How to compute $\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$ $$\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$$
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.
 A: Adding another solution owing to a friend of mine.
Through some algebra, the integral is equivalent to
$$\int_{-\infty}^\infty \exp\left(-\frac1{611}\left((x-x^{-1})-13\right)^2\right)\ dx$$
Then using the following identity
$$\int_{-\infty}^\infty f(x-x^{-1})\ dx = \int_{-\infty}^\infty f(x)\ dx$$
We have 
$$\begin{align}
&\int_{-\infty}^\infty \exp\left(-\frac1{611}\left((x-x^{-1})-13\right)^2\right)\ dx\\
=&\int_{-\infty}^\infty \exp\left(-\frac1{611}(x-13)^2\right)\ dx\\
=&\int_{-\infty}^\infty \exp\left(-\frac1{611}x^2\right)\ dx\\
=&\sqrt{611\pi}
\end{align}$$
A: 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}
A: Let $\displaystyle\;u(x) = \frac{x^2-13x-1}{x}\;$. As $x$ varies over $\mathbb{R}$, we have


*

*u(x) increases monotonically from $-\infty$ at $-\infty$ to $+\infty$ at $0^{-}$.

*u(x) increases monotonically from $-\infty$ at $0^{+}$ to $+\infty$ at $+\infty$.


This means as $x$ varies, $u(x)$ covered $(-\infty,\infty)$ twice.
Let $x_1(u) < 0$ and $x_2(u) > 0$ be the two roots of the equation for a given $u$:
$$u = u(x) = \frac{x^2-13x-1}{x} \quad\iff\quad x^2 - (13+u)x - 1 = 0$$
we have
$$x_1(u) + x_2(u) = 13 + u
\quad\implies\quad
\frac{dx_1}{du} + \frac{dx_2}{du} = 1.
$$
From this, we find
$$\begin{align}
\int_{-\infty}^\infty e^{-u(x)^2/611} dx
&= \left( \int_{-\infty}^{0^{-}} + \int_{0^{+}}^{+\infty}\right) e^{-u(x)^2/611} dx\\
&= \int_{-\infty}^{\infty} e^{-u^2/611}\left(\frac{dx_1}{du} + \frac{dx_2}{du}\right) du\\
&= \int_{-\infty}^{\infty} e^{-u^2/611} du\\
&= \sqrt{611\pi}
\end{align}
$$
A: HINT:
For $a$ fixed 
$\int_{-\infty}^\infty\exp\left(-\frac{(x^2+sx-b)^2}{a x^2}\right)\ dx$
is constant in $s$ and $b\ge 0$.
$\bf{Added:}$ 
The function $\frac{x^2 + s x - b}{x} = x - \frac{b}{x} + s$ invariates the Lebesgue measure as @achille hui: showed in his answer.   
Let $n \in \mathbb{N}$  $\alpha_1$, $\ldots$, $\alpha_n$ distinct real numbers, $\rho_1$, $\ldots$, $\rho_n$ $ >0$ and $\beta \in \mathbb{R}$.  The function 
$$\phi(x) = x - \sum_{i=1}^n \frac{\rho_i}{x- \alpha_i} -\beta $$
invariates the Lebesgue measure  on $\mathbb{R}$.
Lemma: For any $a \in \mathbb{R}$ the equation 
\begin{eqnarray*}
 x- \sum_{i=1}^n \frac{\rho_i}{x- \alpha_i} -\beta = u
\end{eqnarray*}
has $n+1$ distinct real root with sum $u + \sum_i \alpha_i + \beta $. 
Use Viete. 
Lemma: Let $I$ an interval in $\mathbb{R}$ of length $l$. Then the preimage $\phi^{-1} (I)$ is a union of $n+1$ disjoint intervals of total length $l$. 
Consequence:
$$\int_{\mathbb{R}} (f\circ \phi)\, d\,\mu = \int_{\mathbb{R}} f\  d\mu$$
Composing two rational maps that invariate the measure gets a third one.  They will have singularities in general. 
For $f(x) = e^{-\frac{x^2}{a}}$ the composition  $f (\phi(x))$  is still smooth due to the rapid decay at $\infty$ of $e^{-\frac{x^2}{a}}$.
