Q: Suppose $\alpha>0$ and $|\beta|<\pi/2$, show that \begin{align*} \textbf{(1)} \; \int_0^{\infty}e^{-\alpha x^2 \cos \beta} \cos(\alpha x^2 \sin \beta) dx &= \frac 1 2 \sqrt{\pi/\alpha}\cos(\beta/2)\\ \textbf{(2)} \; \int_0^{\infty}e^{-\alpha x^2 \cos \beta} \sin(\alpha x^2 \sin \beta) dx &= \frac 1 2 \sqrt{\pi/\alpha}\sin(\beta/2) \end{align*}

How can I integrate the above with the method of contour?

The integral can be changed into $\displaystyle \int_0^{\infty}e^{-\alpha x^2 \cos \beta} e^{i(\alpha x^2 \sin \beta) }dx = \int_0^{\infty} e^{x^2\alpha e^{i (\pi - \beta)}}dx$. This is similar to $\displaystyle \int_0^{\infty} e^{-x^2}dx$ which has been discussed here except that it has complex coefficients. How do I modify it?

  • $\begingroup$ Use the result $ \int_{0}^{\infty} e^{-bx^2}dx = \frac{\sqrt{\pi}}{2\sqrt{b}}. $ $\endgroup$ May 19, 2013 at 20:48
  • $\begingroup$ @MhenniBenghorbal I am looking for evaluating this with residue theorem. $\endgroup$ May 19, 2013 at 20:49
  • $\begingroup$ Here is a related problem. $\endgroup$ May 19, 2013 at 20:51

1 Answer 1


Use a wedge contour $C$ of angle $-\beta/2$, i.e., below the real axis. That is, consider

$$\oint_c dz \, e^{-a e^{i \beta} z^2} = \int_O^R dx \, e^{-a e^{i \beta} x^2} + i R \int_0^{-\beta/2} d\theta e^{-a R^2 e^{i (\beta+2 \theta)}} + e^{-i \beta/2} \int_R^0 dx \, e^{-a x^2}$$

Note that by using this contour, we get a pure Gaussian integrand along the sloped line to the origin.

That the second integral vanishes in the limit as $R \to \infty$ may be seen by noting that $\cos{(\beta+2 \theta)} \gt 0$ within the integration interval. Therefore,

$$\int_0^{\infty} dx \, e^{-a e^{i \beta} x^2} = e^{-i \beta/2} \int_0^{\infty} dx \, e^{-a x^2} = \frac12 e^{-i \beta/2} \sqrt{\frac{\pi}{a}}$$

The stated answers come from taking real and imaginary parts of the above.


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