1
$\begingroup$

I was solving a problem of mean values, and I would like to solve and evaluate this integral: $$ \langle x^2\rangle=\int_{-\infty}^{\infty}\left(\frac{2\alpha}{\pi}\right)^{1/2}x^2 e^{-2\alpha(x-\beta)^2} dx $$ I have tried integrating by parts, but I get stuck there. The integration must be done with the constants $\alpha, \beta$, because it is a general case. Thank you in advance.

$\endgroup$
1
$\begingroup$

A statistical approach. You are trying to find $\mathbb{E}(X^2)$ where $X$ is a normal random variable with mean $\mathbb{E}(X)=\beta$ and variance $\mathbb{V}(X)={1\over 4\alpha}.$ Therefore $$\mathbb{E}(X^2)=\mathbb{V}(X)+\mathbb{E}(X)^2={1\over 4\alpha}+\beta^2.$$

$\endgroup$
1
$\begingroup$

Hint: Expand $x^2$ along the powers of $x-\beta$ first: $$x^2=(x-\beta)^2+2\beta(x-\beta)+\beta^2$$ and split into three integrals. Use integration parts for $$\int_{-\infty}^{\infty}(x-\beta)^2 \mathrm e^{-2\alpha(x-\beta)^2}\mathrm d\mkern1mu x.$$

$\endgroup$
1
$\begingroup$

$$ \langle x^2\rangle=\int_{-\infty}^{\infty}\left(\frac{2\alpha}{\pi}\right)^{1/2}x^2 e^{-2\alpha(x-\beta)^2} dx $$ $$ \langle x^2\rangle=\left(\frac{2\alpha}{\pi}\right)^{1/2}\int_{-\infty}^{\infty}(y+\beta)^2 e^{-2\alpha y^2} dy $$ $$ \langle x^2\rangle=2\left(\frac{2\alpha}{\pi}\right)^{1/2}\left(\int_{0}^{\infty}y^2 e^{-2\alpha y^2} dy + \beta^2\int_{0}^{\infty}e^{-2\alpha y^2} dy\right) $$ $$ \langle x^2\rangle=2\left(\frac{2\alpha}{\pi}\right)^{1/2}\left(-\dfrac12I'(\alpha) + \beta^2I(\alpha)\right), $$ where $$ I(\alpha) = \int_{0}^{\infty}e^{-2\alpha y^2} dy. $$ $$ I^2(\alpha) = \int_{0}^{\infty}e^{-2\alpha x^2} dx\cdot \int_{0}^{\infty}e^{-2\alpha y^2} dy = \int_{0}^{\infty} \int_{0}^{\infty}e^{-2\alpha (x^2 + y^2)} dxdy $$ Passing to polar coordinates: $$ I^2(\alpha) = \int_{0}^{\infty} \int_{0}^{2\pi}e^{-2\alpha r^2}r\,d\phi\, dr $$ $$ I^2(\alpha) = -\dfrac{2\pi}{4\alpha}e^{-2\alpha r^2}\biggr|_0^{\infty} = \dfrac{\pi}{2\alpha}. $$ $$ I(\alpha) = \left(\dfrac{\pi}{2\alpha}\right)^\frac12 $$ $$ I'(\alpha) = \left(\dfrac{\pi}{2\alpha}\right)^{\frac12}\left(-\dfrac1{2\alpha}\right) $$ $$ \langle x^2\rangle = 2\left(-\dfrac1{2\alpha}+\beta^2\right) $$ $$ \boxed{\langle x^2\rangle = 2\beta^2-\dfrac1\alpha} $$

$\endgroup$
0
$\begingroup$

Hint:

The integral $$ \int x^2e^{-2\alpha(x-\beta)^2}dx $$ with the substitution $$ \sqrt{2\alpha}(x-\beta)=t \quad \Rightarrow \quad dx=\frac{dt}{\sqrt{2\alpha}} $$ $$ x^2=\frac{t^2}{2\alpha}+\frac{2\beta t}{\sqrt{2\alpha}}+\beta^2 $$ becomes: $$ \frac{1}{\sqrt{(2\alpha)^3}}\int t^2e^{-t^2}dt+\frac{\beta}{\alpha}\int t e^{-t^2}dt+\frac{\beta^2}{\sqrt{2\alpha}} \int e^{-t^2}dt $$

The second integral can be easily solved by substitution: $$ \int t e^{-t^2}dt=-\frac{e^{-t^2}}{2} $$ and from this, integrating by parts the first integral we have: $$ \int t^2e^{-t^2}dt=\int t d\left(-e^{-t^2}\right)=\frac{-te^{-t^2}}{2}+\frac{1}{2} \int e^{-t^2}dt $$

Now we have to find the integral $\int e^{-t^2}dt$ that cannot be done with elementary functions. We use the error function, defined as: $$ \frac{d}{dx} \mbox{erf}(x)=\frac{2}{\sqrt{\pi}}e^{-x^2} $$ so: $$ \int e^{-t^2} dt=\frac{\sqrt{\pi}}{2}\mbox{erf}(t) $$ and $$ \int t^2e^{-t^2}dt=\frac{\sqrt{\pi}}{4}\mbox{erf}(t) -\frac{te^{-t^2}}{2} $$ Now you can back substitute and find the primitive of your function.

For the limits of the given integral you can use: $$ \lim_{x\to \infty}\mbox{erf}(x)=1 \qquad \lim_{x\to -\infty}\mbox{erf}(x)=-1 $$ that is a consequence of the Gaussian integral.

$\endgroup$

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.