TITLE: Integral of exponential $\int_0^{+\infty} \exp\left(-a^2 x\left(\frac{x-6}{x-2}\right)^2\right) \frac{dx}{\sqrt{x}}$
QUESTION [7 upvotes]: Calculate the following integral
$$\int_0^{+\infty} \exp\left(-a^2 x\left(\dfrac{x-6}{x-2}\right)^2\right) \dfrac{dx}{\sqrt{x}}$$
I think the relation between this integral and function gamma is clear, But I do not know what variables change I do. Any Hint Pliss!

REPLY [2 votes]: Let $I$ be the integral in the question. First set $u^2=x$ so that:
\begin{align}
I=\int^\infty_0 \exp\left(-a^2x\left(1-\frac{4}{x-2} \right)^2\right)\,\frac{dx}{\sqrt[]{x}} =2\int^\infty_0 \exp\left(-a^2u^2\left(1-\frac{4}{u^2-2} \right)^2\right)\,du
\end{align}
Now notice that: $$I=\int^\infty_{-\infty} \exp\left(-a^2\left(u-\frac{4u}{u^2-2} \right)^2\right)\,du=\int^\infty_{-\infty} \exp\left(-a^2\left(u-\frac{2}{u+\sqrt[]{2}}-\frac{2}{u-\sqrt[]{2}} \right)^2\right)\,du$$
By Glasser's Master Theorem we get the result, namely:
$$I=\int^\infty_{-\infty} \exp(-a^2u^2)\,du=\frac{\sqrt[]{\pi}}{|a|}$$
And we finally see the result Lucian suggested to be true three years ago.