How is the Gamma function used in solving for the expectation value of momentum?

[jg370]

Homework Statement

Hi,

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

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

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

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

Homework Equations

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

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

to

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

The Attempt at a Solution

I have reviewed the $$\Gamma function$$ 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

 

[dextercioby]

The function under the integral sign is symmetric wrt the origin 0, so the integral is twice the value it has at 0. So

$$I = \int_{-\infty}^{\infty} x^2 e^{-x^2} {} dx = 2 \int_{0}^{\infty} x^2 e^{-x^2} {} dx$$

But

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

Now make te substitution $x^2 = t$. You'll immediately get your result.

 

[phyzguy]

From the definition of the gamma function:
$$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt$$
Now make the substitution t=u^2, dt = 2u du, then
$$\Gamma(z) = \int_0^\infty u^{2(z-1)} e^{-u^2} 2 u du = 2\int_0^\infty u^{2z-1} e^{-u^2} du$$
Now, plugging in z=3/2, and recognizing that the resulting integral is even gives:
$$\Gamma(\frac{3}{2}) = 2\int_0^\infty u^2 e^{-u^2} du = \int_{-\infty}^\infty u^2 e^{-u^2} du$$

This is what you wanted to show.