A buddy and I are hung up on this integral.
Prove that:
$\displaystyle \int_{0}^{\infty}e^{-ax^{2}} \sin(b/x^{2})dx=-\text{Im}\int_{0}^{\infty}e^{-ax^{2}-ib/x^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\frac{\sin(\sqrt{2ab})}{e^{\sqrt{2ab}}}$
$\displaystyle\int_{0}^{\infty}e^{-ax^{2}-b/x^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\frac{1}{e^{\sqrt{2ab}}}$ was used.
The thing is, this goes under the assumption that $b$ is imaginary. Isn't this quite a leap to make without justification?
Does anyone know of a good way to prove this?