Poisson Integral: Real vs Complex a

[LagrangeEuler]

$\int^{\infty}_{-\infty}dxe^{-ax^2}=\sqrt{\frac{\pi}{a}}$
Is it correct also when $a$ is complex?

 

[vanhees71]

It's correct, if $\mathrm{Re} \; a>0$.

 

[voko]

For a complex $a$, there are two different roots. Which root is the correct one?

 

[vanhees71]

That's in fact a very good question. I've never thought about this before. I'm not sure whether the following is mathematically rigorous.

I'd start with the simple case $a=1$. The standard way to evaluate the integral is by setting
$$I=\int_{-\infty}^{\infty} \exp(-x^2)>0.$$
Then
$$I^2=\int_{\mathbb{R}^2} \mathrm{d} x \mathrm{d} y exp(-x^2-y^2).$$
Then we use polar coordinates in the $xy$ plane to get
$$I^2=2 \pi \int_0^{\infty} \mathrm{d} r r \exp(-r^2)=-\pi \exp(-r^2)|_{r=0}^{\infty}=\pi.$$
Since $I>0$ we uniquely get
$$I=\sqrt{\pi}$$
with the usual positive square root for a positve real number.

For real $a>0$ then you get by substitution $y=\sqrt{a} x$
$$I_a=\int_{-\infty}^{\infty} \mathrm{d} x \exp(-a x^2) = \frac{1}{\sqrt{a}} \int_{-\infty}^{\infty} \exp(-y^2)=\frac{\pi}{\sqrt{a}},$$
where again, I used the usual positive square root of a positive real number. You could as well have substituted $y=-\sqrt{a} x$, but then the infinite boundaries change signs, and thus the integral also flips signs, so that you get the same positive result as it must be.

Now, for $a \in \mathbb{C}$, for convergence you obviously should have $\mathrm{Re} \; a > 0$. Again we can take both roots in the substitution above, but it's most convenient to write
$$a=|a| \exp(\mathrm{i} \varphi),$$
where $\varphi \in (-\pi/2,\pi/2)$ (this is one of many possible choices of the argument for a complex number with positive real part) and then use the square root as
$$\sqrt{a}=\sqrt{|a|} \exp(\mathrm{i} \varphi/2), \quad \sqrt{|a|}>0. \qquad (*)$$
Now consider the Integral $I_{a}$ as an integral in the complex $x$ plane along the real axis. Then again we substitute $x=\sqrt{a} t$ with the meaning of $\sqrt{a}$ given by (*).

The real integration path in the complex $x$ plane then maps to an integration path in the complex $t$ plane, which is a straight line through the origina, running from the lower left quadrant into the upper right (for $\varphi>0$) or from the upper left to the lower right quadrant (for $\varphi<0$). In both cases, you can define a closed path by adding the real axis (run from right to left in both cases) and two vertical parts at infinity. If you integrate $\exp(-t^2)$ along that closed path you get 0 due to Cauchy's integral theorem, and this means that instead to integrate along the straight line in the $t$ plane you can as well integrate along the real axis (in the normal positive sense). Thus we find
$$I_a=\frac{\sqrt{\pi}}{|a|} \exp(-\mathrm{i} \varphi/2),$$
where the angle $\varphi \in (-\pi/2,\pi/2)$.

 

[voko]

(*) selects one the roots arbitrarily. Would that not mean that subsequent derivation inherits the arbitrariness?

 

[vanhees71]

You can of course choose the other root, i.e.,
$$\sqrt{a}=-\sqrt{|a|} \exp(\mathrm{i} \varphi/2), \quad \varphi \in (-\pi/2,\pi/2).$$
Then in substituting
$$t=-\sqrt{|a|} \exp(\mathrm{i} \varphi/2)x$$
the integration paths in the $t$ plane are straight lines through the origin either running from the upper right to the lower left quadrant (for $\varphi>0$) or from the lower right to the upper left quadrant (for $\varphi<0$).

Then closing the path with the line running along the real axis runs from $-\infty$ to $+\infty$. Thus the different sign in $-1/\sqrt{|a|} \exp(-\mathrm{i} \varphi/2)$ is compensated by the sign of the path along the real axis in the integral along the closed loop, giving 0 due to Cauchy's integral theorem. This shows that the integral is uniquely defined, no matter which root of $a$ you choose in the above substitution.