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Could someone give me some hint or show me how to calculate this integration? $$\huge{\displaystyle\int_{-\infty}^\infty}z^2e^{-\frac{z^2}{2}}\ dz$$Thanks in advance.

Echo0831
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3 Answers3

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Try integration by parts, differentiating $z$ and integrating $ze^{-\frac{z^2}{2}}$. You'll need the fact that $$ \int_{-\infty}^{\infty}e^{-\frac{z^2}{2}}\;dz=\sqrt{2\pi}$$

carmichael561
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Let the function $I(a)$ be the integral given by

$$\begin{align} I(a)&=\int_{-\infty}^\infty e^{-az^2}\,dz\\\\ &=\sqrt{\frac{\pi}{a}} \end{align}$$

Then, note that $I'(1/2)$ is

$$\begin{align} I'(1/2)&=-\int_{-\infty}^\infty z^2e^{-\frac12 z^2}\,dz\\\\ &=-\sqrt{2\pi} \end{align}$$

Therefore, we find

$$\int_{-\infty}^\infty z^2e^{-\frac12 z^2}\,dz=\sqrt{2\pi}$$

NOTE: Legitimacy of Differentiating Under the Integral Sign

Here, we legitimize the differentiation under the integral. First, we form the difference quotient

$$\frac{I(a+h)-I(a)}{h}=2\int_0^\infty \left(-z^2e^{-az^2}\right)\frac{1-e^{-hz^2}}{hz^2}\,dz$$

Case 1: $h>0$

For $h>0$, we have

$$\left|\frac{1-e^{-hz^2}}{hz^2}\right|\le 1$$

Therefore, the Dominated Convergence Theorem guarantees that

$$\begin{align} \lim_{h\to 0^+}\frac{I(a+h)-I(a)}{h}&=2\int_0^\infty \left(-z^2e^{-az^2}\right)\lim_{h\to 0^+}\left(\frac{1-e^{-hz^2}}{hz^2}\right)\,dz\\\\ &=-2\int_0^\infty (-z^2e^{-az^2})\,dz \end{align}$$

Case 2: $h<0$

For $h<0$, we first split the integral of interest as

$$\frac{I(a+h)-I(a)}{h}=\int_0^{1\sqrt{|h|}} \left(-z^2e^{-az^2}\right)\frac{1-e^{-hz^2}}{hz^2}\,dz+\int_{1/\sqrt{|h|}}^\infty \left(-z^2e^{-az^2}\right)\frac{1-e^{-hz^2}}{hz^2}\,dz$$

For $z<1\sqrt{|h|}$, $\left|\frac{1-e^{-hz^2}}{hz^2}\right|\le e-1$ while for $z>1\sqrt{|h|}$ and $|h|<a/2$, $\left|\frac{1-e^{-hz^2}}{hz^2}\right|\le e^{\frac a2 z^2}$. Then, since

$$\left|\frac{1-e^{-hz^2}}{hz^2}\right|\le g(z)=\begin{cases}e-1&,z<1\sqrt{|h|}\\\\e^{\frac a2 z^2}&,z>1/\sqrt{|h|}\end{cases}$$

with $\int_0^\infty g(z)(-z^2e^{-az^2})\,dz<\infty$

the Dominated Convergence Theorem guarantees that

$$\begin{align} \lim_{h\to 0^-}\frac{I(a+h)-I(a)}{h}&=2\int_0^\infty \left(-z^2e^{-az^2}\right)\lim_{h\to 0^-}\left(\frac{1-e^{-hz^2}}{hz^2}\right)\,dz\\\\ &=-2\int_0^\infty (-z^2e^{-az^2})\,dz \end{align}$$ -______________________________

Inasmuch as the limits from right-hand and left-hand sides are equal, then

$$I'(a)=-\int_{-\infty}^\infty z^2e^{-az^2}\,dz$$

Mark Viola
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  • You're welcome! My pleasure. – Mark Viola May 30 '16 at 23:32
  • The integral is $\sqrt{2\pi}$ – Vivek Kaushik May 30 '16 at 23:33
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    @VivekKaushik I've edited – Mark Viola May 30 '16 at 23:35
  • -1. Differentiating under an improper integral implies exchanging three limits. Doing so without justification as if it was trivial is bad form, in my opinion. – Martin Argerami May 31 '16 at 06:50
  • Yes, I spend countless hours trolling the site and downvoting answers. – Martin Argerami May 31 '16 at 15:32
  • @MartinArgerami OK. Now, that was funny! – Mark Viola May 31 '16 at 15:40
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    Seriously, I think it is a bad answer if you don't justify it. It might be obvious to you that dominated convergence works; but in a question with no measure theory tag, I don't think it is reasonable to expect the OP to have a working knowledge of dominated convergence. And, more importantly, I think we shouldn't encourage "happy-go-lucky" math; here differentiation under the integral works, but it doesn't work all the time. It's like those cases where people apply L'Hôpital without caring to see if the conditions for it are satisfied. – Martin Argerami May 31 '16 at 15:52
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    @MartinArgerami I've taken your suggestion seriously and added an appendix in which I show the legitimacy of differentiating under the integral. I do appreciate your suggestion - only wish you would have given me a chance to amend before the down vote. -Mark – Mark Viola May 31 '16 at 15:58
  • Nice work (by the way, downvotes can be reversed, and I already did). Still, I wouldn't call that "straightforward..." ;) One minor caveat, is that dominated convergence requires sequences. This is not a big deal here, because if all sequential limits of Newton quotients converge to the same value, then the Newton quotient will converge to that value.. But, again, a "not-straightforward" part of the argument. – Martin Argerami May 31 '16 at 16:15
  • @MartinArgerami Martin, much appreciative! And yes, it isn't as straightforward as I had thought when originally posting. -Mark – Mark Viola May 31 '16 at 16:23
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Consider $$I=\int_{-\infty}^{\infty}x^2e^{\frac{-x^2}{2}}dx$$ $$I^2=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x^2y^2e^{\frac{-x^2-y^2}{2}}dydx$$

Convert to polar $x=r\cos(\theta),y=r\sin(\theta)$ so that $\frac{\partial(x,y)}{\partial(r,\theta)}=r$

The double integral becomes: $$I^2=\int_{0}^{2\pi}\int_{-\infty}^{\infty}r^5(\cos(\theta)\sin(\theta))^2e^{\frac{-r^2}{2}}drd\theta=\int_{0}^{2\pi}(\cos(\theta)\sin(\theta))^2\left(\int_{-\infty}^{\infty}r^5e^{\frac{-r^2}{2}}dr\right)d\theta$$

For the inner integral use integration by parts $u=r^4,dv=re^{\frac{-r^2}{2}}dr$. You will have to do this repeatedly to get the antiderivative. Then evaluate the antiderivative at the endpoints to get $8$ for the inner integral.

Now we have: $$I^2=\int_{0}^{2\pi}8(\cos(\theta)\sin(\theta))^2 d\theta=\int_{0}^{2\pi}2(\sin(2\theta))^2 d\theta=\int_{0}^{2\pi}1-\cos(4\theta) d\theta=2\pi$$

So $$I=\sqrt{2\pi}$$

Vivek Kaushik
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