# Complex Gaussian Integral - $\int_{-\infty}^\infty e^{-p(t+c)^2}dt = \sqrt{\frac{\pi}{p}}$?

I found some formulas on books, especially the complex gaussian integral formula: $$\int_{-\infty}^\infty e^{-p(t+c)^2}dt = \sqrt{\frac{\pi}{p}}$$ for $p,c\in\mathbb C$. Then if $p=i=\sqrt{-1}$, the integral may have two different values, since $\sqrt{i}$ has two different values on the complex plane. Then how can I justify the above integral formula?

• Are you sure that the integral converges for $p=i$? Am I missing something here? Commented Sep 8, 2014 at 6:22
• @karvens For $p=i$ you will get Fresnel-like integrals since $e^{ix^2}=\cos x^2+i\sin x^2$. Commented Sep 8, 2014 at 6:29
• Well, according to wolframalpha, $p=i$ results in the integral taking the principle value of $\sqrt{i}$. Commented Sep 8, 2014 at 6:30

That is true when $p$ has positive real part. For $p$ with non-positive real part, the integral diverges.
For real $c$, the proof of the real Gaussian integral still largely works.
For $p$ with non-zero real part, a case can be made to choose the square root with positive real part, based on the analycity of $f(z) = \int_{-\infty}^{\infty} e^{-zu^2} du$ around the positive real semi-axis. Purely imaginary $p$ is more complicated.
For non-real $c$, the Fourier transform of the Gaussian gives the same result. Showing that the derivative with respect to $c$ is zero should also work.