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?

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

1 Answer 1


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.

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    $\begingroup$ Answers on MSE should be as self-contained as possible, and with an eye towards semi-permanence. I notice that two of your links are off-site, which means that they could someday suffer from linkrot. Could you please take a few minutes to discuss (in your answer) the relevant details found on the linked pages? $\endgroup$
    – Xander Henderson
    Commented Aug 1, 2018 at 2:09

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