A calculus problem by ....

With $a>0,b>0$ consider

$\int_0^{\infty } e^{-a x^2} \sin \left(b x^2\right) \, dx$

We do contour integration. Let $A=\sqrt{a^2+b^2}$ .

Consider a positively oriented circular sector contour $\Gamma$ of radius R with angle $\omega$ , which we define as follows. Note that there exists a unique $0< \theta< \pi/2$ such that

$\cos(\theta)=a/A, \sin(\theta)=b/A$

Let $\omega = \theta / 2$ .

Hence

$\cos(\omega)=\cos(\theta / 2)=\sqrt{\frac{1+a/A}{2}}$

and we can compute $\sin(\theta)$ similarly.

Integrate $f(z)=e^{-A z^2}$ over the contour. Note that this integral is 0 by Cauchy's Theorem.

Along the x-axis, the integral approaches

$\int_0^{\infty } e^{-A x^2} \, dx=\frac{\sqrt{\pi }}{2 \sqrt{A}}$

We can parametrize the circular part as $z=R e^{i t}$ for t from 0 to $\omega$ . So the integral along the circular part is

as R approaches infinity, since $A>0$ and $\frac{\pi }{4}>\omega$ .

Finally, parametrize the integral along the radial part by $z=e^{i \omega} t$ for t from R to 0. We get

Edit: The above equation should not start with 0 but rather $-\frac{\sqrt{\pi }}{2 \sqrt{A}}$ .

Using the above formulas for $\cos(\omega), \sin(\omega)$ and taking real and imaginary part, we have a system of equations and then can just solve for I1 and I2. I2 is what we want. Then just plug in the values for a and b and compute.

Anybody have a different/simpler solution?

Also, can we generalize this to not just involving $x^2$ but rather $x^n$ (I think so ...)?