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I am taking an online class that has the following homework question. I don't want anyone to solve it for me, but it's been a long time since I've done any integration, and I can't figure out how to start solving it. If someone can point out a technique, I'd appreciate it.

The original question is

$\int_{-\infty}^\infty \tau^2 \lvert \frac{1}{\sqrt {2\pi}} e^{\frac{-\tau^2}{2}} \rvert^2 d\tau$

which I thought I correctly simplified to, but I make the silliest algebra errors at times...

$\frac{1}{2\pi} \int_{-\infty}^\infty \tau^2 e^{-\tau^2} d\tau$

The answer is one of $\frac{1}{2\sqrt{\pi}}$ or $\frac{1}{4\sqrt{\pi}}$, and I can't even figure out why there is a $\sqrt{\pi}$ in there if the original function was squaring the $\frac{1}{\sqrt{2\pi}}$

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    $\begingroup$ The integral of $e^{-x^2}$ over the real line is something like $\frac{1}{\sqrt{2\pi}}$, so that's probably where your root-pi is coming from. But you've got that awkward $\tau^2$ in there. I suggest you try integrating by parts, with $u = \tau$ and $dv = \tau e^{-\tau^2}$. $\endgroup$ – John Hughes Jan 19 '14 at 18:05
  • $\begingroup$ If you have not seen the trick of converting this single integral to double integral then I am not sure how you can integrate it. However, if you have studied normal distributions, then the integral is the formula for the variance with a normalizing factor of $\sqrt{2 \pi}$. If you have never seen either of them, then I am not sure there is a simple way to integrate it. $\endgroup$ – user44197 Jan 19 '14 at 18:11
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Getting there. As John said, you need to integrate by parts using his substitutions; that is, for the integral set $u = \tau$ and $dv = \tau e^{-\tau^2} d\tau$. Then, $du = d\tau$ and $v = -\dfrac{1}{2}e^{-\tau^2}$, so for $\int u(dv) = uv - \int v(du)$, we obtain

$$\begin{aligned} \dfrac{1}{2\pi} \int_{-\infty}^{\infty}\tau^2 e^{-\tau^2} d\tau &= \dfrac{1}{2\pi}\left(-\dfrac{1}{2}\tau e^{-\tau^2} + \dfrac{1}{2}\int_{-\infty}^{\infty} e^{-\tau^2}\,d\tau \right) \end{aligned}$$

Thus,

$$\begin{aligned} \dfrac{1}{2\pi}\left(-\dfrac{1}{2}\tau e^{-\tau^2} + \dfrac{1}{2}\int_{-\infty}^{\infty} e^{-\tau^2}\,d\tau \right) &= \left.-\dfrac{1}{4\pi}\tau e^{-\tau^2}\right\vert_{\tau = -\infty}^{\tau = \infty} + \dfrac{1}{4\pi}\int_{-\infty}^{\infty} e^{-\tau^2}\,d\tau\\ &= 0 + \dfrac{1}{4\pi}\sqrt{\pi}\\ &= \dfrac{1}{4\sqrt{\pi}} \end{aligned}$$

Notes:

  • $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$. See here for the derivation.
  • Sometimes, integration can be tricky here, like the one I showed you. Hopefully, you understand how such integral is evaluated.
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  • $\begingroup$ Thank you all so much. I'd started trying integration by parts but set it up incorrectly. I also didn't know $\int_{-\infty}^{\infty} e^{-x^2}dx$ is $\sqrt{\pi}$ $\endgroup$ – user929371 Jan 19 '14 at 18:24
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hey i think i help u some more i have a formula

$\int_{x=-\infty}^\infty{x^2.exp(-\alpha x^2)dx}=\sqrt\frac{pi}{4\alpha^3}$

here $\alpha=1$ now multiply r.h.s by $\frac{1}{2pi}$ & your is $\frac{1}{4\sqrt{pi}}$

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