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Say we have the integral

$ I = \int_{-\infty}^{\infty} e^{-a(k+bi)^2} dk $

I know that this gives $\sqrt{\frac{\pi}{a}}$ though Wolfram and other sources. I am also fairly certain I need to solve this problem using contour integration, however, I have not seen anything similar in form and don't know what contour to choose the correct contour to evaluate it over, I need to be able to show rigorously show $I = \sqrt{\frac{\pi}{a}}$, but how?

A final question of if it isn't contour integration then how should I approach the problem?

Matt South
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2 Answers2

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If you want to do it without contour integration you could use a standard result for the Fourier transform. Expand the integral $$ I=\int_{-\infty}^\infty e^{a b^2}e^{-2i ab k}e^{-a k^2}\;dk\\ I=C\int_{-\infty}^\infty e^{-2 \pi i x k}e^{-a k^2}\;dk\\ $$ where $x=2ab/(2\pi)$. We can identify this as the Fourier transform of a Gaussian which can be found in a standard table $$ I = C \frac{e^{{-\frac{\pi^2 x^2}{a}}} \sqrt{\pi}}{\sqrt{a}}\\ I = e^{a b^2} \frac{e^{{-\frac{\pi^2 (2ab/(2\pi))^2}{a}}} \sqrt{\pi}}{\sqrt{a}} \\ I= e^{a b^2} \frac{e^{{-ab^2}} \sqrt{\pi}}{\sqrt{a}} \\ I= \frac{\sqrt{\pi}}{\sqrt{a}} $$

Benedict W. J. Irwin
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Consider the contour $$ \gamma=[-R,R]\cup[R,R+bi]\cup[R+bi,-R+bi]\cup[-R+bi,-R] $$ The integral $$ \int_\gamma e^{-az^2}\,\mathrm{d}z=0 $$ since there are no singularities of $e^{-az^2}$ inside it (or anywhere). Since the integral along $[R,R+bi]$ and $[-R+bi,-R]$ vanish as $R\to\infty$, we get that the integral along this contour is the difference $$ \int_{-\infty}^\infty e^{-ak^2}\,\mathrm{d}k-\int_{-\infty}^\infty e^{-a(k+bi)^2}\,\mathrm{d}k $$

robjohn
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