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:
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$Im(z)>0$
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$z$ on real axis
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$Im(z)<0$
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?
Best Answer
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 $z$.