Integration...Interesting.

R u sure u r getting the recurrence u have written?

This problem(or anything comparable to this) will not be asked in IITJEE.
That's why i posted this in olympiad section.

I don't think that it is correct either...

Here's my solution:
I(\alpha,\beta)=\int_{0}^{\infty}{e^{-\alpha^2\left(x^2+\frac{\beta^2}{x^2} \right)}dx} =\int_{0}^{\infty}{e^{-\alpha^2\left[\left(x-\frac{\beta}{x} \right)^2+2\beta \right]}dx} \\\\=e^{-2\beta\alpha^2}\int_{0}^{\infty}{e^{-\alpha^2\left(x-\frac{\beta}{x} \right)^2}dx}

Now differentiate both sides wrt \beta
So:

\frac{\partial I}{\partial \beta}=-{2\alpha^2\beta}\int_{0}^{\infty}{\frac{e^{-\alpha^2\left(x^2+\frac{\beta^2}{x^2} \right)}}{x^2}dx}

Now:

I-\frac{1}{2\alpha^2}\frac{\partial I}{\partial \beta}=e^{-2\alpha^2\beta}\int_{0}^{\infty}{\left(1+\frac{\beta}{x^2} \right)e^{-\alpha^2\left(x-\frac{\beta}{x} \right)^2}dx}

Take:

x-\frac{\beta}{x}=\frac{z}{\alpha} \\\\\Rightarrow \left( 1+\frac{\beta}{x^2} \right)dx=\frac{dz}{\alpha}

I-\frac{1}{2\alpha^2}\frac{\partial I}{\partial\beta}=\frac{e^{-2\alpha^2\beta}}{\alpha}\int_{-\infty}^{\infty}{e^{-z^2}dz}=\frac{\sqrt{\pi}e^{-2\alpha^2\beta}}{\alpha}

so we now get a differential equation (linear):

\frac{\partial I}{\partial \beta}-2\alpha^2I=-2\sqrt{\pi}\alpha e^{-2\alpha^2\beta}

Solving this (which is no big deal) we obtain:
I=\frac{\sqrt{\pi}e^{-2\alpha^2\beta}}{2\alpha}+Ce^{2\alpha^2\beta}

I(\beta=0)=\int_{0}^{\infty}{e^{-\alpha^2x^2}dx}=\frac{\sqrt{\pi}}{2\alpha} \\\\\Rightarrow C=0.

So,
I(\alpha,\beta)=\int_{0}^{\infty}{e^{-\alpha^2\left(x^2+\frac{\beta^2}{x^2} \right)}dx}=\frac{\sqrt{\pi}}{2\alpha}e^{-2\alpha^2\beta}