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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?2012-01-05
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    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}}}$2012-01-05
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    For which values of $a$ and $b$?2012-01-05
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    @Cody have you tried the silver bullet for integration i.e. wolphram alpha ? :)2012-01-05
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    It said "prove", so perhaps "Wolfram Alpha" does not qualify.2012-01-05
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    Maybe the answer is that (at least the real part of) the integral is just zero for real $b$. If you write $b=b_1+ib_2$, then you get your second integral, where additionally the integrand is multiplied by $e^{i\frac{-b_1}{x^2}}$. My guess is that for small $x$, this oscillation $e^{i\infty}$ kills the integral and for large $x$ the integrand falls of due to the gauss curve anyway.2012-01-05

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