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How can I solve this integral

$$\int_{-\infty}^{+\infty}\mathrm{e}^{-(\alpha +\beta i)x^{2}}\,\mathrm{d}x$$

knowing that

$$\int_{-\infty}^{+\infty}\mathrm{e}^{ix^{2}}\,\mathrm{d}x=\sqrt{\pi }\mathrm{e}^{i\pi /4}?$$

I have to use Cauchy's integral theorem and integrate over a circular section of $ \pi/4 $ rad.

Daemonium
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  • have you tried Wolfram Alpha? – Dr. Sonnhard Graubner Oct 12 '17 at 19:05
  • is $i$ the imaginary unit? – Dr. Sonnhard Graubner Oct 12 '17 at 19:06
  • Yes, is the imaginary unit. I do not have Wolfram but Maple and it does not work very well. – Daemonium Oct 12 '17 at 19:08
  • i will try Maple too, one moment – Dr. Sonnhard Graubner Oct 12 '17 at 19:09
  • See also: https://ccrma.stanford.edu/~jos/sasp/Gaussian_Integral_Complex_Offset.html –  Oct 12 '17 at 19:10
  • the result is given by $$\frac{\pi}{\sqrt{\alpha+\beta i}}$$ – Dr. Sonnhard Graubner Oct 12 '17 at 19:14
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    Let $F(z) = \int_0^z e^{-s^2}ds$. Then for $\Re(a) > 0$, $\lim_{x \to \infty} F(ax)-F(x) = \lim_{x \to \infty} \int_ x^{ax} e^{-s^2}ds = 0$. Therefore $\int_0^\infty e^{-a x^2}dx =\int_0^\infty a^{-1/2} \frac{d}{dx} F(a^{1/2} x)dx= a^{-1/2}\lim_{x \to \infty} F(a^{1/2} x)= a^{-1/2}\lim_{x \to \infty} F(x)$ – reuns Oct 12 '17 at 19:14
  • @Jack: this problem is not a duplicate of the one you linked to. – Ron Gordon Oct 12 '17 at 20:26
  • @RonGordon: No, not an exact duplicate. –  Oct 12 '17 at 21:35
  • @Jack: no, not even close. Compare my solution below to the one in the link you posted. – Ron Gordon Oct 12 '17 at 21:36
  • I cannot believe that five people closed this question as a duplicate to a COMPLETELY DIFFERENT QUESTION. Please, I beg you, read BOTH questions carefully before deciding to close as a duplicate. In this case, the questions were not even close! The "dup" concerned a shift in the variable of integration to a line above the real axis. This concerns a proof that the familiar Gaussian integral also works for a complex quadratic coefficient in the exponent (something missing in the "dup" problem.) Again, read for yourselves. – Ron Gordon Oct 13 '17 at 02:18

1 Answers1

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This problem looks like that for evaluating Fresnel integrals, but we have to understand the nature of how it converges. For this, we turn to Cauchy's theorem. Here, consider

$$\oint_C dz \, e^{-\gamma z^2} $$

where $\gamma = \alpha + i \beta$ and $C$ is a circular sector in the complex plane, centered at the origin, of radius $R$ and having a wedge angle of $\theta_0 = \frac12 \arctan{(\beta/\alpha)}$ below the real axis. Of course it is assumed that $\alpha \gt 0$.

The above contour integral is equal to

$$\int_0^R dx \, e^{-\gamma x^2} + i R \int_0^{-\theta_0} d\theta \, e^{i \theta} \, e^{-\gamma R^2 e^{i 2 \theta}} + e^{-i \theta_0} \int_R^0 \, dt \, e^{-\gamma e^{-i 2 \theta_0} t^2} = 0$$

To show that the second integral vanishes as $R \to \infty$, we note that $\gamma e^{i 2 \theta} = |\gamma| e^{i 2 (\theta+\theta_0)} $. Accordingly, the magnitude of the second integral takes the form

$$ R \int_0^{\theta_0} d\theta \, e^{-|\gamma| R^2 \cos{(2 \theta_0-2 \theta)}} \le R e^{-|\gamma| \cos{(2 \theta_0)} R^2} \int_0^{\theta_0} d\theta e^{-|\gamma| R^2 \sin{(2 \theta_0)} (4 \theta/\pi)}$$

The latter approximation uses the fact that $\cos{(2 \theta)} \le 1$ and $\sin{(2 \theta)} \le 4 \theta/\pi$ in the integration interval. Thus, the magnitude of the second integral is bounded by $e^{-|\gamma| \cos{(2 \theta_0)} R^2} \frac{\pi}{4 |\gamma| R^2 \sin{(2 \theta_0)}}$, which clearly vanishes as $R \to \infty$.

Thus, we are left with

$$\int_0^{\infty} dx \, e^{-\gamma x^2} = e^{-i \theta_0} \int_0^{\infty} \, dt \, e^{-|\gamma| t^2} $$

because $\gamma e^{-i 2 \theta_0} = |\gamma|$. Therefore,

$$\int_{-\infty}^{\infty} dx \, e^{-\gamma x^2} = e^{-i \theta_0}\sqrt{\frac{\pi}{|\gamma|}} = \sqrt{\frac{\pi}{\gamma}}$$

as was to be shown.

Ron Gordon
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