I'm trying to calculate the integral $\int_{-\infty}^{\infty}e^{-(1-i)x^2} dx$, without invoking some sort of complex integral identify such as erf (ie as elementarily as possible)
I'm thinking to solve this similarly to solving $\int_{-\infty}^{\infty}e^{x^2} dx$, where we can call in a second variable $y$, convert to polar, and take the square root of the result to get $\sqrt \pi$.
My attempt at this is to consider $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(1-i)(x^2+y^2)} dxdy = \int_{0}^{\infty}\int_{0}^{2\pi}e^{(1-i)(-r^2)} d\theta dr$. Normally, as in the case of $e^{-x^2}$, the limit is easy to compute, and we simply get something in terms of area of the unit circle. However, in this case, I'm not sure what to do with the imaginary term.
Am I on the right track? How can I continue?
I should not suggest to use polar coordinates.
Consider instead $$\int_{-\infty}^{+\infty}e^{-ax^2} \,dx=\frac{\sqrt{\pi }}{\sqrt{a}} \quad \text{if} \quad \Re(a) >0$$ $$a=1-i \implies I=\frac{\sqrt{\pi }}{\sqrt{1-i}}=\sqrt{1+i}\,\sqrt{\frac \pi 2}$$