How does one prove this formula?

Question

Show that $$\int_0^{\infty}\frac{1}{\sqrt{x}}\exp\left[-a^2 x\left(\frac{x-6}{x-2}\right)^2\right]\,dx=\frac{\sqrt{\pi}}{a},$$ where $a$ is real.

Answer 1

Let $x=t^2$ and $u=t(t^2-6)/(t^2-2)$. The equation $t^3-ut^2-6t+2u=0$ has the three roots $t_1(u),t_2(u),t_3(u)$ whose sum is $t_1+t_2+t_3=u$, so $t_1'+t_2'+t_3'=1$. Note that each $t_j(u)$ is defined for $-\infty<u<\infty$. Now write the integral (by symmetry) as $$I=\int_{-\infty}^{\infty}\exp[-a^2u^2]]dt=\int_{-\infty}^{-\sqrt{2}}+\int_{-\sqrt{2}}^{\sqrt{2}}+\int_{\sqrt{2}}^{\infty}\exp[-a^2u^2] dt$$ Next change the integration variable to $u$ in each segment to get $$I=\int_{-\infty}^{\infty}(t_1'(u)+t_2'(u)+t_3'(u)) e^{-a^2u^2}du=\int_{-\infty}^{\infty}e^{-a^2u^2}du.$$