How can I calculate the following complex definite integral, where $\alpha > 0$ and $\beta \in \mathbb{R}$?

$$ \int_{-\infty}^{\infty} e^{-\left(\alpha x + i\beta\right)^2}\, dx $$


After a substitution, we can assume that $\alpha = 1$. Then write the integral in the complex form

$$\int_{-\infty + i\beta}^{+\infty + i\beta} e^{-z^2}\,dz,$$

and use Cauchy's integral formula to shift the contour to the real axis,

$$\int_{-\infty + i\beta}^{+\infty + i\beta} e^{-z^2}\,dz = \int_{-\infty}^{+\infty} e^{-z^2}\,dz,$$

since the integrand is entire, and decays very fast as $\lvert\operatorname{Re} z\rvert \to \infty$. The last integral is well-known from real analysis, its value is $\sqrt{\pi}$.

  • $\begingroup$ Thank you. So with the substitution it's $\sqrt{\pi}/\alpha$, right? $\endgroup$ – Evan Aad Sep 20 '14 at 21:04
  • 1
    $\begingroup$ Yes, and $\sqrt{\pi}/\lvert\alpha\rvert$ if we also allow $\alpha < 0$. $\endgroup$ – Daniel Fischer Sep 20 '14 at 21:07

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