Gaussian integral using integration by parts

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  • #1
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Homework Statement


Show in detail that:
[tex]
\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}
[/tex]
where,
[tex]
G_{X,\sigma}(x) = \frac{1}{\sigma \sqrt{2 \pi}}e^{-\frac{(x-X)^{2}}{2\sigma^{2}}}
[/tex]


Homework Equations


[tex]
\int u dv = uv -\int v du
[/tex]



The Attempt at a Solution


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

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

Answers and Replies

  • #2
D H
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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.
 
  • #3
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If the problem is what you asked, you want to integrate by parts twice, taking down the ##x^2## one degree at each step.
 
  • #4
D H
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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\,\,##).
 
  • #5
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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.
 
  • #6
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Working with integration by parts I've tried some different combinations of [itex]u[/itex] and [itex]dv[/itex]:

[tex]
\frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\
u=z^{2}\\
du = 2 z dz\\
dv = e^{-\frac{z^{2}}{2}}dz\\
v = 2.5\\
2.5 z^{2} |_{-\infty}^{\infty} - \int 2.5 *2z dz\\
=0
[/tex]

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

What I would like is a hint about choosing my [itex]u[/itex] and [itex]dv[/itex]. Or it seems like there's something I'm missing here.
 
  • #7
vela
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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.
 
  • #8
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[tex]
\frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^{2}}{2}}dz\\
u = z\\
du = dz\\
\\
dv = z e^{-\frac{z^{2}}{2}} dz\\
v = ?\\
\\
[/tex]
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
[tex]
\int z e^{-\frac{z^{2}}{2}} dz\\
u = z\\
du = dz\\
\\
dv = e^{-\frac{z^{2}}{2}} dz \\
v = \frac{\sqrt{\pi}}{\sqrt{2}}\\
\\
z\frac{\sqrt{\pi}}{\sqrt{2}} - \int \frac{\sqrt{\pi}}{\sqrt{2}} dz = 0\\
[/tex]
plugging back into above
[tex]
u = z\\
du = dz\\
v = 0\\
dv = z e^{-\frac{z^{2}}{2}} dz\\
\\
[/tex]
It seems like everything just keeps going to 0 or infinity.
 
  • #9
vela
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Don't use limits when integrating dv. Calculate the indefinite integral.
 
  • #10
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How would I handle [itex] \int e^{-\frac{z^{2}}{2}} dz[/itex] if its an indefinite integral?
 
  • #11
vela
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You don't.
 
  • #12
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[tex]
\sigma_{x}^{2} = \int_{-\infty}^{\infty} (x-X)^2 \frac{1}{\sigma\sqrt{2\pi}} e^{-(x-X)^{2}/2\sigma^{2}} \\

= \int_{-\infty}^{\infty} y^2 \frac{1}{\sigma\sqrt{2\pi}} e^{-y^{2}/2\sigma^{2}} \\

= \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \frac{y^2}{\sigma^{2}}e^{-y^{2}/2\sigma^{2}} \\

= \frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty} z^{2}e^{-z^{2}/2} \\
[/tex]

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

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

Is there a trick to evaluating [itex]\int_{-\infty}^{\infty} e^{-z^{2}/2}dz[/itex] by hand?
 
  • #13
D H
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Can you see where I'm missing the extra [itex]\sigma[/itex]?
You left out the dx, dy, and dz in your integrals. Don't do that.
 
  • #14
Ray Vickson
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Working with integration by parts I've tried some different combinations of [itex]u[/itex] and [itex]dv[/itex]:

[tex]
\frac{\sigma}{\sqrt{2\pi}} \int_{-\infty}^{\infty} z^{2} e^{-\frac{z^2}{2}} dz\\
u=z^{2}\\
du = 2 z dz\\
dv = e^{-\frac{z^{2}}{2}}dz\\
v = 2.5\\
2.5 z^{2} |_{-\infty}^{\infty} - \int 2.5 *2z dz\\
=0
[/tex]

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

What I would like is a hint about choosing my [itex]u[/itex] and [itex]dv[/itex]. 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.
 
  • #15
61
2
You left out the dx, dy, and dz in your integrals. Don't do that.
And there is my missing sigma! Thank you DH.
 

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