TITLE: Evaluating $\int_0^{\infty}e^{-\alpha x^2 \cos \beta} \cos(\alpha x^2 \sin \beta) dx$
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.