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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.

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    $\begingroup$ When you integrate by parts you should end up with another integral... $\int f \frac{dg}{dx} dx = f g - \int \frac{df}{dx} g d x$ $\endgroup$ – johannesvalks Jul 20 '15 at 11:48
  • $\begingroup$ Okay I see where I messed up now. However, how do you integrate $e^{-x^2}$. $\endgroup$ – qmd Jul 20 '15 at 11:55
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$$ \begin{eqnarray} \int_0^\infty x^2 e^{-x^2} dx &=& \Big[\frac{1}{2}x e^{-x^2}\Big]_0^\infty x^2 + \frac{1}{2} \int_0^\infty e^{-x^2} dx\\ &=& \frac{1}{4} \int_{-\infty}^\infty e^{-x^2} dx\\ &=& \frac{\sqrt{\pi}}{4} \end{eqnarray} $$


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

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  • $\begingroup$ 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 '15 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 '15 at 12:02
  • $\begingroup$ @HasanSaad Thanks! That makes sense. $\endgroup$ – qmd Jul 20 '15 at 12:03
  • $\begingroup$ Thanks for editing your to include the answer to my comment. $\endgroup$ – qmd Jul 20 '15 at 12:06
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    $\begingroup$ @SuH - you are welcome! Thank you for the 'best answere' :) $\endgroup$ – johannesvalks Jul 20 '15 at 12:07
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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 $\frac{\sqrt{\pi}}{4}$.

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  • $\begingroup$ Thanks! This integral was "harder" than I thought. I have never seen the error function before. $\endgroup$ – qmd Jul 20 '15 at 11:59
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    $\begingroup$ @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 '15 at 12:03
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$$\begin{eqnarray*}\int_{0}^{+\infty}-\frac{x}{2}(-2x e^{-x^2})\,dx &=& \left.-\frac{x}{2} e^{-x^2}\right|_{0}^{+\infty}+\frac{1}{2}\int_{0}^{+\infty}e^{-x^2}\,dx\\&=&\frac{1}{2}\cdot\frac{\sqrt{\pi}}{2}=\color{red}{\frac{\sqrt{\pi}}{4}}\end{eqnarray*}$$ or: $$ \begin{eqnarray*}\int_{0}^{+\infty}x^2 e^{-x^2}\,dx = \frac{1}{2}\int_{0}^{+\infty}z^{1/2}e^{-z}\,dz = \frac{\Gamma(3/2)}{2} = \frac{\Gamma(1/2)}{4}=\color{red}{\frac{\sqrt\pi}{4}}.\end{eqnarray*}$$

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  • $\begingroup$ 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 '15 at 12:01
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    $\begingroup$ @SuH: definitely yes. Just look at here. $\endgroup$ – Jack D'Aurizio Jul 20 '15 at 12:06

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