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Use of he Gamma function

  • Thread starter jg370
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  • #1
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Homework Statement


jg370 said:
Hi,

My question relates to the solution to a question regarding the expectation value of momentum, that is [tex]<p^2>[/tex].

As the solution unfold, we have the following two expressions:

[tex]-\frac{m*\omega*\hbar}{\sqrt Pi}\left[\int_{-\infty}^{\infty} \xi^2 *e^-\xi^2 d\xi-\int_{-\infty}^{\infty} e^-\xi^2 d\xi\right][/tex]

[tex]-\frac{m*\omega*\hbar}{\sqrt Pi}\left[\Gamma(\frac{3}{2}) -\sqrt Pi\right][/tex]


Homework Equations


My problem with the above, is that I do not understand how one gets from

[tex] \left[\int_{-\infty}^{\infty} \xi^2 *e^-\xi^2 d\xi\right][/tex]

to

[tex]\left[\Gamma(\frac{3}{2}) \right][/tex]


The Attempt at a Solution



I have reviewed the [tex]\Gamma function[/tex] and tried to make a conversion from the exponential function to the gamma function; this did not lead me to understand the relation ship involved in this case.

I hope that some one can help me with this.

I thank you for your kind assistance

jg370

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
dextercioby
Science Advisor
Homework Helper
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The function under the integral sign is symmetric wrt the origin 0, so the integral is twice the value it has at 0. So

[tex] I = \int_{-\infty}^{\infty} x^2 e^{-x^2} {} dx = 2 \int_{0}^{\infty} x^2 e^{-x^2} {} dx [/tex]

But

[tex] \Gamma\left(\frac{3}{2}\right) = \int_{0}^{\infty} t^{1/2} e^{-t} {} dt [/tex]

Now make te substitution [itex] x^2 = t [/itex]. You'll immediately get your result.
 
  • #3
phyzguy
Science Advisor
4,505
1,447
From the definition of the gamma function:
[tex]\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt[/tex]
Now make the substitution t=u^2, dt = 2u du, then
[tex]\Gamma(z) = \int_0^\infty u^{2(z-1)} e^{-u^2} 2 u du = 2\int_0^\infty u^{2z-1} e^{-u^2} du[/tex]
Now, plugging in z=3/2, and recognizing that the resulting integral is even gives:
[tex]\Gamma(\frac{3}{2}) = 2\int_0^\infty u^2 e^{-u^2} du = \int_{-\infty}^\infty u^2 e^{-u^2} du[/tex]

This is what you wanted to show.
 

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