Gaussian integration for complex phase

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SUMMARY

The forum discussion centers on the evaluation of the Gaussian integral involving complex phases, specifically the integral of the form \(\int dx'\ \frac{1}{\sqrt{AB}}\exp\bigg[i\frac{(x''-x')^{2}}{A}\bigg]\exp\bigg[i\frac{(x'-x)^{2}}{B}\bigg]\). The proposed solution involves completing the square, which is identified as the most straightforward method for this integration. Additionally, a participant raises a concern regarding the correctness of the integral, noting that Gaussian integrals typically include factors of \(\sqrt{2\pi}\).

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spaghetti3451
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I would like to prove that

##\displaystyle{\int dx'\ \frac{1}{\sqrt{AB}}\exp\bigg[i\frac{(x''-x')^{2}}{A}\bigg]\exp\bigg[i\frac{(x'-x)^{2}}{B}\bigg]=\frac{1}{\sqrt{A+B}}\exp\bigg[i\frac{(x''-x)^{2}}{A+B}\bigg]}##

Is there an easy way to do this integration that does not involve squaring the brackets?
 
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failexam said:
I would like to prove that

##\displaystyle{\int dx'\ \frac{1}{\sqrt{AB}}\exp\bigg[i\frac{(x''-x')^{2}}{A}\bigg]\exp\bigg[i\frac{(x'-x)^{2}}{B}\bigg]=\frac{1}{\sqrt{A+B}}\exp\bigg[i\frac{(x''-x)^{2}}{A+B}\bigg]}##

Is there an easy way to do this integration that does not involve squaring the brackets?

Hmm. Completing the square would seem to me by far to be the easiest way to do it. Why do you want a different way?

A second comment: Are you sure that is correct? Gaussian integrals usually involve factors of \sqrt{2\pi}
 

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