Example of an analytic continuation for a function in integral form Given $f(z) = \int_{-\infty}^\infty \frac{exp(-t^2)}{z-t}\,dt$, where $Im(z)>0$.
Find an analytic continuation to the region $Im(z)<0$.
Firstly the solution said that there is a branch cut on the real axis but I fail to see how. I do not see why $f(z)$ is not analytic everywhere. I considered 3 cases:


*

*$Im(z)>0$

*$z$ on real axis

*$Im(z)<0$
Using the semicircle contour in case 1 and 3 we would find, by the residue theorem, the principal value of the integral equaling $2\pi i $ times(residue at the pole t=z). Similarly we can find for case 2, $-\pi i$ (residue at the pole t=z).
Secondly, the solution suggests that to continue $f(z)$ into the lower half plane, one should deform the contour on the real axis such that it includes the pole in lower half plane with a very sharp "spike" circulating the pole. I do not understnad this, since this way are we not just continuing to the region including only this pole?
 A: This is an important issue. First, qualitatively, in general the analytic continuation of a holomorphic function given by an integral is not always equal to the evaluation of the integral. The simplest case is Cauchy's integral formula $f(z)={1\over 2\pi i}\int_\gamma {f(w)\over w-z}dw$ for a circle $\gamma$ enclosing $z$. The integral makes sense also for $z$ outside the circle, where it gives $0$, which is rarely the analytic continuation of $f$ (from inside the circle).
Similarly, in the present example, for $f(z)$ given by the formula for $z$ in the upper half-plane, the analytic continuation to the lower half-plane is not given by the same formula, as will be indicated in a moment, even though the integral behaves perfectly well there. And, visibly, the integral has problems for $z$ on the real line. (No, excuses about principal value integrals do not resolve the issue.)
One specific way to understand the situation is a regularization device, such as
$$
f(z) \;=\;\int_{-\infty}^\infty {e^{-t^2}\over t-z} - {e^{-z^2}(z+i)\over (t-z)(t+i)}dt + \int_{-\infty}^\infty {e^{-z^2}(z+i)\over (t-z)(t+i)}dt
\;=\;\hbox{holo}+2\pi i e^{-z^2}
$$
The factor $(z+i)/(t+i)$ is just for convergence. The regularized integral, that is, with the subtraction so that it no longer blows up at $t=z$, is holomorphic, now having a removable singularity at $t=z$. 
Thus, the regularized expression still does make sense, and gives the analytic continuation, for $z$ in the lower half-plane (as well as for $z$ real).
For $z$ in the lower half-plane, we can "un-regularize", by moving the subtracted/regularizing term out of the integral. However, it will not cancel the $2\pi ie^{-z^2}$, because for $z$ in the lower half-plane the poles of $1/(t-z)(t+i)$ are both in the lower half-plane, and by moving the contour upward we see that the integral is $0$. Thus, for $z$ in the lower half-plane, the analytic continuation has values
$$
f(z) \;=\;\int_{-\infty}^\infty {e^{-t^2}\over t-z}dt 
+ 2\pi i e^{-z^2}
$$
In particular, it does not quite agree with the original formula evaluated at such $z$.
A: Observe that f is even-symmetric. Thus the bottom-half-plane case is determined by the upper one. 
