Imaginary Gaussian integral

Question

I want to do the integral $$ \int dx e^{ -\frac{a}{2}\pi i x^2}dx = \sqrt{\frac{2\pi}{ia}} $$ for $a = 1$. But the above result is conditional on Im${(a)}>0$. Therefore it is not as simple to take the limit to the limit $a \to 1$. So I wonder what is the right approach here. Mathematica spits out $$\frac{1-i}{\sqrt{2}}$$ but I am not sure how this is obtained or why it is the correct one. Ok, maybe the solution is trivial, I do not know it though.

Answer 1

You can make an analytic continuation in the complex plane $z = x + i y$. Consider the closed contour as shown in the figure

Since the function has no poles in the domain enclosed by the contour $c$, then

$$ \oint_c {\rm d}z~ e^{-\frac{1}{2}\pi i z^2} = 0 $$

Moreover, once you decompose the integral the problem reduces to

$$ \oint_c {\rm d}z~ e^{-\frac{1}{2}\pi i z^2} = \int_{c_2} {\rm d}z~ e^{-\frac{1}{2}\pi i z^2} + \int_{c_4} {\rm d}z~ e^{-\frac{1}{2}\pi i z^2} = 0 \tag{1} $$

The integral over $c_1$ and $c_3$ vanishes, in the limit when the contour stretches to infinity. $c_4$ is the integral you want to calculate, the only term that is left is the one associated with $c_2$, for that one consider the parametrization

$$ x = t, ~~~ y = -i t, ~~~~ z = x + iy = (1 - i)t $$

so that

\begin{eqnarray} \int_{c_4} {\rm d}z~ e^{-\frac{1}{2}\pi i z^2} &=& \int_{+\infty}^{-\infty}{\rm d}t ~(1-i)e^{-\frac{1}{2}\pi i (1-i)^2t^2} \\ &=& -(1-i)\int_{-\infty}^{+\infty}e^{\frac{1}{2}\pi t^2} \\ &=& -(1-i) \tag{2} \end{eqnarray}

Replacing (2) in (1):

$$ \int_{c_2} {\rm d}z~ e^{-\frac{1}{2}\pi i z^2} = i - 1 $$

But over $c_2$ we have $z = x$, so that

$$ \int_{-\infty}^{+\infty} {\rm d}x~ e^{-\frac{1}{2}\pi i x^2} = i - 1 $$

That matches the result from WolframAlpha