Gaussian integral $\int dx \,e^{-ax^2}$ for complex $a$.

Asked Nov 05 '16 at 17:27

Active Nov 05 '16 at 17:52

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I want to prove the Gaussian integral formula

$$\int_{-\infty}^\infty e^{-ax^2}\,dx = \sqrt{\pi/a}$$

for $a \in \mathbb{C}$ where $ \Re(a)>0$. In the exercise it says to 'shift the contour of integration' but I don't see why this can't be done explicitly, exactly how one would do with real $a$, with the one difference that the positive real part is necessary for convergence after going to polar coordinates.

Anyway, I'd like to do it as the book wants, so I'd appreciate a tip which contour to use, and for which function. I tried $e^{-az^2}$ on a rectangle from $-R$ to $R$, and of "height" $\varepsilon \rightarrow 0$, but this just gives me an identity (would be useful if I had to prove invariance of the integral under the shift $x \rightarrow x + i\varepsilon $, but not here).

edited Nov 05 '16 at 17:52

Michael Hardy

asked Nov 05 '16 at 17:27

Spine Feast

4,770

2

Look here – Giuseppe Negro Nov 05 '16 at 17:33
$$\int_{-R}^R e^{-ax^2},dx=\frac{1}{\sqrt{a}}\int_{-\sqrt{a}R}^{\sqrt{a}R}e^{-z^2},dz$$Then, using Cauchy's Integral Theorem, deform the contour from $-\sqrt{a}R$ to $\sqrt{a}R$ back onto the real axis. – Mark Viola Nov 05 '16 at 17:44
Also you should realize that for $a > 0$ it is trivial, and by analytic continuation (or the identity theorem) the formula stays true for $Re(a) > 0$ (where the LHS and the RHS are analytic) – reuns Nov 05 '16 at 18:08
@Dr.MV the square root is not unique though, isn't that a problem? – Spine Feast Nov 05 '16 at 20:34
@SpineFeast Cut the plane to define it uniquely. – Mark Viola Nov 05 '16 at 20:38

Gaussian integral $\int dx \,e^{-ax^2}$ for complex $a$.

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