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}}. $ $\endgroup$ – Mhenni Benghorbal May 19 '13 at 20:48
  • $\begingroup$ @MhenniBenghorbal I am looking for evaluating this with residue theorem. $\endgroup$ – Mula Ko Saag May 19 '13 at 20:49
  • $\begingroup$ Here is a related problem. $\endgroup$ – Mhenni Benghorbal May 19 '13 at 20:51

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.