7.3: Gamma Function

Another function that often occurs in the study of special functions is the Gamma function. We will need the Gamma function in the next section on Bessel functions.

For \(x>0\) we define the Gamma function as

\[\Gamma(x)=\int_{0}^{\infty} t^{x-1} e^{-t} d t, \quad x>0 \label{7.36} \]

The Gamma function is a generalization of the factorial function. In fact, we have

\[\Gamma(1)=1 \nonumber \]

and

\[\Gamma(x+1)=x \Gamma(x). \nonumber \]

The reader can prove this identity by simply performing an integration by parts. (See Problem 7.7.) In particular, for integers \(n \in Z^{+}\), we then have

\[\Gamma(n+1)=n \Gamma(n)=n(n-1) \Gamma(n-2)=n(n-1) \cdots 2 \Gamma(1)=n !. \nonumber \]

We can also define the Gamma function for negative, non-integer values of \(x\). We first note that by iteration on \(n \in Z^{+}\), we have

\[\Gamma(x+n)=(x+n-1) \cdots(x+1) x \Gamma(x), \quad x<0, \quad x+n>0 \text {. } \nonumber \]

Solving for \(\Gamma(x)\), we then find

\[\Gamma(x)=\dfrac{\Gamma(x+n)}{(x+n-1) \cdots(x+1) x}, \quad-n<x<0 \nonumber \]

Note that the Gamma function is undefined at zero and the negative integers.

Example 7.7. We now prove that 

\[\Gamma\left(\dfrac{1}{2}\right)=\sqrt{\pi} . \nonumber \]

This is done by direct computation of the integral:

\[\Gamma\left(\dfrac{1}{2}\right)=\int_{0}^{\infty} t^{-\dfrac{1}{2}} e^{-t} d t \nonumber \]

Letting \(t=z^{2}\), we have

\[\Gamma\left(\dfrac{1}{2}\right)=2 \int_{0}^{\infty} e^{-z^{2}} d z \nonumber \]

Due to the symmetry of the integrand, we obtain the classic integral

\[\Gamma\left(\dfrac{1}{2}\right)=\int_{-\infty}^{\infty} e^{-z^{2}} d z \nonumber \]

which can be performed using a standard trick. Consider the integral

\[I=\int_{-\infty}^{\infty} e^{-x^{2}} d x \nonumber \]

Then,

\[I^{2}=\int_{-\infty}^{\infty} e^{-x^{2}} d x \int_{-\infty}^{\infty} e^{-y^{2}} d y \nonumber \]

Note that we changed the integration variable. This will allow us to write this product of integrals as a double integral:

\[I^{2}=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d x d y . \nonumber \]

This is an integral over the entire \(x y\)-plane. We can transform this Cartesian integration to an integration over polar coordinates. The integral becomes

\[I^{2}=\int_{0}^{2 \pi} \int_{0}^{\infty} e^{-r^{2}} r d r d \theta \nonumber \]

This is simple to integrate and we have \(I^{2}=\pi\). So, the final result is found by taking the square root of both sides:

\[\Gamma\left(\dfrac{1}{2}\right)=I=\sqrt{\pi} . \nonumber \]

We have seen that the factorial function can be written in terms of Gamma functions. One can write the even and odd double factorials as

\[(2 n) ! !=2^{n} n !, \quad(2 n+1) ! !=\dfrac{(2 n+1) !}{2^{n} n !} \nonumber \]

In particular, one can write

\[\Gamma\left(n+\dfrac{1}{2}\right)=\dfrac{(2 n-1) ! !}{2^{n}} \sqrt{\pi} . \nonumber \]

Another useful relation, which we only state, is

\[\Gamma(x) \Gamma(1-x)=\dfrac{\pi}{\sin \pi x} . \nonumber \]