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Gaussian Integral

  • #1
532
2
I see that the formula for this general integral is

[tex] \int^{+\infty}_{-\infty} x^{2}e^{-Ax^{2}}dx=\frac{\sqrt{\pi}}{2A^{3/2}}[/tex]

However, I am not getting this form with my function. I transformed the integral using integration by parts so that I could use another gaussian integral that I knew at the time.

[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx [/tex]

Let [tex] u = x^{2} \rightarrow du = 2x dx [/tex]

and

[tex] dv = e^{\frac{-2amx^{2}}{\hbar}}dx \rightarrow v = -\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}} [/tex]

Therefore,


[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx = \left x^{2}\left(-\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}}\right)\right|^{+\infty}_{-\infty}-\int^{+\infty}_{-\infty}\left(-\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}}\right)2xdx[/tex]


The middle term equals zero, so letting [tex]z =\left(\sqrt{2am/\hbar}\right)x \rightarrow dx= \left(\sqrt{\hbar/2am}\right)dz[/tex] gives


[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx = \frac{\hbar}{2am}\int^{+\infty}_{-\infty}e^{\frac{-2amx^{2}}{\hbar}}\right)dx=\left(\frac{\hbar}{2am}\right)^{3/2}\int^{+\infty}_{-\infty}e^{-z^{2}}dz =\left(\frac{\hbar}{2am}\right)^{3/2}\sqrt{\pi}[/tex]

which is not in the appropriate form--missing a factor of 1/2. I can't see where I am going wrong. Any thoughts?
 
Last edited:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
Take your function v and differentiate to get dv. It's not what you say it is. Use the product rule. You usually handle this problem by differentiating the integral of exp(-Ax^2) with respect to A.
 
  • #3
532
2
SOLVED

Yes, this is exactly right. Thanks dick. I forgot to consider the 'x'.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955
I see that the formula for this general integral is

[tex] \int^{+\infty}_{-\infty} x^{2}e^{-Ax^{2}}dx=\frac{\sqrt{\pi}}{2A^{3/2}}[/tex]

However, I am not getting this form with my function. I transformed the integral using integration by parts so that I could use another gaussian integral that I knew at the time.

[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx [/tex]

Let [tex] u = x^{2} \rightarrow du = 2x dx [/tex]

and

[tex] dv = e^{\frac{-2amx^{2}}{\hbar}}dx \rightarrow v = -\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}} [/tex]
No, this is incorrect. It looks like you are taking the derivative rather than the anti-derivative. [itex]e^{-x^2}[/itex] does NOT have an elementary anti-derivative.

I would recommend taking u= x, [itex]dv= x e^{\frac{-2amx^{2}}{\hbar}}dx [/quote] instead.

Therefore,


[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx = \left x^{2}\left(-\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}}\right)\right|^{+\infty}_{-\infty}-\int^{+\infty}_{-\infty}\left(-\frac{\hbar}{4amx}e^{\frac{-2amx^{2}}{\hbar}}\right)2xdx[/tex]


The middle term equals zero, so letting [tex]z =\left(\sqrt{2am/\hbar}\right)x \rightarrow dx= \left(\sqrt{\hbar/2am}\right)dz[/tex] gives


[tex] \int^{+\infty}_{-\infty} x^{2}e^{\frac{-2amx^{2}}{\hbar}}dx = \frac{\hbar}{2am}\int^{+\infty}_{-\infty}e^{\frac{-2amx^{2}}{\hbar}}\right)dx=\left(\frac{\hbar}{2am}\right)^{3/2}\int^{+\infty}_{-\infty}e^{-z^{2}}dz =\left(\frac{\hbar}{2am}\right)^{3/2}\sqrt{\pi}[/tex]

which is not in the appropriate form--missing a factor of 1/2. I can't see where I am going wrong. Any thoughts?
 

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