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?
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Are you sure that the integral converges for $p=i$? Am I missing something here? – karvens Sep 08 '14 at 06:22
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1@karvens For $p=i$ you will get Fresnel-like integrals since $e^{ix^2}=\cos x^2+i\sin x^2$. – Tunk-Fey Sep 08 '14 at 06:29
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Well, according to wolframalpha, $p=i$ results in the integral taking the principle value of $\sqrt{i}$. – Silynn Sep 08 '14 at 06:30
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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.
Xander Henderson
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1Answers 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? – Xander Henderson Aug 01 '18 at 02:09