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

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

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Bernard Lacey
Let $$\displaystyle{u}{\left({x}\right)}={\frac{{{x}^{{2}}-{13}{x}-{1}}}{{{x}}}}$$. As x varies over $$\displaystyle{\mathbb{{{R}}}}$$, we have
u(x) increases monotonically from $$\displaystyle-\infty$$ at $$\displaystyle-\infty$$ to $$\displaystyle+\infty$$ at $$\displaystyle{0}^{{-{}}}$$.
u(x) increases monotonically from $$\displaystyle-\infty$$ at $$\displaystyle{0}^{+}$$ to $$\displaystyle+\infty$$ at $$\displaystyle+\infty$$
This means as x varies, u(x) covered $$\displaystyle{\left(-\infty,\infty\right)}$$ twice.
Let $$\displaystyle{x}_{{1}}{\left({u}\right)}{ < }{0}$$ and $$\displaystyle{x}_{{2}}{\left({u}\right)}{>}{0}$$ be the two roots of the equation for a given u:
$$\displaystyle{u}={u}{\left({x}\right)}={\frac{{{x}^{{2}}-{13}{x}-{1}}}{{{x}}}}\Leftrightarrow{x}^{{2}}-{\left({13}+{u}\right)}{x}-{1}={0}$$
we have
$$\displaystyle{x}_{{1}}{\left({u}\right)}+{x}_{{2}}{\left({u}\right)}={13}+{u}\Rightarrow{x}^{{2}}-{\left({13}+{u}\right)}{x}-{1}={0}$$
we have
$$\displaystyle{x}_{{1}}{\left({u}\right)}+{x}_{{2}}{\left({u}\right)}={13}+{u}\Rightarrow{\frac{{{\left.{d}{x}\right.}_{{1}}}}{{{d}{u}}}}+{\frac{{{\left.{d}{x}\right.}_{{2}}}}{{{d}{u}}}}={1}$$
From this, we find
$$\displaystyle{\int_{{-\infty}}^{\infty}}{e}^{{-{u}\frac{{\left({x}\right)}^{{2}}}{{611}}}}{\left.{d}{x}\right.}={\left({\int_{{-\infty}}^{{{0}^{{-}}}}}+{\int_{{{0}^{+}}}^{{+\infty}}}{e}^{{-{u}\frac{{\left({x}\right)}^{{2}}}{{611}}}}{\left.{d}{x}\right.}\right.}$$
$$\displaystyle={\int_{{-\infty}}^{\infty}}{e}^{{-{u}\frac{{\left({x}\right)}^{{2}}}{{611}}}}{\left({\frac{{{\left.{d}{x}\right.}_{{1}}}}{{{d}{u}}}}+{\frac{{{\left.{d}{x}\right.}_{{2}}}}{{{d}{u}}}}\right)}{d}{u}$$
$$\displaystyle={\int_{{-\infty}}^{\infty}}{e}^{{-{u}\frac{{\left({x}\right)}^{{2}}}{{611}}}}{d}{u}$$
$$\displaystyle=\sqrt{{{611}\pi}}$$
###### Not exactly what youâ€™re looking for?
Jonathan Burroughs
Adding another solution owing to a friend of mine.
Through some algebra, the integral is equivalent to
$$\displaystyle{\int_{{-\infty}}^{{\infty}}}{\exp{{\left(-{\frac{{{1}}}{{{611}}}}{\left({\left({x}-{x}^{{-{1}}}\right)}-{13}\right)}^{{2}}\right)}}}{\left.{d}{x}\right.}$$
Then using the following identity
$$\displaystyle{\int_{{-\infty}}^{{\infty}}}{f{{\left({x}-{x}^{{-{1}}}\right)}}}{\left.{d}{x}\right.}={\int_{{-\infty}}^{{\infty}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$
We have
$$\displaystyle{\int_{{-\infty}}^{\infty}}{\exp{{\left(-{\frac{{{1}}}{{{611}}}}{\left({\left({x}-{x}^{{-{1}}}\right)}-{13}\right)}^{{2}}\right)}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\int_{{-\infty}}^{{\infty}}}{\exp{{\left(-{\frac{{{1}}}{{{611}}}}{\left({x}-{13}\right)}^{{2}}\right)}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{-\infty}}^{{\infty}}}{\exp{{\left(-{\frac{{{1}}}{{{611}}}}{x}^{{2}}\right)}}}{\left.{d}{x}\right.}$$
$$\displaystyle=\sqrt{{{611}\pi}}$$
star233

Using identity:
$$\int_{-\infty}^{\infty}f(x-x^{-1})dx=\int_{-\infty}^{\infty} f(x)dx$$
The problem can be generalised to evaluate
$$\int_{-\infty}^{\infty}exp(-\frac{(x^2-bx-1)^2}{ax^2})dx=\sqrt{a\pi}$$
Proof:
It's easy to see that $$\frac{(x^2-bx-1)^2}{ax^2}=\frac{1}{a}(x-x^{-1}-b)^2$$, then
$$\int_{-\infty}^{\infty} exp(-\frac{(x^2-bx-1)^2}{ax^2})dx \\=\int_{-\infty}^{\infty} exp(-\frac{(x-x^{-1}-b)^2}{a})dx \\=\infty_{-\infty}^{\infty} exp(-\frac{(x-b)^2}{a})dx \\=\int_{-\infty}^{\infty} exp(-\frac{y^2}{a})dy \\=\sqrt{a}\int_{-\infty}^\infty exp(-z^2)dz \\=\sqrt{a\pi}$$
Therefore
$$\int_{-\infty}^{\infty} exp(-\frac{(x^2-13x-1)^2}{611x^2})dx=\sqrt{611\pi}$$