Suppose $a$ is a complex number with positive real part (so $a$ is not necessarily just real). I am considering the integral
$$ \int_{-\infty}^{\infty} e^{-x^{2}/(2a)} \, dx. $$
So far I was able to show (using the famous trick by Gauss)
$$ \int_{-\infty}^{\infty} e^{-x^{2}/(2a)} \, dx = \pm“\sqrt{2\pi a}" $$
where $“\sqrt{2\pi a}"$ stands for the square-root with positive real part. My last step is to show the above equation must use the $+$ sign. This is obvious if $a$ is real and the LHS integral is entirely real, but I am not sure how to proceed if $a$ is complex with $\text{Re}(a) > 0$.
Am I missing something obvious? Any suggestions?
Work So Far
Okay, here is one attempt I am pursuing. Write $a = |a|e^{i\phi}$. Then
\begin{align}
\exp\left(-\frac{x^{2}}{2a}\right) &= \exp\left(-\frac{x^{2}}{2|a|}e^{-i\phi}\right) = \exp\left(-\frac{x^{2}}{2|a|}(\cos\phi – i\sin\phi)\right) \\
&= \exp\left(-\frac{x^{2}}{2|a|}\cos\phi\right)\exp\left(\frac{x^{2}}{2|a|}i\sin\phi\right) \\
&= e^{-x^{2}\cos\phi/(2|a|)}\left( \cos\left(\frac{x^{2}}{2|a|}\sin\phi\right) + i\sin\left(\frac{x^{2}}{2|a|}\sin\phi\right) \right).
\end{align}
Thus the real part of the main integral is of the form
$$ A = \int_{-\infty}^{\infty} e^{-\alpha x^{2}}\cos(\beta x^{2}) \, dx $$
where $\alpha, \beta$ are real constants.
Now I won't go into detail here (unless someone asks for more details), but we can assume $\alpha > 0$. The possibility that $\phi = 0$ corresponds to the case where $a$ is real, which is trivial, so we assume $\phi\ne 0$. Thus, $\beta\ne 0$. If $\beta$ is negative, we can WLOG replace it by $-\beta$, because it appears inside a cosine function, which is an even function. Thus, we will assume $\alpha, \beta > 0$.
The integrand of $A$ is even in $x$, so we can write it as
$$ A = 2\int_{0}^{\infty}e^{-\alpha x^{2}}\cos(\beta x^{2}) \, dx. $$
Do a substitution of variables:
$$ x = \sqrt{y}\cdot \frac{1}{\sqrt{\beta}}, \qquad dx = \frac{1}{2\sqrt{y}}\frac{dy}{\sqrt{\beta}}. $$
This gives
$$ A = 2\cdot\frac{1}{2\sqrt{\beta}}\int_{0}^{\infty} \frac{1}{\sqrt{y}}e^{-\gamma y}\cos(y) \, dy $$
where $\gamma = \alpha/\beta$ is some positive constant.
The outside constant factor is irrelevant, so it all comes down to showing that
$$ I = \int_{0}^{\infty} \frac{1}{\sqrt{y}}e^{-\gamma y}\cos(y) \, dy $$
is positive, where $\gamma > 0$. Is there any way to see that $I$ must be positive?
Best Answer
Both sides are holomorphic functions of $a$ in the strict right-half plane (the left side due to Leibniz's integral rule, or Morera+Fubini) and the right by elementary considerations. They are clearly equal on the positive real axis, so must be equal everywhere in the right-half plane ('identity theorem').
Note that in order to obtain this equality, we must exploit the regularity of $a\mapsto\int_{\Bbb{R}}e^{-x^2/2a}\,dx$, because if in general all you know is that a function satisfies $[f(a)]^2=2\pi a$, then there are two possibilities for $f(a)$, and it may even happen that $f(a_1)$ and $f(a_2)$ are given by two different branches of the square root. Above, I exploited analyticity, but continuity should suffice as well.