3
$\begingroup$

How does one evaluate $$\int_{-\infty}^\infty x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} \ dx ?$$

The result is $1$ and it corresponds to $E[X^2]$, where $X$ is a random variable with $X\sim\mathcal{N}(0,1)$. I have tried to do some substituions and I've tried integration by parts but didn't succeed to integrate it. With the integration by parts I ended up with a harder integral in both cases and I couldn't find a good substitution.

$\endgroup$
6
  • $\begingroup$ Do you have available the special value $\int_{-\infty}^{\infty} e^{-t^2} dt = \sqrt{\pi}$? $\endgroup$ Jan 4, 2016 at 16:11
  • $\begingroup$ No. But if it shouldn't work otherwise you may use it. It would be nice to know how it works that way anyways. $\endgroup$
    – ndrizza
    Jan 4, 2016 at 16:13
  • 4
    $\begingroup$ What about $Var(X)=E(X^2)-\left( E(X) \right) ^2$ ? $\endgroup$ Jan 4, 2016 at 16:14
  • $\begingroup$ @callculus yes you may use that fact. I see, that would make things very simple: $1=E(X^2)-0^2 \ \Longrightarrow E(X^2) = 1$. Thanks for that one! $\endgroup$
    – ndrizza
    Jan 4, 2016 at 16:20
  • $\begingroup$ That was my intention. $\endgroup$ Jan 4, 2016 at 16:21

6 Answers 6

5
$\begingroup$

Notice, integral can be usually computed using Laplace transform as follows, $$\int_{-\infty}^{\infty}x^2\frac{1}{\sqrt{2\pi }}e^{-\frac{x^2}{2}}\ dx $$ By symmetry of even function: $f(-x)=f(x)$, $$=2\frac{1}{\sqrt{2\pi }}\int_{0}^{\infty}x^2e^{-\frac{x^2}{2}}\ dx $$$$=\sqrt{\frac{2}{\pi }}\int_{0}^{\infty}x^2e^{-\frac{x^2}{2}}\ dx $$ let $\frac{x^2}{2}=u\implies x\ dx=du$ or $dx=\frac{1}{\sqrt{2u}}\ du=\frac{1}{\sqrt2}\frac{ du}{\sqrt u}$, $$=\sqrt{\frac{2}{\pi }}\int_{0}^{\infty}(2u)e^{-u} \frac{1}{\sqrt2}\frac{ du}{\sqrt u}$$ $$=\frac{2}{\sqrt {\pi}}\int_{0}^{\infty}u^{1/2}e^{-u}\ du$$ using Laplace transform: $\color{blue}{\int_0^{\infty}t^ne^{-st}\ dt=\frac{\Gamma(n+1)}{s^{n+1}}}$, $$=\frac{2}{\sqrt {\pi}}\left[\frac{\Gamma\left(1+\frac{1}{2}\right)}{s^{1+\frac{1}{2}}}\right]_{s=1}$$ $$=\frac{2}{\sqrt {\pi}}\left[\frac{\frac{1}{2}\Gamma\left(\frac{1}{2}\right)}{(1)^{3/2}}\right]$$ $$=\frac{2}{\sqrt {\pi}}\left[\frac{1}{2}\sqrt \pi\right]=\color{red}{1}$$

$\endgroup$
4
  • $\begingroup$ Thanks. But I don't know how to do a Laplace transform. Maybe there is not even a way to do it without it. $\endgroup$
    – ndrizza
    Jan 4, 2016 at 16:19
  • $\begingroup$ There may be other methods of doing this but it is one of the easiest & simpliest methods $\endgroup$ Jan 4, 2016 at 16:20
  • 1
    $\begingroup$ @ndrizza Laplace Transforms typically are covered in a differential equations course, you could take a look at that or simplicity, you can google "Laplace Transform Table" $\endgroup$
    – Adam
    Jan 4, 2016 at 16:22
  • $\begingroup$ Alright. I'll have a look at it. Thank you! $\endgroup$
    – ndrizza
    Jan 4, 2016 at 16:22
2
$\begingroup$

Here is another solution

starting with $$\int_{-\infty}^{\infty} e^{-t^2} dt = \sqrt{\pi}$$

and with simple substitution generalizing to (for $u>0$)

$$f(u) = \int_{-\infty}^{\infty} e^{-ut^2} dt = \sqrt{\frac{\pi}u}$$

Now, take the derivative w.r.t. u and evaluate at $u=1.$

$$\frac{d f(u)}{du} \biggr\rvert_{u=1} = \int_{-\infty}^{\infty} t^2e^{-t^2} dt = \frac12 \sqrt{\frac{\pi}{u^3}} = \frac{\sqrt{\pi}}2 $$

Or, evaluating at $u=\frac12$ will give

$$ \int_{-\infty}^{\infty} t^2e^{-\frac12 t^2} dt = \sqrt{2\pi} $$

$\endgroup$
1
$\begingroup$

We know that, if $F'=f$, then $$\int_a^bxf(x)dx=bF(b)-aF(a)-\int_a^b F(x)dx$$

Now, note that if $F(x)=e^{-x^2/2}$ then $F'(x)=f(x)=-xe^{-x^2/2}$.

So, $$\int_{-\infty}^\infty x^2e^{-\frac{x^2}2}dx=-\lim_{x\to\infty}\left(2xe^{-\frac{x^2}2}-\int_{-x}^xF(t)dt\right)=\sqrt{2\pi}$$

$\endgroup$
1
$\begingroup$

Here are two solutions that exploit the same two tools in different orders:

  1. It's a standard trick using elementary multivariable calculus and converting to polar coordinates (see, e.g., this answer) to show that the Gaussian integral has value $$\require{cancel}\int_{-\infty}^{\infty} e^{-t^2} dt = \sqrt{\pi} .$$ Now, applying integration by parts with $u = e^{-t^2}, dv = dt$ to the above integral gives $du = -2 t e^{-t^2} dt, v = t$ and hence $$\sqrt{\pi} = \cancelto{0}{\left.\left(e^{-t^2}\right)(t)\right\vert_{-\infty}^{\infty}} - \int_{-\infty}^{\infty} (t)(-2t e^{-t^2}) dt = 2 \int_{-\infty}^{\infty} t^2 e^{-t^2} dt,$$ and so $$\int_{-\infty}^{\infty} t^2 e^{-t^2} dt = \frac{\sqrt{\pi}}{2} .$$ Now, substituting $t = \frac{x}{\sqrt{2}}, dt = \frac{dx}{\sqrt{2}}$ in the integral gives $$\frac{1}{2 \sqrt{2}} \int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2}} dx = \frac{\sqrt{\pi}}{2},$$ and multiplying by $\frac{2}{\sqrt{\pi}}$ gives the desired result: $$\color{#bf0000}{\boxed{\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} x^2 e^{-\frac{x^2}{2}} dx = 1}} .$$
  2. We can apply the "multivariable polar" trick directly to the integral at hand rather than to the Gaussian integral: If one denotes $$I := \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} x^2 e^{-\frac{x^2}{2}} dx ,$$ we have $$ I^2 = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} x^2 e^{-\frac{x^2}{2}} dx \cdot \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} y^2 e^{-\frac{y^2}{2}} dy = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x^2 y^2 e^{-\frac{x^2 + y^2}{2}} dx \, dy . $$ Now, converting to polar coordinates gives \begin{align} I^2 &= \frac{1}{2 \pi} \int_0^{2 \pi} \int_0^{\infty} (r \cos \theta)^2 (r \sin \theta)^2 e^{-\frac{r^2}{2}} \cdot r \,dr \,d\theta \\ &= \frac{1}{2 \pi} \int_0^{2 \pi} \int_0^{\infty} r^5 e^{-\frac{r^2}{2}} \sin^2 \theta \cos^2 \theta \,dr \,d\theta \\ &= \frac{1}{2 \pi} \int_0^{2 \pi} \sin^2 \theta \cos^2 \theta \,d\theta \int_0^{\infty} r^5 e^{-\frac{r^2}{2}} dr . \end{align} Now, the first integral is a standard exercise that can be handled with double-angle identities and has value $\frac{\pi}{4}$. The second integral can be evaluated readily with the substitution $u = -\frac{r^2}{2}, du = -r \,dr$ and has value $8$, so $$I^2 = \frac{1}{2 \pi} \left(\frac{\pi}{4}\right) (8) = 1 .$$ On the other hand, the integrand in $I$ is everywhere nonnegative, and so again $I$ has value $$\color{#bf0000}{\boxed{\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} x^2 e^{-\frac{x^2}{2}} dx = 1}}$$ as claimed.
$\endgroup$
1
$\begingroup$

Let's assume an integral $\int_{-\infty}^{\infty}x^2\frac{1}{\sqrt{2\pi }}e^{-{\alpha x^2}}\ dx$ you have $\alpha=1/2$, then $$\int_{-\infty}^{\infty}x^2\frac{1}{\sqrt{2\pi }}e^{-{\alpha x^2}}\ dx=-\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi }}\frac{ \partial }{ \partial \alpha}e^{-\alpha x^2}\ dx=-\frac{1}{\sqrt{2\pi }}\frac{ \partial }{ \partial \alpha}\int_{-\infty}^{\infty}e^{-\alpha x^2}\ dx.$$ The letter is the well known Euler–Poisson integral, which is equal to $\sqrt{\frac{\pi}{\alpha}}$, so: $$-\frac{1}{\sqrt{2\pi }}\frac{ \partial }{ \partial \alpha}\int_{-\infty}^{\infty}e^{-\alpha x^2}\ dx=-\frac{1}{\sqrt{2\pi }}\frac{ \partial }{ \partial \alpha}\sqrt{\frac{\pi}{\alpha}}=\frac{1}{2\sqrt{2}}\alpha^{-3/2}.$$ After setting $\alpha=1/2$ you'll get the correct answer $\int_{-\infty}^{\infty}x^2\frac{1}{\sqrt{2\pi }}e^{-\frac{x^2}{2}}\ dx=1$.

$\endgroup$
1
$\begingroup$

\begin{align*} \int_{-\infty}^\infty x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} \; \mathrm{d}x&\overset{IBP}=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}x^2} \; \mathrm{d}x\\ &=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty} e^{-x^2} \; \mathrm{d}x\\ &= \frac{\sqrt{\pi}}{\sqrt{\pi}}\\ &=\boxed{1} \end{align*} Where $u=\frac{x}{\sqrt{2 \pi}}$ and $dv=xe^{-\frac{1}{2}x^2} \implies v=-e^{-\frac{1}{2}x^2}$

$\endgroup$

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.