Evaluation/Estimation of a Gaussian integral

Question

Is there a closed form expression for the following definite integral: $$ F(u) = \frac{1}{2}\int_{-u}^u e^{-\frac{\alpha^2}{x^2}-\beta^2 x^2}\,dx = e^{-2\alpha\beta} \int_0^u e^{-\left(\frac{\alpha}{x}-\beta x\right)^2}\,dx? $$ Graphing the function makes it clear this quantity is somehow at least close to the standard Gaussian, and I've seen computations that indicate that the value at infinity corresponds to the standard Gaussian one (that is, $\lim_{u\rightarrow\infty} F(u) = \sqrt{\pi}/\beta$) although I've never been able to compute this directly myself. Since $F$ seems comparable to Gaussian, I expect that the answer to my question is negative, at least for finite $u$. But I would still be very interested in (1) the trick for evaluating at $u=\infty$, and (2) estimates on $F$, in particular, lower bounds.

Answer 1

Consider $I=\mathrm e^{2\alpha\beta}F(\infty)$ and the change of variable $\beta z=\alpha x^{-1}$, then $$ I=\int_0^\infty\exp\left(-\left(\alpha x^{-1}-\beta x\right)^2\right)\mathrm dx=\int_0^\infty\exp\left(-\left(\alpha z^{-1}-\beta z\right)^2\right)(\alpha/\beta)z^{-2}\mathrm dz, $$ that is $\beta I=\alpha J$ with $$ J=\int_0^\infty\exp\left(-\left(\alpha x^{-1}-\beta x\right)^2\right)x^{-2}\mathrm dx. $$ Now, $\mathrm d\left(\alpha x^{-1}-\beta x\right)=-\left(\alpha x^{-2}+\beta\right)\mathrm dx$ hence $$ \beta I+\alpha J=\int_0^\infty\exp\left(-\left(\alpha x^{-1}-\beta x\right)^2\right)\left(\alpha x^{-2}+\beta\right)\mathrm dx=\int_{-\infty}^\infty\mathrm e^{-t^2}\mathrm dt=\sqrt\pi, $$ hence $\beta I=\alpha J=\frac12\sqrt\pi$ and $$ F(\infty)=\frac{\sqrt\pi}{2\beta\mathrm e^{\alpha\beta}}. $$