# Fourier transform of function involving $\log$

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!

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If you're only interested in the final answer, than you can go to here. I don't know how to do it by hand, but since the answer is pretty complicated you probably need some integral handbook... – yohBS Jan 12 '12 at 23:10

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) $$f(a,t)= (1-e^{-a t}) \frac{H(t)}{t}.$$

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Something cannot be correct. If I follow your argument, I find that $f(a,t) = c_1(t) + a c_2(t)$ with two constants $c_1(t)$ and $c_2(t)$. – Fabian Jan 12 '12 at 23:13
Not quite, I think: $\int_{\mathbb R} \frac{d\omega}{2\pi}\ \frac{i}{\omega+ia}\ e^{-i\omega t} = e^{-at}$ for $t > 0$ and $0$ for $t < 0$, assuming $a > 0$. – Robert Israel Jan 12 '12 at 23:21
@Robert: Oups I exchanged $a$ and $\omega$ somewhat I fear. I'll edit all that but numerical verification is rather difficult here... – Raymond Manzoni Jan 12 '12 at 23:31
@Robert: ok, now I understand (it is not that difficult after all). I get $f= c(t)- e^{-a t} H(t)/t$. But now, we need to find $c(t)$. – Fabian Jan 12 '12 at 23:44
@Fabian: Numerical integration seem too to favor the formula provided. – Raymond Manzoni Jan 13 '12 at 0:40