$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 10.3: Gamma Function

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

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

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

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}

The first term is zero, and we obtain $\Gamma(\nu) = (\nu-1)\Gamma(\nu-1)$

From this we conclude that

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

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

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

Exercise $$\PageIndex{1}$$

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

$$2!=2$$, $$10!=3628800$$, $$1.65\times0.9001168163=1.485192746$$.
We also would like to determine the $$\Gamma$$ function for $$\nu<1$$. One can invert the recursion relation to read $\Gamma(\nu-1) = \frac{\Gamma(\nu)}{\nu-1},$ $$\Gamma(0.7) = \Gamma(1.7)/0.7=0.909/0.7=1.30$$.
What is $$\Gamma(\nu)$$ for $$\nu<0$$? Let us repeat the recursion derived above and find $\Gamma(-1.3) = \frac{\Gamma(-0.3)}{-1.3} = \frac{\Gamma(0.7)}{-1.3\times -0.3} = \frac{\Gamma(1.7)}{0.7\times-0.3\times -1.3}=3.33\;.$ 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 $$\epsilon$$ $\Gamma(\pm\epsilon)= \frac{\Gamma(1\pm\epsilon)}{\pm\epsilon} =\pm \frac{1}{\epsilon}\rightarrow \pm \infty.$ 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 $$\Gamma(1/2)$$, $\Gamma(\frac{1}{2}) = \int_0^\infty e^{-t}t^{-1/2} dt = 2\int_0^\infty e^{-t}dt^{1/2} = 2\int_0^\infty e^{-x^2}dx.$ 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} (See the discussion of polar coordinates in Sec. 7.1.) We thus find $\Gamma(1/2)=\sqrt{\pi},\;\;\Gamma(3/2)=\frac{1}{2}\sqrt{\pi},\;\;{\rm etc.}$