I found the following problem which I am unable to solve.
Calculate the following integral $\int_{\mathbb{R}} \frac{d\omega}{2\pi} \log (1 + i a/\omega ) e^{-i \omega t}$ for $a>0$ and $t\in\mathbb{R}$.
Any help is appreciated!
I found the following problem which I am unable to solve.
Calculate the following integral $\int_{\mathbb{R}} \frac{d\omega}{2\pi} \log (1 + i a/\omega ) e^{-i \omega t}$ for $a>0$ and $t\in\mathbb{R}$.
Any help is appreciated!
Let's consider $t \in \mathbb{R}$ constant and define $f(a)=\int_{\mathbb{R}} \frac{d\omega}{2\pi} \log (1 + i a/\omega ) e^{-i \omega t}$ then
$\displaystyle \frac{df}{da}=\int_{\mathbb{R}} \frac{d\omega}{2\pi} \frac{i}{\omega +i a} e^{-i \omega t}$
EDIT: Let's rather try this directly..
$\displaystyle \frac{df}{da}=e^{-at}\int_{\mathbb{R}} \frac{d\omega}{2\pi} \frac{i}{\omega +i a} e^{-i (\omega +i a)t}= e^{-a t} H(t)$
(we should prove that the shift of the path of $ia$ doesn't change the integral)
After integration we get $f(a)= c(t)- e^{-a t} \frac{H(t)}{t}$ as both of you found faster! :-)
$c(t)=\frac{H(t)}{t}$ seems to return 0 as in the case of $\lim_{a\to 0} f(a)$ so that we 'could have' (it's late here) $ f(a,t)= (1-e^{-a t}) \frac{H(t)}{t}.$