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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?

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It is a bit hard to guess what precisely you are asking. Could you please make this clearer? – Rasmus Jan 5 '12 at 12:44
I'm sorry. Prove that $\displaystyle\int_{0}^{\infty}e^{-ax^{2}}sin(b/x^{2})dx=\frac{1}{2}\sqrt{\frac{‌​\pi}{a}}\frac{sin(\sqrt{2ab})}{e^{\sqrt{2ab}}}$ – Cody Jan 5 '12 at 13:50
For which values of $a$ and $b$? – Rasmus Jan 5 '12 at 14:50
@Cody have you tried the silver bullet for integration i.e. wolphram alpha ? :) – NikBels Jan 5 '12 at 14:52
It said "prove", so perhaps "Wolfram Alpha" does not qualify. – GEdgar Jan 5 '12 at 15:43

Consider the formula $$ \int_{0}^{\infty}e^{-ax^{2}-b/x^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}\frac{1}{e^{\sqrt{2ab}}} $$ In the domain of pairs $a,b$ of complex numbers where it converges, the left-hand side is an analytic function of $a$ and $b$. There is a sensible domain where the right-hand side is an analytic function of $a$ and $b$. Therefore, if the two are equal on a large enough set (for example, all positive real $a,b$), then it is true on the entire domain of convergence. So, we only need to check that there is a sensible domain for this that includes imaginary $b$.

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Thanks for the input. We'll see how things turn out and get back. – Cody Jan 6 '12 at 0:06

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