1
$\begingroup$

I want to check if the following improper integral converges or diverges:

$$\int_0^{\infty} x^2e^{-x^2}\space dx$$

Rewriting the integrand:

$$\int_0^{\infty}-\frac{1}{2}x(-2x e^{-x^2}) \space dx$$

Integrating by parts:

$u=-\frac{1}{2}x; \space u'=-\frac{1}{2}; \space v=e^{-x^2}; \space v'=-2xe^{-x^2}$

$$\implies\int_0^{\infty} -\frac{1}{2}x(-2x e^{-x^2}) \space dx=((-\frac{1}{2}xe^{-x^2})_0^\infty-(-\frac{1}{2}e^{-x^2})_0^{\infty}) $$

$$=(\lim_{N \to \infty}(Ne^{-N^2})-0)-(\lim_{N \to \infty}(-\frac{1}{2}e^{-N^2})-(-\frac{1}{2}))=\color{red}{-\frac{1}{2}}$$

As far as I know the answer should be $\frac{1}{4}$ but I can't figure out where I am making a mistake.

$\endgroup$
2
  • 4
    $\begingroup$ When you integrate by parts you should end up with another integral... $\int f \frac{dg}{dx} dx = f g - \int \frac{df}{dx} g d x$ $\endgroup$ Jul 20, 2015 at 11:48
  • $\begingroup$ Okay I see where I messed up now. However, how do you integrate $e^{-x^2}$. $\endgroup$
    – qmd
    Jul 20, 2015 at 11:55

3 Answers 3

5
$\begingroup$

$$ \begin{eqnarray} \int_0^\infty x^2 e^{-x^2} dx &=& \Big[\frac{1}{2}x e^{-x^2}\Big]_0^\infty x^2 + \frac{1}{2} \int_0^\infty e^{-x^2} dx\\ &=& \frac{1}{4} \int_{-\infty}^\infty e^{-x^2} dx\\ &=& \frac{\sqrt{\pi}}{4} \end{eqnarray} $$


Your question in the comment - how to integrate $e^{-x^2}$... $$ \begin{eqnarray} \int_{-\infty}^\infty e^{-x^2} dx &=& \sqrt{ \left( \int_{-\infty}^\infty e^{-x^2} dx \right)^2 }\\ &=& \sqrt{ \int_{-\infty}^\infty e^{-x^2} dx \int_{-\infty}^\infty e^{-y^2} dy}\\ &=& \sqrt{ \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2-y^2} dx dy}\\ &=& \sqrt{ \int_{0}^\infty \int_0^{2\pi} e^{-r^2} r d\phi dr }\\ &=& \sqrt{ \pi \int_{0}^\infty e^{-r^2} d\phi d(r^2) }\\ &=& \sqrt{ \pi } \end{eqnarray} $$

$\endgroup$
5
  • $\begingroup$ Thanks. Can you explain how you went from $\frac{1}{2}\int_0^{\infty}e^{-x^2}$ to $\frac{1}{4}\int_{-\infty}^{\infty}e^{-x^2}$? $\endgroup$
    – qmd
    Jul 20, 2015 at 12:00
  • $\begingroup$ @SuH $e^{-x^2}$ is an even function so $\int_{-a}^{a} e^{-x^2}=2\int_0^a e^{-x^2}$ $\endgroup$
    – Hasan Saad
    Jul 20, 2015 at 12:02
  • $\begingroup$ @HasanSaad Thanks! That makes sense. $\endgroup$
    – qmd
    Jul 20, 2015 at 12:03
  • $\begingroup$ Thanks for editing your to include the answer to my comment. $\endgroup$
    – qmd
    Jul 20, 2015 at 12:06
  • 1
    $\begingroup$ @SuH - you are welcome! Thank you for the 'best answere' :) $\endgroup$ Jul 20, 2015 at 12:07
3
$\begingroup$

There is a small mistake in the integration by parts.

$$F(x) =\int x^2e^{-x^2}\space dx=-\frac12e^{-x^2} + \int \frac12 e^{-x^2}\space dx=-\frac12e^{-x^2} + \frac{\sqrt{\pi} \cdot \mathrm{erf}(x)}{4}+C$$

Since we have

$$\lim_{x\to\infty} -\frac12e^{-x^2} + \frac{\sqrt{\pi} \cdot \mathrm{erf}(x)}{4} = \lim_{x\to\infty} -\frac12e^{-x^2} + \frac{\sqrt{\pi}}{4} \lim_{x\to\infty} \mathrm{erf}(x) = \frac{\sqrt{\pi}}{4}$$

and $F(0)=0$, the result is $\frac{\sqrt{\pi}}{4}$.

$\endgroup$
2
  • $\begingroup$ Thanks! This integral was "harder" than I thought. I have never seen the error function before. $\endgroup$
    – qmd
    Jul 20, 2015 at 11:59
  • 1
    $\begingroup$ @SuH It is good to know some properties of the error function and the incomplete gamma functions once you get further in calculus. $\endgroup$
    – wythagoras
    Jul 20, 2015 at 12:03
1
$\begingroup$

$$\begin{eqnarray*}\int_{0}^{+\infty}-\frac{x}{2}(-2x e^{-x^2})\,dx &=& \left.-\frac{x}{2} e^{-x^2}\right|_{0}^{+\infty}+\frac{1}{2}\int_{0}^{+\infty}e^{-x^2}\,dx\\&=&\frac{1}{2}\cdot\frac{\sqrt{\pi}}{2}=\color{red}{\frac{\sqrt{\pi}}{4}}\end{eqnarray*}$$ or: $$ \begin{eqnarray*}\int_{0}^{+\infty}x^2 e^{-x^2}\,dx = \frac{1}{2}\int_{0}^{+\infty}z^{1/2}e^{-z}\,dz = \frac{\Gamma(3/2)}{2} = \frac{\Gamma(1/2)}{4}=\color{red}{\frac{\sqrt\pi}{4}}.\end{eqnarray*}$$

$\endgroup$
2
  • $\begingroup$ Thank you. I didn't/don't know how to integrate $e^{-x^2}$. Is that some sort of special integral I should know? $\endgroup$
    – qmd
    Jul 20, 2015 at 12:01
  • 1
    $\begingroup$ @SuH: definitely yes. Just look at here. $\endgroup$ Jul 20, 2015 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.