# Convergence of $\int_{-\infty}^\infty e^{-\pi (a+z)^2} \, da$ where $z\in\mathbb C$

The integral $$\displaystyle\int_{-\infty}^\infty e^{-\pi (a+z)^2}\, da$$ should converge for every $$z\in\mathbb C$$, but in my proof, it seems that it diverges when $$\operatorname{Re}z=\alpha=0$$: \begin{align*}\left|e^{-\pi (a+z)^2}\right|&=\left|e^{-\pi a^2}e^{-2\pi a\overbrace{z}^{\alpha +\beta i}}e^{-\pi \overbrace{z^2}^{(\alpha +\beta i)^2}}\right|\\&=e^{-\pi a^2-2\pi a\alpha -\pi \alpha ^2+\pi\beta ^2}\underbrace{\left|e^{-\beta i(2\pi a+2\pi \alpha)}\right|}_{1}\end{align*} \begin{align*}\left|\displaystyle\int_{-\infty}^\infty e^{-\pi (a+z)^2}\, da \right|&\le \displaystyle\int_{-\infty}^\infty \left|e^{-\pi (a+z)^2}\right|\, da\\&=e^{-\pi \left(\alpha ^2-\beta ^2\right)}\displaystyle\int_{-\infty}^\infty e^{-\pi \left(a^2+2a\alpha\right)}\, da\\&\overset{\alpha ^2+2a\alpha =t^2}{=}e^{-\pi \left(\alpha ^2-\beta ^2\right)}\dfrac{\sqrt{\alpha (\alpha +2a)}}{\alpha}\displaystyle\int_{-\infty}^\infty e^{-\pi t^2}\, dt\\&\overset{t=m/\sqrt{\pi}}{=}\dfrac{e^{-\pi \left(\alpha ^2-\beta ^2\right)}\sqrt{\alpha (\alpha +2a)}}{\alpha}\displaystyle\int_{-\infty}^\infty e^{-m^2}\, \dfrac{dm}{\sqrt{\pi}}\\&=\dfrac{e^{-\pi \left(\alpha ^2-\beta ^2\right)}\sqrt{\alpha (\alpha +2a)}}{\alpha}.\end{align*} There must be a mistake somewhere, but I can't find it.

The change of variables $$\alpha^2+2a\alpha=t^2$$ is meaningless when $$\alpha=0$$ so you can't reach any conclusions in that case. You'll need to handle it separately. Fortunately the integral is easily seen to be convergent then.
• I tried to find the antiderivative, and then I evaluated it at the limits: $\int_{-\infty}^\infty e^{-\pi (a+z)^2}\, da=\frac{1}{2}\left[\operatorname{erf}\sqrt{\pi}(a+z)\right]_{-\infty}^\infty=\lim_{a\to\infty}\operatorname{erf}\sqrt{\pi}(a+z)=1$, though I don't know how would I rigorously prove the result of the limit. – Tu1 Aug 27 '19 at 12:20
• Riemann sphere came to my mind, namely that $z+\infty =\infty$. Can this be used in this context to conclude that $\lim_{a\to\infty}\operatorname{erf}\sqrt{\pi}(a+z)=\lim_{a\to \infty}\operatorname{erf}a=1?$ – Tu1 Aug 27 '19 at 12:28