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(Reposted from math stack exchange)

I have searched and failed to find a rigorous proof showing that $$\int_{0}^\infty e^{-ax^2}\ \mathrm{d}x = \frac{\sqrt{\pi}}{2\sqrt{a}}$$ is true for $\Re(a)=0$ and $\Im(a)\neq0$. For example, why does $$ \int_{-\infty}^\infty e^{-ix^2}\ \mathrm{d}x = \sqrt{\frac{\pi}{i}} $$ and does one need to invoke contour integration to show this? Is it just a coincidence this happens to work? Or is the following technique valid:

Suppose $\Re (b)>0$ and $a$ is complex; then $$ \int_{-\infty}^\infty e^{-ax^2}\ \mathrm{d}x \stackrel{?}{=} \lim_{b\to0}\int_0^\infty e^{-(a+ b)x^2}\ \mathrm{d}x = \lim_{b\to0}\frac{\sqrt{\pi}}{2\sqrt{a+b}}. $$

Another example is $$ \int_0^\infty x^{s-1}e^{-ix}\ \mathrm{d}x = i^{-s}\Gamma(s) $$ which follows from $$ \int_0^\infty x^{s-1}e^{-ax}\ \mathrm{d}x = a^{-s}\Gamma(s), $$ but why is it that both of these are true? I fail to believe it is just a coincidence. Is there an explanation (not a calculation) that can justify these results?

Thank you.

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  • $\begingroup$ Look up Fresnel integrals for an answer to your first question. $\endgroup$
    – Michael Engelhardt
    Commented Dec 10, 2020 at 2:56
  • $\begingroup$ I am already familiar with the Fresnel integrals. I am looking for a reason as to why the Fresnel integrals can be evaluated the way I have done in the OP. $\endgroup$
    – user161698
    Commented Dec 10, 2020 at 3:02
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    $\begingroup$ Write $\int_0^\infty = \int_0^1 + \int_1^\infty$. The first part is easily seen to be a continuous function of $a \in \bf C$. The second part does not converge absolutely, so integrate by parts once or twice to get an equivalent integral that's absolutely convergent (plus an elementary function coming from the end point $x=1$) and that you can prove is a continuous function of $a$ in the closed upper half-plane. Finally undo the integration by parts and recombine the two integrals. $\endgroup$
    – Noam D. Elkies
    Commented Dec 10, 2020 at 3:13
  • $\begingroup$ What do you mean by $\sqrt a$? $\endgroup$
    – markvs
    Commented Dec 10, 2020 at 4:13
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    $\begingroup$ [correction to my earlier comment: $a$ in the closed right half-plane . . .] $\endgroup$
    – Noam D. Elkies
    Commented Dec 10, 2020 at 4:18

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