# Use of he Gamma function

• jg370

## Homework Statement

jg370 said:
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

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.

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.