10.3: Gamma Function

For $\nu$ not an integer the recursion relation for the Bessel function generates something very similar to factorials. These quantities are most easily expressed in something called a Gamma-function, defined as

$$\Gamma(\nu) = \int_0^\infty e^{-t}t^{\nu-1} dt,\;\;\;\nu>0. \nonumber$$

Some special properties of $\Gamma$ function now follow immediately:

$$\begin{aligned} \Gamma(1) & = & \int_0^\infty e^{-t}dt = \left.-e^{-1}\right|^\infty_0 =1-e^{-\infty}=1\nonumber\\ \Gamma(\nu) & = & \int_0^\infty e^{-t} t^{\nu-1} dt = -\int_0^\infty \frac{de^{-t}}{dt} t^{\nu-1} dt\nonumber\\ &=& \left.- e^{-t} t^{\nu-1} \right|^\infty_0 +(\nu-1) \int_0^\infty e^{-t} t^{\nu-2} dt \end{aligned} \nonumber$$

The first term is zero, and we obtain

$$\Gamma(\nu) = (\nu-1)\Gamma(\nu-1) \nonumber$$

From this we conclude that

$$\Gamma(2)=1\cdot1=1,\;\Gamma(3)=2\cdot1\cdot1=2,\; \Gamma(4)=3\cdot2\cdot1\cdot1=2,\;\Gamma(n) = (n-1)!. \nonumber$$

Thus for integer argument the $\Gamma$ function is nothing but a factorial, but it also defined for other arguments. This is the sense in which $\Gamma$ generalises the factorial to non-integer arguments. One should realize that once one knows the $\Gamma$ function between the values of its argument of, say, 1 and 2, one can evaluate any value of the $\Gamma$ function through recursion. Given that $\Gamma(1.65) = 0.9001168163$ we find

$$\Gamma(3.65) = 2.65\times1.65\times0.9001168163 = 3.935760779 . \nonumber$$

Exercise $\PageIndex{1}$

Evaluate $\Gamma(3)$, $\Gamma(11)$, $\Gamma(2.65)$.

Answer

$2!=2$, $10!=3628800$, $1.65\times0.9001168163=1.485192746$.

We also would like to determine the $\Gamma$ function for $\nu<1$. One can invert the recursion relation to read

$$\Gamma(\nu-1) = \frac{\Gamma(\nu)}{\nu-1}, \nonumber$$ $\Gamma(0.7) = \Gamma(1.7)/0.7=0.909/0.7=1.30$.

What is $\Gamma(\nu)$ for $\nu<0$? Let us repeat the recursion derived above and find

$$\Gamma(-1.3) = \frac{\Gamma(-0.3)}{-1.3} = \frac{\Gamma(0.7)}{-1.3\times -0.3} = \frac{\Gamma(1.7)}{0.7\times-0.3\times -1.3}=3.33\;. \nonumber$$ This works for any value of the argument that is not an integer. If the argument is integer we get into problems. Look at $\Gamma(0)$. For small positive $\epsilon$ $$\Gamma(\pm\epsilon)= \frac{\Gamma(1\pm\epsilon)}{\pm\epsilon} =\pm \frac{1}{\epsilon}\rightarrow \pm \infty. \nonumber$$ Thus $\Gamma(n)$ is not defined for $n\leq0$. This can be easily seen in the graph of the $\Gamma$ function, Fig. $\PageIndex{1}$.

Figure $\PageIndex{1}$: A graph of the $\Gamma$ function (solid line). The inverse $1/\Gamma$ is also included (dashed line). Note that this last function is not discontinuous.

Finally, in physical problems one often uses $\Gamma(1/2)$,

$$\Gamma(\frac{1}{2}) = \int_0^\infty e^{-t}t^{-1/2} dt = 2\int_0^\infty e^{-t}dt^{1/2} = 2\int_0^\infty e^{-x^2}dx. \nonumber$$ This can be evaluated by a very smart trick, we first evaluate $\Gamma(1/2)^2$ using polar coordinates $$\begin{aligned} \Gamma(\frac{1}{2})^2 &=& 4 \int_0^\infty e^{-x^2}dx\int_0^\infty e^{-y^2}dy \nonumber\\ &=& 4 \int_0^\infty \int_0^{\pi/2} e^{-\rho^2}\rho d\rho d\phi = \pi.\end{aligned} \nonumber$$ (See the discussion of polar coordinates in Sec. 7.1.) We thus find $$\Gamma(1/2)=\sqrt{\pi},\;\;\Gamma(3/2)=\frac{1}{2}\sqrt{\pi},\;\;{\rm etc.} \nonumber$$