+\infty}\frac{1-\cos(wt)}{t}e^{-t}\,dt =\int_{0}^{+\infty}t \sin(wt) e^{-t}\,dt \tag{3}$$
where the last integral is easy to compute by integration by parts.
A: $\int \frac {(1−\cos\omega t)e^{−t}}{t}dt$ 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} = $$\int_0^{\infty} -(1−\cos\omega t)e^{−st}dt\\
 \frac 1s e^{-st} + \frac {-s\cos\omega te^{−st} + \omega \sin\omega t e^{−st}}{s^2+\omega^2} |_0^\infty\\
 -\frac 1s + \frac {s}{s^2+\omega^2}$
$F(\infty) - F(1) = \int_1^{\infty}-\frac 1s + \frac {s}{s^2+\omega^2}\\
F(\infty) - F(1) = -\ln s + \frac 12 (s^2 + \omega^2)|_1^{\infty}$
I am going to leave it to you to prove to yourself that
$\lim_\limits{s\to\infty} \ln s + \frac 12 (s^2 + \omega^2) = 0$
$F(\infty) - F(1) = -\frac 12 \ln(1 + \omega^2)$
Going back to the definition of $F$, it should be clear that $F(\infty) = 0$
$F(1) = \frac 12 \ln(1 + \omega^2)$
A: 
Here is an approach that does not use Feynman's trick or the Generalized Frullani's Theorem, but rather uses integration by parts ("IBP") along with the well-known integral (See the NOTE at the end of THIS ANSWER$$\int_0^\infty \log(t)e^{-t}\,dt=-\gamma\tag1$$where $\gamma$ is the Euler-Mascheroni constant.  It is easy to see from $(1)$ that for $\text{Re}(s)>0$ $$\int_0^\infty \log(t)e^{-st}\,dt=-\frac{\gamma-\log(s)}s\tag2$$


First, by enforcing the substitution $x= t/w$ reveals that 
$$\int_0^\infty \frac{1-\cos(wx)}{x}e^{-x}\,dx=\int_0^\infty \frac{1-\cos(t)}{t}e^{-(1/w)t}\,dt$$

Letting $s=1/w$, we have  
$$\begin{align}
\int_0^{\infty} \left(\frac{1- \cos t}{t}\right)e^{-st}dt&=\int_0^{\infty} (1- \cos t)\left(\frac{d\log(t)}{dt}\right)e^{-st}dt\\\\
&=\int_0^{\infty} \left(e^{-st}- \frac12 e^{-(s+i)t}-\frac12 e^{-(s-i)t}\right)\left(\frac{d\log(t)}{dt}\right)\,dt\\\\
&\overbrace{=}^{IBP}-\int_0^{\infty} \log(t)\frac{d}{dt}\left(e^{-st}- \frac12 e^{-(s+i)t}-\frac12 e^{-(s-i)t}\right) dt\\\\
&=s\int_0^{\infty} \log(t)e^{-st}dt\\\\
&-\frac12(s+i)\int_0^{\infty} \log(t)e^{-(s+i)t}dt\\\\
&-\frac12(s-i)\int_0^{\infty} \log(t)e^{-(s-i)t}dt\\\\
&\overbrace{=}^{\text{Using} (2)}s\left(\frac{-\gamma-\log(s)}{s}\right)\\\\
&-\frac12 (s+i)\left(\frac{-\gamma-\log(s+i)}{s+i}\right)\\\\&-\frac12 (s-i)\left(\frac{-\gamma-\log(s-i)}{s-i}\right)\\\\
&=\frac12 \log\left(\frac{s^2+1}{s^2}\right)
\end{align}$$

Finally, replacing $s$ with $1/w$ yields the coveted result
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{1-\cos(wx)}{x}e^{-x}\,dx=\frac12\log\left(1+w^2\right)}$$
