Need help to understand calculating fourier transform of 1 this issue is related to a physics problem, but since it is mathematical I will post it here.
When calculating the following Fourier transform
$$
-i\int_\infty^\infty dt~ e^{i\omega t}\theta(t)e^{-i\xi_k t} = -i\int_0^\infty dt~e^{i(\omega -\xi_k)t},
$$
where $\theta(t)$ is the heaviside step function. The solution then states that in order to make the integral converge in the upper limit, we let $\omega \rightarrow \omega + i\eta$, where $\eta = 0^+$. Then they just write out the result which is
$$
\frac{1}{\omega - \xi_k + i\eta}.
$$
However, when I try to do the calculation more detailed, I fail to understand why the addition of the positive infinitesimal helps us, and how do I get 1 in the numerator? So my question is: How does the addition of the positive infinitesimal make the integral converge?
 A: The addition of $i\eta$ helps because what you end up with at that point is an exponential of the form $e^{i(\omega+i\eta-\xi_k)t} = e^{i(\omega-\xi_k)t}e^{-\eta t}$. Notice then that you have exponential decay due to $\eta$ which allows integrals to converge nicely. Otherwise your integral, as it stands, is not convergent in the usual way (since $e^{it}$ oscillates). Then the integral is easy to do:
$$-i\int_0^{\infty} e^{(i\omega-i\xi_k-\eta)t}\,dt = \left.-\frac{i}{i\omega-i\xi_k-\eta}e^{(i\omega-i\xi-\eta)t}\right|_0^{\infty} = \frac{1}{\omega-\xi_k+i\eta}.$$
Personally I don't like "regularization" like this. I prefer to treat such integrals distributionally as it's is a more rigorous approach but both roads lead you to the same answer. However regularizing gives answers more readily. If you do it distributionally, you often have to do quite a bit of work to get an answer. Each has its merits and pitfalls.
A: The problem here is that $$
  \int_0^\infty e^{i\alpha t}
$$
does not converge, because $e^{i\alpha t} = \cos \alpha t + i\sin \alpha t$ doesn't go to zero for $t \to \infty$. But if you add an additional exponential factor $e^{-\eta t}$, you get $$
  \int_0^\infty e^{-\eta t} e^{i\alpha t} \,dt = \int_0^\infty e^{i(\alpha +i\eta)t} \,dt
$$
which does converge, because $e^{-\eta t}$ goes to zero fast enough.
What you end up doing here is computing the laplace transform $F(s)$ of $f(t) = -i\theta(x)e^{-i\xi_k t}$, i.e. computing $$
  F(s) = \int_{0}^\infty f(t) e^{st} \,dt \text{,}
$$
which exists for $\textrm{Re } s < 0$. If the fourier transform exists, the fourier transform of $f$ is simply $\tilde F(\omega) = F(i\omega)$. But even if the fourier transform does not exists, you can often extend $F$ analytically to a large part of the imaginary axis, and thus get a "kind-of" fourier transform. Here, $F$ exists for all $s$ to the left of the imaginary axis, so we simply set $$
  \tilde F(\omega = \lim_{v \to 0^+} F(i\omega - v) \text{,}
$$
and call $\tilde F$ the fourier transform of $f$.
