# 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.

• I have a strong feeling that you can use Feynman's trick on this, but I am still trying to find the correct one. – UserX Sep 29 '14 at 7:20
• How can you be sure that $\sqrt{a\pi}$ is not just a good approximation ? Could you produce some values with more digits ? – Claude Leibovici Sep 29 '14 at 7:34
• @ClaudeLeibovici I'm not entirely sure if it's exactly $\sqrt{a\pi}$, but for every value of $a$ I've tried, WolframAlpha always gives a number very close to $\sqrt{a\pi}$. I was hoping this fact might help with solving the integral. – user137794 Sep 29 '14 at 7:40
• Working with very high precision, I can confirm that the result is exactly $\sqrt{a\pi}$ what you would get with $$\int_{-\infty}^\infty\exp\left(-\frac{x^2}{a}\right)\ dx=\sqrt{a\pi}$$ – Claude Leibovici Sep 29 '14 at 8:12

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}

• This is a new technique for me (+1), I think it can be generalised. If so, what is the condition? Say $u(x)=\dfrac{px^2+qx+r}{sx}$ – Anastasiya-Romanova 秀 Sep 30 '14 at 10:44
• @Anastasiya-Romanova algebraically, $pr < 0$. Geometrically, this ensure the monotonicity of the $u(x)$ and $u$ cover $(-\infty,\infty)$ twice. – achille hui Sep 30 '14 at 12:47
• +1 for its potent to be generalized (to a higher degree, maybe?). – Sangchul Lee Sep 30 '14 at 19:43
• @sos440 the generalization is essentially the "Added" part of orangeskid answer. – achille hui Sep 30 '14 at 19:59

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

• That is remarkably simple and straightforward. +1 – Tom-Tom Sep 29 '14 at 9:56
• Would you be able to elaborate on how you concluded this? – user137794 Sep 29 '14 at 11:45
• @user137794: Added more detail. – Orest Bucicovschi Sep 30 '14 at 6:43
• @Tom-Tom: Thanks! Added more detail. – Orest Bucicovschi Sep 30 '14 at 16:36
• For $f(x) = e^{-\frac{x^2}{a}}$ the function $f(\phi(x))$ is still smooth. – Orest Bucicovschi Sep 30 '14 at 20:19

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}

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

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

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

• Nice generalization! It's interesting, that $b$ does not matter, on the other hand $-1$ is how important! – user153012 Oct 1 '14 at 12:07