Gaussian integral using integration by parts

[mbigras]

Homework Statement

Show in detail that:
$$\sigma_{x}^{2} = \int_{-\infty}^{\infty} (x -\bar{x})^{2} \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-X)^{2}}{2\sigma^{2}}} = \sigma^{2}$$
where,
$$G_{X,\sigma}(x) = \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-X)^{2}}{2\sigma^{2}}}$$

Homework Equations

$$\int u dv = uv -\int v du$$

The Attempt at a Solution

There are some hints that are given. Replace $\bar{x}$ with $X$ (according to the text, this is because they are equal after many trials). make the substitutions $x-X=y$ and $y/\sigma=z$. Integrate by parts to obtain the result.

After following the hints I get the following integral:
$$\frac{\sigma}{\sqrt{2\pi}} \int_{\infty}^{\infty} z^{2}e^{-\frac{z^{2}}{2}}dz$$
How would one go about using integration by parts for this integral?

 

[D H]

What have you tried?

BTW, your problem statement doesn't look right. That $(x-\bar x)$ should be $(x-\bar x)^2$. You also have a problem in your final integral.

 

[brmath]

If the problem is what you asked, you want to integrate by parts twice, taking down the $x^2$ one degree at each step.

 

[D H]

You only need to integrate by parts once if you know the value of $\int_{-\infty}^{\infty} \exp(x^2) \, dx$ (or of $\int_{-\infty}^{\infty} \exp(x^2/2) \, dx\,\,$).

 

[brmath]

You only need to integrate by parts once if you know the value of $\int_{-\infty}^{\infty} \exp(x^2) \, dx$ (or of $\int_{-\infty}^{\infty} \exp(x^2/2) \, dx\,\,$).

as it happens I don't know, altho I think I have done them over and over -- you'd think I would learn.

 

[mbigras]

Working with integration by parts I've tried some different combinations of $u$ and $dv$:

$$\frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\<br /> u=z^{2}\\<br /> du = 2 z dz\\<br /> dv = e^{-\frac{z^{2}}{2}}dz\\<br /> v = 2.5\\<br /> 2.5 z^{2} |_{-\infty}^{\infty} - \int 2.5 *2z dz\\<br /> =0$$

Trying to recognize some symetry:
$$\frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\<br /> \frac{2\sigma}{\sqrt{2\pi}} \int_{0}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\<br /> u = z^{2}\\<br /> dv = e^{-\frac{z^{2}}{2}} dz$$
seems to blow up to infinity as does setting $u=z$ and $dv = ze^{-\frac{z^{2}}{2}}dz$

What I would like is a hint about choosing my $u$ and $dv$. Or it seems like there's something I'm missing here.

 

[vela]

Don't use limits when integrating dv. The latter choice of u=z and $dv=ze^{-z^2/2}\,dz$ is the way to go.

 

[mbigras]

$$\frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^{2}}{2}}dz\\<br /> u = z\\<br /> du = dz\\<br /> \\<br /> dv = z e^{-\frac{z^{2}}{2}} dz\\<br /> v = ?\\<br /> \\$$
working to find v and using a strange result that I found on the boards mentioned by nicksauce near the middle of the page link
$$\int z e^{-\frac{z^{2}}{2}} dz\\<br /> u = z\\<br /> du = dz\\<br /> \\<br /> dv = e^{-\frac{z^{2}}{2}} dz \\<br /> v = \frac{\sqrt{\pi}}{\sqrt{2}}\\<br /> \\<br /> z\frac{\sqrt{\pi}}{\sqrt{2}} - \int \frac{\sqrt{\pi}}{\sqrt{2}} dz = 0\\$$
plugging back into above
$$u = z\\<br /> du = dz\\<br /> v = 0\\<br /> dv = z e^{-\frac{z^{2}}{2}} dz\\<br /> \\$$
It seems like everything just keeps going to 0 or infinity.

 

[vela]

Don't use limits when integrating dv. Calculate the indefinite integral.

 

[mbigras]

How would I handle $\int e^{-\frac{z^{2}}{2}} dz$ if its an indefinite integral?

 

[vela]

You don't.

 

[mbigras]

$$\sigma_{x}^{2} = \int_{-\infty}^{\infty} (x-X)^2 \frac{1}{\sigma\sqrt{2\pi}} e^{-(x-X)^{2}/2\sigma^{2}} \\<br /> <br /> = \int_{-\infty}^{\infty} y^2 \frac{1}{\sigma\sqrt{2\pi}} e^{-y^{2}/2\sigma^{2}} \\<br /> <br /> = \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{y^2}{\sigma^{2}}e^{-y^{2}/2\sigma^{2}} \\<br /> <br /> = \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty} z^{2}e^{-z^{2}/2} \\$$

then IBP and a u-sub to find $v$ by evaluating $dv$ as an indefinite integral.
$$u = z\\<br /> du = dz\\<br /> \\<br /> v = -e^{-\frac{z^{2}}{2}}\\<br /> dv = z e^{-\frac{z^{2}}{2}} dz\\<br /> \\<br /> \\<br /> -ze^{-z^2/2} |_{-\infty}^{\infty} + \int_{-\infty}^{\infty} e^{-z^{2}/2}dz\\<br /> = \sqrt{2\pi}$$
going back up to the top
$$\sigma_{x}^{2} = \sigma$$

Can you see where I'm missing the extra $\sigma$? Also, what is the point of doing integration by parts if I just end up with the integral $\int_{-\infty}^{\infty} e^{-z^{2}/2}dz=\sqrt{2\pi}$, but I had to use wolfram to evaluate it, when at the very beginning I could have used wolfram to get $\int_{-\infty}^{\infty} z^{2}e^{-z^{2}/2}=\sqrt{2\pi}$.

Is there a trick to evaluating $\int_{-\infty}^{\infty} e^{-z^{2}/2}dz$ by hand?

 

[D H]

Can you see where I'm missing the extra $\sigma$?

You left out the dx, dy, and dz in your integrals. Don't do that.

 

[Ray Vickson]

Working with integration by parts I've tried some different combinations of $u$ and $dv$:

$$\frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\<br /> u=z^{2}\\<br /> du = 2 z dz\\<br /> dv = e^{-\frac{z^{2}}{2}}dz\\<br /> v = 2.5\\<br /> 2.5 z^{2} |_{-\infty}^{\infty} - \int 2.5 *2z dz\\<br /> =0$$

Trying to recognize some symetry:
$$\frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\<br /> \frac{2\sigma}{\sqrt{2\pi}} \int_{0}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\<br /> u = z^{2}\\<br /> dv = e^{-\frac{z^{2}}{2}} dz$$
seems to blow up to infinity as does setting $u=z$ and $dv = ze^{-\frac{z^{2}}{2}}dz$

What I would like is a hint about choosing my $u$ and $dv$. Or it seems like there's something I'm missing here.

Using $u = z$ and $dv = z\: \exp(-z^2/2) \, dz$ is exactly the way to go. I cannot get it to "blow up" as you claim.

 

[mbigras]

You left out the dx, dy, and dz in your integrals. Don't do that.

And there is my missing sigma! Thank you DH.