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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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    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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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.2012-01-06