Gamma function notes

The above is called the Euler representation. Or if we want it deﬁned in complex domain, the above becomes

Since the above is deﬁned only for right half plane, there is way to extend this to left half plane, using what is called analytical continuation. More on this below. First, some relations involving \(\Gamma \left ( x\right ) \)

\begin{align*} \Gamma \left ( z\right ) & =\left ( z-1\right ) \Gamma \left ( z-1\right ) \qquad \operatorname {Re}\left ( z\right ) >1\\ \Gamma \left ( 1\right ) & =1\\ \Gamma \left ( 2\right ) & =1\\ \Gamma \left ( 3\right ) & =2\\ \Gamma \left ( 4\right ) & =3!\\ \Gamma \left ( n\right ) & =\left ( n-1\right ) !\\ \Gamma \left ( n+1\right ) & =n!\\ \Gamma \left ( \frac {1}{2}\right ) & =\sqrt {\pi }\\ \Gamma \left ( z+1\right ) & =z\Gamma \left ( z\right ) \qquad \text {recursive formula}\\ \Gamma \left ( \bar {z}\right ) & =\overline {\Gamma \left ( z\right ) }\\ \Gamma \left ( n+\frac {1}{2}\right ) & =\frac {1\cdot 3\cdot 5\cdots \left ( 2n-1\right ) }{2^{n}}\sqrt {\pi }\end{align*}

\(\blacksquare \) To extend \(\Gamma \left ( z\right ) \) to the left half plane, i.e. for negative values. Let us deﬁne, using the above recursive formula

\[ \bar {\Gamma }\left ( -\frac {3}{2}\right ) =\frac {\bar {\Gamma }\left ( -\frac {3}{2}+1\right ) }{-\frac {3}{2}}=\left ( \frac {1}{-\frac {3}{2}}\right ) \bar {\Gamma }\left ( -\frac {1}{2}\right ) =\left ( \frac {1}{-\frac {3}{2}}\right ) \left ( \frac {1}{-\frac {1}{2}}\right ) \Gamma \left ( \frac {1}{2}\right ) =\left ( \frac {1}{-\frac {3}{2}}\right ) \left ( \frac {1}{-\frac {1}{2}}\right ) \sqrt {\pi }=\frac {4}{3}\sqrt {\pi }\]

And so on. Notice that for \(x<0\) the functions \(\Gamma \left ( x\right ) \) are not deﬁned for all negative integers \(x=-1,-2,\cdots \) it is also not deﬁned for \(x=0\)

\(\blacksquare \) The above method of extending (or analytical continuation) of the Gamma function to negative values is due to Euler. Another method to extend Gamma is due to Weierstrass. It starts by rewriting from the deﬁnition as follows, where \(a>0\)

\begin{align} \Gamma \left ( z\right ) & =\int _{0}^{\infty }t^{z-1}e^{-t}dt\nonumber \\ & =\int _{0}^{a}t^{z-1}e^{-t}dt+\int _{a}^{\infty }t^{z-1}e^{-t}dt \tag {1}\end{align}

\begin{align*} \int _{0}^{a}t^{z-1}e^{-t}dt & =\int _{0}^{a}t^{z-1}\left ( 1+\left ( -t\right ) +\frac {\left ( -t\right ) ^{2}}{2!}+\frac {\left ( -t\right ) ^{3}}{3!}+\cdots \right ) dt\\ & =\int _{0}^{a}t^{z-1}\left ( 1+\left ( -t\right ) +\frac {\left ( -t\right ) ^{2}}{2!}+\frac {\left ( -t\right ) ^{3}}{3!}+\cdots \right ) dt\\ & =\int _{0}^{a}t^{z-1}\sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}t^{n}}{n!}dt\\ & =\int _{0}^{a}\sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}t^{n+z-1}}{n!}dt\\ & =\sum _{n=0}^{\infty }\int _{0}^{a}\frac {\left ( -1\right ) ^{n}t^{n+z-1}}{n!}dt\\ & =\sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}}{n!}\int _{0}^{a}t^{n+z-1}dt\\ & =\sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}}{n!}\left [ \frac {t^{n+z}}{n+z}\right ] _{0}^{a}\\ & =\sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}}{n!\left ( n+z\right ) }a^{n+z}\end{align*}

This takes care of the ﬁrst integral in (1). Now, since the lower limits of the second integral in (1) is not zero, then there is no problem integrating it directly. Remember that in the Euler deﬁnition, it had zero in the lower limit, that is why we said there \(\operatorname {Re}\left ( z\right ) >1\). Now can can choose any value for \(a\). Weierstrass choose \(a=1\). Hence (1) becomes

\begin{align} \Gamma \left ( z\right ) & =\int _{0}^{a}t^{z-1}e^{-t}dt+\int _{a}^{\infty }t^{z-1}e^{-t}dt\nonumber \\ & =\sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}}{n!\left ( n+z\right ) }+\int _{1}^{\infty }t^{z-1}e^{-t}dt \tag {2}\end{align}

Notice the term \(a^{n+z}\) now is just \(1\) since \(a=1\). The second integral above can now be integrated directly. Let us now verify that Euler continuation \(\bar {\Gamma }\left ( z\right ) \) for say \(z=-\frac {1}{2}\) gives the same result as Weierstrass formula. From above, we found that \(\bar {\Gamma }\left ( z\right ) =-2\sqrt {\pi }\). Equation (2) for \(z=-\frac {1}{2}\) becomes

\begin{equation} \bar {\Gamma }\left ( -\frac {1}{2}\right ) =\sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}}{n!\left ( n-\frac {1}{2}\right ) }+\int _{1}^{\infty }t^{-\frac {3}{2}}e^{-t}dt \tag {3}\end{equation}

\[ \sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}}{n!\left ( n-\frac {1}{2}\right ) }=-2\sqrt {\pi }+2\sqrt {\pi }\left ( 1-\operatorname {erf}\left ( 1\right ) \right ) -2\frac {1}{e}\]

\begin{align*} \bar {\Gamma }\left ( -\frac {1}{2}\right ) & =\left ( -2\sqrt {\pi }+2\sqrt {\pi }\left ( 1-\operatorname {erf}\left ( 1\right ) \right ) -2\frac {1}{e}\right ) +\left ( -2\sqrt {\pi }+2\sqrt {\pi }\operatorname {erf}\left ( 1\right ) +\frac {2}{e}\right ) \\ & =-2\sqrt {\pi }\end{align*}

Which is the same as using Euler method. Let us check for \(z=-\frac {2}{3}\). We found above that \(\bar {\Gamma }\left ( -\frac {3}{2}\right ) =\frac {4}{3}\sqrt {\pi }\) using Euler method of analytical continuation. Now we will check using Weierstrass method. Equation (2) for \(z=-\frac {3}{2}\) becomes

\[ \bar {\Gamma }\left ( -\frac {3}{2}\right ) =\sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}}{n!\left ( n-\frac {3}{2}\right ) }+\int _{1}^{\infty }t^{-\frac {5}{2}}e^{-t}dt \]

\[ \sum _{n=0}^{\infty }\frac {\left ( -1\right ) ^{n}}{n!\left ( n-\frac {3}{2}\right ) }=\frac {4\sqrt {\pi }}{3}-\frac {4\sqrt {\pi }\left ( 1-\operatorname {erf}\left ( 1\right ) \right ) }{3}+\frac {2}{3e}\]

\begin{align*} \bar {\Gamma }\left ( -\frac {3}{2}\right ) & =\left ( \frac {4\sqrt {\pi }}{3}-\frac {4\sqrt {\pi }\left ( 1-\operatorname {erf}\left ( 1\right ) \right ) }{3}+\frac {2}{3e}\right ) +\left ( -\frac {4\sqrt {\pi }\operatorname {erf}\left ( 1\right ) }{3}+\frac {4\sqrt {\pi }}{3}-\frac {2}{3e}\right ) \\ & =\frac {4}{3}\sqrt {\pi }\end{align*}

Which is the same as using the Euler method. Clearly the Euler method for analytical continuation of the Gamma function is simpler to compute.

\begin{align*} \Gamma \left ( x\right ) \Gamma \left ( 1-x\right ) & =\int _{0}^{\infty }\frac {t^{x-1}}{1+t}dt\qquad 0<x<1\\ & =\frac {\pi }{\sin \left ( \pi x\right ) }\end{align*}

Where contour integration was used to derive the above. See Mary Boas text book, page 607, second edition, example 5 for full derivation.

\(\blacksquare \) \(\Gamma \left ( z\right ) \) has singularities at \(z=0,-1,-2,\cdots \) and \(\Gamma \left ( 1-z\right ) \) has singularities at \(z=1,2,3,\cdots \) so in the above reﬂection formula, the zeros of \(\sin \left ( \pi x\right ) \) cancel the singularities of \(\Gamma \left ( x\right ) \) when it is written as

\(\blacksquare \) There are other representations for \(\Gamma \left ( x\right ) \). One that uses products by Euler also is

\begin{align*} \Gamma \left ( z\right ) & =\frac {1}{z}\Pi _{n=1}^{\infty }\frac {\left ( 1+\frac {1}{n}\right ) ^{z}}{1+\frac {z}{n}}\\ & =\lim _{n\rightarrow \infty }\frac {n!\left ( n+1\right ) ^{z}}{z\left ( z-1\right ) \cdots \left ( z+n\right ) }\end{align*}

\begin{align*} \Gamma \left ( z\right ) & =\frac {e^{-\gamma z}}{z}\Pi _{n=1}^{\infty }\frac {e^{\frac {z}{n}}}{1+\frac {z}{n}}\\ & =e^{-\gamma z}\lim _{n\rightarrow \infty }\frac {n!\exp \left ( z\left ( 1+\frac {1}{2}+\cdots +\frac {1}{n}\right ) \right ) }{z\left ( z+1\right ) \left ( z+2\right ) \cdots \left ( z+n\right ) }\end{align*}

16 Gamma function notes