# Gamma function for mathphys course

1. Nov 27, 2005

### relatively_me

The given problem is this:

Solve using the gamma function

$\int_0^{\infty}\sqrt{x}\exp{^{-x}}{ dx}$

My problem is that I don't know how to use the gamma function. It doesn't make sense to me...any insight would be helpful.

Last edited: Nov 27, 2005
2. Nov 27, 2005

### relatively_me

I don't know what's wrong with my code...the upper limit should be

$\infty$

3. Nov 27, 2005

### Physics Monkey

The gamma function can be defined as
$$\Gamma(z) = \int^\infty_0 x^{z-1} e^{-x} \, dx,$$
so find the value z that makes this integral look like yours. I assume this is what is meant by "solve using the gamma function". You can then look up the result in a table.

4. Nov 27, 2005

### shmoe

You might want to review what you do know about the gamma function.

You should be able to write that integral as Gamma(s) for some value of s. What properties of Gamma do you know? Have you evaluated Gamma at any non-integral points before?

5. Nov 27, 2005

### benorin

Hint: $$\int_0^{\infty}\sqrt{x}\exp{^{-x}}{ dx}=\int_0^{\infty}x^{\frac{3}{2}-1}\exp{^{-x}}{ dx}$$

6. Nov 27, 2005

### relatively_me

Well, if z=3/2, then, according to the table I found,

$$\Gamma(\frac{3}{2}) = \int_0^{\infty}x^{\frac{3}{2}-1}\exp{^{-x}}{ dx} = 8.386226 \times 10^{-1}$$

7. Nov 27, 2005

### relatively_me

No need for tables, you can find $$\Gamma(3/2)$$ exactly with the relation:
$$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin \pi s}$$