Find the analytic continuation of the function $f(z)$ defined by

$ f(z) = \int_{0}^{\infty} \frac{\exp(-zt)}{1+t^2} dt$ , $ |\arg(z)| < \pi/2$

to the domain $ -\pi/2 < \arg(z) < \pi$

I have no clue how to approach this problem.


What follows is definitely not quite orthodox but it may help get started. Note that $$f(z) = \int_0^\infty \frac{e^{-zt}}{1+t^2} dt$$ has a nice structure that makes it possible to expand it into a power series around a point $w$, which goes like this: $$f(z) = \int_0^\infty \frac{e^{-(z-w)t}\times e^{-wt}}{1+t^2} dt = \int_0^\infty \left(\sum_{q\ge 0} \frac{(-1)^q}{q!} (z-w)^q t^q\right) \frac{e^{-wt}}{1+t^2} dt \\ = \sum_{q\ge 0} \frac{(-1)^q}{q!} (z-w)^q \int_0^\infty t^q\frac{e^{-wt}}{1+t^2} dt.$$ Note that the integral forming part of the coefficient converges when $\Re(w)>0.$

Next we re-write said integral in order to examine the order of growth as a function of $q$. Put $wt=u$. We obtain $$ \int_0^\infty t^q\frac{e^{-wt}}{1+t^2} dt = \int_0^\infty (u/w)^q \frac{e^{-u}}{1+(u/w)^2} \frac{1}{w} du \\ = \int_0^\infty u^q \frac{1}{w^{q+1}} \frac{e^{-u}}{1+(u/w)^2}du = \frac{1}{w^{q-1}} \int_0^\infty u^q \frac{e^{-u}}{u^2+w^2} du.$$ At this point it becomes apparent that for $u$ sufficiently large the integrand behaves like $u^{q-2}$ which gives $$ \int_0^\infty u^{q-2} e^{-u} du = \int_0^\infty u^{(q-1)-1} e^{-u} du = \Gamma(q-1) = (q-2)!$$ It follows that the coefficient on $(z-w)^q$ in the power series is asymptotic to $$\frac{(-1)^q}{q!} \frac{1}{w^{q-1}} (q-2)! = \frac{1}{w^{q-1}} \frac{(-1)^q}{q(q-1)}.$$ Now either by inspection or by the ratio test we find that the radius of convergence of the series is $$\lim_{q\to\infty} \left|\frac{(-1)^q/q/(q-1)/w^{q-1}}{(-1)^{q+1}/(q+1)/q/w^q} \right| = \lim_{q\to\infty} \left|w\frac{q+1}{q-1}\right| = |w|.$$ The conclusion is that we may choose $w$ extremely close to the imaginary axis in the right half plane and thereby obtain a power series that converges in the circle of radius $|w|$ centered at $w$. Elementary geometric reasoning now shows that in this way taking $|w|$ large we can cover the right half plane (where the original integral converges) and the left half plane with the triangular region enclosed by two rays from the origin above and below the negative real axis removed. This gives an analytic continuation to $$-\pi < \arg(z) < \pi.$$

Post Scriptum. Using the integral for $f(z)$ we find that $$f(5i+1/20) \approx 0.01307910528846 - 0.2200394976659 i.$$ Taking $w= 6i+1/30$ the series gives $$f(5i+1/20) \approx 0.01307910528847 - 0.2200394976659 i.$$ These two agree as they ought to.

On the other hand the value $$f(5i-1/20) \approx 0.008062325160153 - 0.2210974509631 i$$ cannot be calculated from the integral and needs the series expansion, yet it shows continuity across the imaginary axis.

This continuity is even more evident in the two values $$f(5i+1/500) \approx 0.01068426529406 - 0.2205730068288i \\ \quad\text{and}\quad \\ f(5i-1/500) \approx 0.01048357716278 - 0.2206153425701i$$ where we point out one more time that the value to the left of the imaginary axis cannot be calculated from the integral formula.

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