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I need to prove the following equation, and I found a very smart answer, but the only problem is when he changes the variable, the limit changes also. We need to find another suitable contour and invoke the residue theorem to change the limit back to what it should be. Any ideas how to find a suitable contour to achieve this goal? enter image description here

  • $\begingroup$ If $t = \sqrt{a}x + b/2\sqrt{a}$, then as $x \to \infty$, $t \to \infty$. So I'm not sure what your issue is? Your result just says the Fourier transform of $e^{- \pi x^{2}}$ is itself. $\endgroup$ – mattos Sep 30 '16 at 20:41
  • $\begingroup$ You talk about "suitable contour" and stuff, yet the solution you quoted is purely real analysis: no complex integration done here. $\endgroup$ – DonAntonio Sep 30 '16 at 20:41
  • $\begingroup$ @DonAntonio yes that's the issue. Does that mean this method is incorrect for this problem? $\endgroup$ – J.doe Sep 30 '16 at 20:52
  • $\begingroup$ @Mattos My issue is I used real analysis in the method, but this is a complex analysis problem? $\endgroup$ – J.doe Sep 30 '16 at 20:52
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    $\begingroup$ I think maybe it is, @DonAntonio. You have made the transformation $t=\sqrt{\pi} x + i\sqrt{\pi}\eta^2$ to the integral goes from $t= $"$-\infty + i\sqrt{\pi}\eta^2$" to "$+\infty + i\sqrt{\pi}\eta^2$". We have to justify equating that to an intergral along the real axis. That is not hard, because the contributions from the vertical path segments an large $|t|$ are exponentially small. $\endgroup$ – Mark Fischler Sep 30 '16 at 20:57

Here's a another proof that avoids contour integration all together. Consider \begin{align} F(\xi) := \int^\infty_{-\infty}e^{-\pi x^2}e^{2\pi ix\xi}\ dx, \end{align} then it follows \begin{align} F'(\xi) =&\ \int^\infty_{-\infty} e^{-\pi x^2} \frac{d}{d\xi}e^{2\pi i x\xi}\ dx = 2\pi i\int^\infty_{-\infty} xe^{-\pi x^2} e^{2\pi ix\xi}=\ -i\int^\infty_{-\infty} \frac{d}{dx} e^{-\pi x^2} e^{2\pi ix\xi}\ dx. \end{align} Using integration by parts, we have that \begin{align} F'(\xi) = i\int^\infty_{-\infty} e^{-\pi x^2} \frac{d}{dx} e^{2\pi i x\xi}\ dx = -2\pi \xi \int^\infty_{-\infty}e^{-\pi x^2}e^{2\pi ix\xi}\ dx = -2\pi \xi F(\xi). \end{align} Solving the ode \begin{align} F'(\xi)= -2\pi \xi F(\xi) \end{align} yields the solution \begin{align} F(\xi) = F(0)e^{-\pi \xi^2}. \end{align} But \begin{align} F(0) = \int^\infty_{-\infty} e^{-\pi x^2}\ dx = 1 \end{align} which means \begin{align} F(\xi) = e^{-\pi \xi^2}. \end{align}


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