How to compute $\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$

Question

$$\int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$$

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.

Answer 1

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

$$\begin{align} \int_{-\infty}^\infty e^{-u(x)^2/611} dx &= \left( \int_{-\infty}^{0^{-}} + \int_{0^{+}}^{+\infty}\right) e^{-u(x)^2/611} dx\\ &= \int_{-\infty}^{\infty} e^{-u^2/611}\left(\frac{dx_1}{du} + \frac{dx_2}{du}\right) du\\ &= \int_{-\infty}^{\infty} e^{-u^2/611} du\\ &= \sqrt{611\pi} \end{align} $$

Answer 2

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 $e^{-\frac{x^2}{a}}$.

Answer 3

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

$$\begin{align} &\int_{-\infty}^\infty \exp\left(-\frac1{611}\left((x-x^{-1})-13\right)^2\right)\ dx\\ =&\int_{-\infty}^\infty \exp\left(-\frac1{611}(x-13)^2\right)\ dx\\ =&\int_{-\infty}^\infty \exp\left(-\frac1{611}x^2\right)\ dx\\ =&\sqrt{611\pi} \end{align}$$

Answer 4

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}

Answer 5

Here is a solution without using Glasser's master theorem, which some of the other answers utilized.

$$I = \int_{-\infty}^\infty\exp\left(-\frac{(x^2-13x-1)^2}{611x^2}\right)\ dx$$

$$ = \int_{-\infty}^\infty \exp\left(-\frac{(x-\frac{1}{x}-13)^2}{611}\right)\ dx $$

$$ = \int_{-\infty}^0 \exp\left(-\frac{(x-\frac{1}{x}-13)^2}{611}\right)\ dx + \int_{0}^{\infty} \exp\left(-\frac{(x-\frac{1}{x}-13)^2}{611}\right)\ dx $$

$$ \overset{\color{red} x\color{red}\rightarrow \color{red}-\color{red}1\color{red}/\color{red}t }{=} \int_{0}^\infty \exp\left(-\frac{(-\frac{1}{t}+t-13)^2}{611}\right)\ \frac{dt}{t^2} + \int_{-\infty}^{0} \exp\left(-\frac{(-\frac{1}{t}+t-13)^2}{611}\right)\ \frac{dt}{t^2} $$

$$ = \int_{-\infty}^\infty \exp\left(-\frac{(-\frac{1}{t}+t-13)^2}{611}\right)\ \frac{dt}{t^2} $$

Therefore, $$ 2 I = \int_{-\infty}^\infty \exp\left(-\frac{(x-\frac{1}{x}-13)^2}{611}\right) (1+\frac{1}{x^2}) dx $$ $$ = \int_{-\infty}^{-1} + \int_{-1}^{0} + \int_{0}^1 + \int_{1}^\infty \left[\exp\left(-\frac{(x-\frac{1}{x}-13)^2}{611}\right) (1+\frac{1}{x^2}) dx\right]$$

$$ \overset{\color{red}( \color{red} x \color{red}-\color{red}1\color{red}/\color{red}x \color{red}) \rightarrow \color{red} z}{=} \int_{-\infty}^{0} \exp\left(-\frac{(z-13)^2}{611}\right) dz + \int_{0}^{\infty} \exp\left(-\frac{(z-13)^2}{611}\right) dz + \int_{-\infty}^{0} \exp\left(-\frac{(z-13)^2}{611}\right) dz + \int_{0}^{\infty} \exp\left(-\frac{(z-13)^2}{611}\right) dz $$

$$ = 2 \int_{-\infty}^{\infty} \exp\left(-\frac{(z-13)^2}{611}\right) dz $$

$$ = 2 \sqrt{611 \pi}$$

Hence, $$I = \sqrt{611 \pi}.$$

Here, we could replace $611$ with any positive number.