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

Mary Reyes 2022-01-05 Answered
How to compute
\(\displaystyle{\int_{{-\infty}}^{\infty}}{\exp{{\left(-{\frac{{{\left({x}^{{2}}-{13}{x}-{1}\right)}^{{2}}}}{{{611}{x}^{{2}}}}}\right)}}}{\left.{d}{x}\right.}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Bernard Lacey
Answered 2022-01-06 Author has 97 answers
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 you’re looking for?
Ask My Question
0
 
Jonathan Burroughs
Answered 2022-01-07 Author has 2167 answers
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
\(\displaystyle{\int_{{-\infty}}^{\infty}}{\exp{{\left(-{\frac{{{1}}}{{{611}}}}{\left({\left({x}-{x}^{{-{1}}}\right)}-{13}\right)}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{-\infty}}^{{\infty}}}{\exp{{\left(-{\frac{{{1}}}{{{611}}}}{\left({x}-{13}\right)}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle{\int_{{-\infty}}^{{\infty}}}{\exp{{\left(-{\frac{{{1}}}{{{611}}}}{x}^{{2}}\right)}}}{\left.{d}{x}\right.}\)
\(\displaystyle=\sqrt{{{611}\pi}}\)
0
star233
Answered 2022-01-11 Author has 0 answers

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

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2022-01-03
\(\displaystyle{\int_{{-\infty}}^{\infty}}{e}^{{-{\frac{{{1}}}{{{2}}}}{x}^{{2}}}}{\left.{d}{x}\right.}\) and \(\displaystyle{\int_{{-\infty}}^{\infty}}{x}^{{2}}{e}^{{-{\frac{{{1}}}{{{2}}}}{x}^{{2}}}}{\left.{d}{x}\right.}\)
how i compute these integrals via Gauss Integral?
asked 2021-12-30
How to evaluate:
\(\displaystyle{\int_{{-\infty}}^{\infty}}{\frac{{{\ln{{\left({x}^{{2}}+{1}\right)}}}}}{{{x}^{{2}}+{1}}}}{\left.{d}{x}\right.}\)
asked 2022-01-02
How can I find the principal value of the following?
\(\displaystyle{P}{V}{\int_{{-\infty}}^{\infty}}{\frac{{{e}^{{{i}{x}}}}}{{{x}{\left({x}^{{2}}+{x}+{1}\right)}}}}{\left.{d}{x}\right.}\)
asked 2022-01-03
How can I evaluate
\(\displaystyle{\int_{{-\infty}}^{\infty}}{\frac{{{\cos{{x}}}}}{{{\text{cosh}{{x}}}}}}{\left.{d}{x}\right.}\) and \(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{\sin{{x}}}}}{{{e}^{{x}}-{1}}}}{\left.{d}{x}\right.}\)
asked 2022-01-07
Compute the following integral:
\(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{e}^{{x}}{\sin{{\left({x}\right)}}}}}{{{x}}}}{\left.{d}{x}\right.}\)
asked 2022-01-05
Evaluate the integral \(\displaystyle{\int_{{-\infty}}^{{\infty}}}{\frac{{{\cos{{\left({x}\right)}}}}}{{{x}^{{2}}+{1}}}}{\left.{d}{x}\right.}\)
asked 2021-08-20
a) Write the sigma notation formula for the right Riemann sum \(\displaystyle{R}_{{{n}}}\) of the function \(\displaystyle{f{{\left({x}\right)}}}={4}-{x}^{{{2}}}\) on the interval \(\displaystyle{\left[{0},\ {2}\right]}\) using n subintervals of equal length, and calculate the definite integral \(\displaystyle{\int_{{{0}}}^{{{2}}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}\) as the limit of \(\displaystyle{R}_{{{n}}}\) at \(\displaystyle{n}\rightarrow\infty\).
(Reminder: \(\displaystyle{\sum_{{{k}={1}}}^{{{n}}}}{k}={n}\frac{{{n}+{1}}}{{2}},\ {\sum_{{{k}={1}}}^{{{n}}}}{k}^{{{2}}}={n}{\left({n}+{1}\right)}\frac{{{2}{n}+{1}}}{{6}}{)}\)
b) Use the Fundamental Theorem of Calculus to calculate the derivative of \(\displaystyle{F}{\left({x}\right)}={\int_{{{e}^{{-{x}}}}}^{{{x}}}}{\ln{{\left({t}^{{{2}}}+{1}\right)}}}{\left.{d}{t}\right.}\)
...