TITLE: How to compute $\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$
QUESTION [26 upvotes]: $$\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$$
WolframAlpha gives a numerical answer of $43.8122$, which appears to be $\sqrt{611\pi}$. And playing with that, it seems that replacing $611$ with $a$ just gives $\sqrt{a\pi}$. My trouble is that the stuff in the exponential always seems to be just a big mess, and I haven't been able to get it into a form I can understand or deal with.
I would greatly appreciate seeing a method for solving this integral.

REPLY [6 votes]: Using identity in @user137794's answer:
\begin{equation}
\int_{-\infty}^\infty f(x-x^{-1})\ dx = \int_{-\infty}^\infty f(x)\ dx
\end{equation}
where the complete proof can be seen here. The problem can be generalised to evaluate

\begin{equation}
\int_{-\infty}^\infty \exp\left(-\frac{(x^2-bx-1)^2}{ax^2}\right)\ dx = \sqrt{a\pi}
\end{equation}

Proof:
It's easy to see that $\dfrac{(x^2-bx-1)^2}{ax^2}=\dfrac{1}{a}\left(x-x^{-1}-b\right)^2$, then
\begin{align}
\int_{-\infty}^\infty \exp\left(-\frac{(x^2-bx-1)^2}{ax^2}\right)\ dx &=\int_{-\infty}^\infty \exp\left(-\frac{(x-x^{-1}-b)^2}{a}\right)\ dx\\
&=\int_{-\infty}^\infty \exp\left(-\frac{(x-b)^2}{a}\right)\ dx\\
&=\int_{-\infty}^\infty \exp\left(-\frac{y^2}{a}\right)\ dy\\
&=\sqrt{a}\int_{-\infty}^\infty \exp\left(-z^2\right)\ dz\\
&=\sqrt{a\pi}
\end{align}
Therefore
\begin{equation}
\int_{-\infty}^\infty \exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx = \sqrt{611\pi}
\end{equation}