Special Functions – Hypergeometric functions – Properties

Sometimes the most general hypergeometric function \({}_pF_q\) is called a generalized hypergeometric function. Then the words "hypergeometric function" refer to the special case

\[{}_2F_1(a,b;\,c;\,z)={}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)=\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_n}\cdot\frac{z^n}{n!},\quad c\notin\{0,-1,-2,\ldots\}.\]

Note that if \(a=-N\) with \(N\in\{0,1,2,\ldots\}\), then we have \((a)_n=(-N)_n=(-N)(-N+1)(-N+2)\cdots(-N+n-1)=0\) for \(n=N+1,N+2,N+3,\ldots\). Hence

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{-N,b}{c}\,;\,z\right)=\sum_{n=0}^N\frac{(-N)_n(b)_n}{(c)_n}\cdot\frac{z^n}{n!},\quad N\in\{0,1,2,\ldots\}.\]

Otherwise, the series converges for \(|z|<1\) and also for \(|z|=1\) if \(\text{Re}(c-a-b)>0\).

Many elementary functions have representations as hypergeometric series. An example is

\[\ln(1+z)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}z^{n+1} =\sum_{n=0}^{\infty}\frac{(1)_n(1)_n}{(2)_n}\cdot\frac{(-1)^nz^{n+1}}{n!}=z\,{}_2F_1\left(\genfrac{}{}{0pt}{}{1,1}{2}\,;\,-z\right),\]

since \((1)_n=n!\) and \((2)_n=(n+1)!\). We also have

\[\arctan z=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}z^{2n+1} =\sum_{n=0}^{\infty}\frac{(1/2)_n(1)_n}{(3/2)_n}\cdot\frac{(-1)^nz^{2n+1}}{n!}=z\,{}_2F_1\left(\genfrac{}{}{0pt}{}{1/2,1}{3/2}\,;\,-z^2\right),\]

since

\[\frac{(1/2)_n}{(3/2)_n}=\frac{\frac{1}{2}\cdot\frac{3}{2}\cdots\frac{2n-1}{2}} {\frac{3}{2}\cdot\frac{5}{2}\cdots\frac{2n+1}{2}}=\frac{\frac{1}{2}}{\frac{2n+1}{2}}=\frac{1}{2n+1}.\]

Note that the example \(\ln(1-z)=-z\,{}_2F_1(1,1;\,2;\,z)\) shows that although the series converges for \(|z|<1\), it has an analytic continuation as a single-valued function in the complex plane except for the (half) line joining \(1\) and \(\infty\). This describes the general situation; a \({}_2F_1\) function has an analytic continuation in the complex plane with branch points at \(1\) and \(\infty\).

Special cases lead to other elementary functions, such as

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{b}\,;\,z\right)={}_1F_0\left(\genfrac{}{}{0pt}{}{a}{-}\,;\,z\right) =\sum_{n=0}^{\infty}\frac{(a)_n}{n!}z^n=\sum_{n=0}^{\infty}\binom{-a}{n}(-z)^n=(1-z)^{-a},\quad|z|<1,\]

which is called the binomial theorem, and

\[{}_0F_0\left(\genfrac{}{}{0pt}{}{-}{-}\,;\,z\right)=\sum_{n=0}^{\infty}\frac{z^n}{n!}=e^z,\quad z\in\mathbb{C}.\]

Note that we have

\[\lim\limits_{b\to\infty}{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{b}\,;\,\frac{z}{b}\right) =\lim\limits_{b\to\infty}\sum_{n=0}^{\infty}\frac{(a)_nz^n}{(c)_n\,n!}\cdot\frac{(b)_n}{b^n} =\sum_{n=0}^{\infty}\frac{(a)_n}{(c)_n}\cdot\frac{z^n}{n!}={}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)\]

and

\[\lim\limits_{c\to\infty}{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,cz\right) =\lim\limits_{c\to\infty}\sum_{n=0}^{\infty}\frac{(a)_n(b)_nz^n}{n!}\cdot\frac{c^n}{(c)_n} =\sum_{n=0}^{\infty}(a)_n(b)_n\cdot\frac{z^n}{n!}={}_2F_0\left(\genfrac{}{}{0pt}{}{a,b}{-}\,;\,z\right).\]

For the hypergeometric function \({}_2F_1\) we have an integral representation due to Euler:

Theorem: For \(\text{Re}(c)>\text{Re}(b)>0\) we have

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}\,dt\]

for all \(z\) in the complex plane cut along the real axis from \(1\) to \(\infty\). Here it is understood that \(\arg(t)=\arg(1-t)=0\) and \((1-zt)^{-a}\) has its principal value.

Proof: First suppose that \(|z|<1\), then the binomial theorem implies that

\[(1-zt)^{-a}=\sum_{n=0}^{\infty}\frac{(a)_n}{n!}z^nt^n.\]

This implies that

\[\int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}\,dt=\sum_{n=0}^{\infty}\frac{(a)_n}{n!}z^n\int_0^1t^{n+b-1}(1-t)^{c-b-1}\,dt.\]

The latter integral is a beta integral which equals

\[\int_0^1t^{n+b-1}(1-t)^{c-b-1}\,dt=B(n+b,c-b)=\frac{\Gamma(n+b)\Gamma(c-b)}{\Gamma(n+c)}.\]

Now we use the fact that

\[\frac{\Gamma(n+b)}{\Gamma(b)}=b(b+1)(b+2)\cdots(b+n-1)=(b)_n,\quad n=0,1,2,\ldots\]

to obtain

\[\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}\,dt =\frac{\Gamma(c)}{\Gamma(b)}\sum_{n=0}^{\infty}\frac{\Gamma(n+b)}{\Gamma(n+c)}\frac{(a)_n}{n!}z^n =\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_n}\cdot\frac{z^n}{n!}={}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right),\]

which proves the theorem for \(|z|<1\). Since the integral is analytic in the cut plane \(\mathbb{C}\setminus(1,\infty)\), the theorem holds in that region as well.

Euler's integral representation can be used to prove Gauss's summation formula:

Theorem: For \(\text{Re}(c-a-b)>0\) we have

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,1\right)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.\]

Proof: If we take the limit \(z\to1\) in Euler's integral representation we obtain

\begin{align*} {}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,1\right)&=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1t^{b-1}(1-t)^{c-a-b-1}\,dt=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}B(b,c-a-b)\\[2.5mm] &=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\frac{\Gamma(b)\Gamma(c-a-b)}{\Gamma(c-a)} =\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} \end{align*}

for \(\text{Re}(c)>\text{Re}(b)>0\) and \(\text{Re}(c-a-b)>0\). The condition \(\text{Re}(c)>\text{Re}(b)>0\) can be removed by using analytic continuation.

If \(a=-n\) with \(n\in\{0,1,2,\ldots\}\) Gauss's summation theorem reduces to a finite summation theorem named after Chu-Vandermonde:

Theorem: For \(c\neq0,-1,-2,\ldots\) we have

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{-n,b}{c}\,;\,1\right)=\frac{(c-b)_n}{(c)_n},\quad n=0,1,2,\ldots.\]

Proof: For \(a=-n\) with \(n\in\{0,1,2,\ldots\}\) we have

\[\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)} =\frac{\Gamma(c)\Gamma(c-b+n)}{\Gamma(c+n)\Gamma(c-b)}=\frac{(c-b)_n}{(c)_n}.\]

In view of the binomial theorem Euler's integral representation can also be written as

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1t^{b-1}(1-t)^{c-b-1}{}_1F_0\left(\genfrac{}{}{0pt}{}{a}{-}\,;\,zt\right)\,dt\]

for \(\text{Re}(c)>\text{Re}(b)>0\). This can be generalized to

\[{}_{p+1}F_{q+1}\left(\genfrac{}{}{0pt}{}{a_1,a_2,\ldots,a_p,a_{p+1}}{b_1,b_2,\ldots,b_q,b_{q+1}}\,;\,z\right) =\frac{\Gamma(b_{q+1})}{\Gamma(a_{p+1})\Gamma(b_{q+1}-a_{p+1})}\int_0^1t^{a_{p+1}-1}(1-t)^{b_{q+1}-a_{p+1}-1} {}_pF_q\left(\genfrac{}{}{0pt}{}{a_1,a_2,\ldots,a_p}{b_1,b_2,\ldots,b_q}\,;\,zt\right)\,dt\]

for \(\text{Re}(b_{q+1})>\text{Re}(a_{p+1})>0\).

As an application of Euler's integral representation we will prove Pfaff's transformation formula for the \({}_2F_1\):

Theorem: \(\displaystyle{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)=(1-z)^{-a}{}_2F_1\left(\genfrac{}{}{0pt}{}{a,c-b}{c}\,;\,\frac{z}{z-1}\right)\).

Proof: We start with Euler's integral represenation

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}\,dt\]

for \(\text{Re}(c)>\text{Re}(b)>0\). We use the substitution \(t=1-s\) in the integral to obtain

\[\int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}\,dt=-\int_1^0(1-s)^{b-1}s^{c-b-1}(1-z+zs)^{-a}\,ds =(1-z)^{-a}\int_0^1s^{c-b-1}(1-s)^{b-1}\left(1-\frac{zs}{z-1}\right)^{-a}\,ds,\]

which proves Pfaff's transformation formula for \(\text{Re}(c)>\text{Re}(b)>0\). These conditions can be removed by using analytic continuation.

Another result is Euler's transformation formula for the \({}_2F_1\):

Theorem: \(\displaystyle{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)=(1-z)^{c-a-b}{}_2F_1\left(\genfrac{}{}{0pt}{}{c-a,c-b}{c}\,;\,z\right)\).

Proof: Apply Pfaff's transformation formula twice:

\begin{align*} {}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)&=(1-z)^{-a}{}_2F_1\left(\genfrac{}{}{0pt}{}{a,c-b}{c}\,;\,\frac{z}{1-z}\right) =(1-z)^{-a}\cdot\left(1-\frac{z}{z-1}\right)^{b-c}{}_2F_1\left(\genfrac{}{}{0pt}{}{c-a,c-b}{c}\,;\,\frac{\frac{z}{z-1}}{\frac{z}{z-1}-1}\right)\\[2.5mm] &=(1-z)^{c-a-b}{}_2F_1\left(\genfrac{}{}{0pt}{}{c-a,c-b}{c}\,;\,z\right). \end{align*}

Note that Euler's transformation formula can also be written in the form

\[(1-z)^{-c+a+b}{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)={}_2F_1\left(\genfrac{}{}{0pt}{}{c-a,c-b}{c}\,;\,z\right) =\sum_{n=0}^{\infty}\frac{(c-a)_n(c-b)_n}{(c)_n}\cdot\frac{z^n}{n!}.\]

The left-hand side can be written as

\[(1-z)^{-c+a+b}{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)=\sum_{j=0}^{\infty}\frac{(c-a-b)_j}{j!}\,z^j \sum_{k=0}^{\infty}\frac{(a)_k(b)_k}{(c)_k}\cdot\frac{z^k}{k!} =\sum_{n=0}^{\infty}\sum_{k=0}^n\frac{(a)_k(b)_k(c-a-b)_{n-k}}{(c)_k\,k!\,(n-k)!}\,z^n.\]

Comparing the coefficients of \(z^n\) we conclude that

\[\sum_{k=0}^n\frac{(a)_k(b)_k(c-a-b)_{n-k}}{(c)_k\,k!\,(n-k)!}=\frac{(c-a)_n(c-b)_n}{(c)_n\,n!},\quad n=0,1,2,\ldots.\]

Note that

\[\frac{n!}{(n-k)!}=n(n-1)\cdots(n-k+1)=(-1)^k(-n)(-n+1)\cdots(-n+k-1)=(-1)^k(-n)_k\]

and

\[(c-a-b)_{n-k}=\frac{(c-a-b)_n}{(c-a-b+n-k)(c-a-b+n-k+1)\cdots(c-a-b+n-1)}=\frac{(c-a-b)_n}{(-1)^k(1+a+b-c-n)_k}\]

for \(k\in\{0,1,2,\ldots,n\}\) and \(n=0,1,2,\ldots\). This implies that

\[\sum_{k=0}^n\frac{(-n)_k(a)_k(b)_k}{(c)_k(1+a+b-c-n)_k\,k!}=\frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n},\quad n=0,1,2,\ldots.\]

This is the Pfaff-Saalschütz summation formula for a terminating \({}_3F_2\):

Theorem: \(\displaystyle{}_3F_2\left(\genfrac{}{}{0pt}{}{-n,a,b}{c,1+a+b-c-n}\,;\,1\right)=\frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n},\quad n=0,1,2,\ldots\).

Note that the limit case for \(n\to\infty\) of the Pfaff-Saalschütz summation formula reduces to Gauss's summation formula for the \({}_2F_1\). In fact we have

\[(a)_n=a(a+1)(a+2)\cdots(a+n-1)=\frac{\Gamma(a+n)}{\Gamma(a)}.\]

This implies that

\[\frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n}=\frac{\Gamma(c-a+n)\Gamma(c-b+n)\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)\Gamma(c+n)\Gamma(c-a-b+n)} =\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}\cdot\frac{\Gamma(c-a+n)\Gamma(c-b+n)}{\Gamma(c+n)\Gamma(c-a-b+n)}\]

Now we have by Stirling's asymptotic formula

\[\frac{\Gamma(c-a+n)\Gamma(c-b+n)}{\Gamma(c+n)\Gamma(c-a-b+n)}\sim n^{c-a+c-b-c-c+a+b}=n^0=1\quad\text{for}\quad n\to\infty.\]

This implies that

\[\lim\limits_{n\to\infty}\frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n}=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.\]

Hence

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,1\right)=\lim\limits_{n\to\infty}{}_3F_2\left(\genfrac{}{}{0pt}{}{-n,a,b}{c,1+a+b-c-n}\,;\,1\right) =\lim\limits_{n\to\infty}\frac{(c-a)_n(c-b)_n}{(c)_n(c-a-b)_n}=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.\]

The confluent hypergeometric function is given by

\[{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)=\lim\limits_{b\to\infty}{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,\frac{z}{b}\right) =\sum_{n=0}^{\infty}\frac{(a)_n}{(c)_n}\,\frac{z^n}{n!},\quad c\notin\{0,-1,-2,\ldots\}.\]

Based on Euler's integral representation for the \({}_2F_1\), one might expect that the confluent hypergeometric function satisfies

\[{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)=\lim\limits_{b\to\infty}{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,\frac{z}{b}\right) =\lim\limits_{b\to\infty}\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1t^{a-1}(1-t)^{c-a-1}\left(1-\frac{zt}{b}\right)^{-b}\,dt.\]

Now we have

\[\lim\limits_{b\to\infty}\left(1-\frac{zt}{b}\right)^{-b}=e^{zt},\]

which leads to

Theorem: For \(\text{Re}(c)>\text{Re}(a)>0\) we have \(\displaystyle{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right) =\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1t^{a-1}(1-t)^{c-a-1}e^{zt}\,dt\).

Proof: Note that we have

\[\int_0^1t^{a-1}(1-t)^{c-a-1}e^{zt}\,dt=\sum_{n=0}^{\infty}\frac{z^n}{n!}\,\int_0^1t^{n+a-1}(1-t)^{c-a-1}\,dt\]

and for \(\text{Re}(a)>0\) and \(\text{Re}(c-a)>0\)

\[\int_0^1t^{n+a-1}(1-t)^{c-a-1}\,dt=B(n+a,c-a)=\frac{\Gamma(n+a)\Gamma(c-a)}{\Gamma(n+c)} =\frac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\,\frac{(a)_n}{(c)_n},\quad n=0,1,2,\ldots.\]

This implies that

\[\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1t^{a-1}(1-t)^{c-a-1}e^{zt}\,dt =\sum_{n=0}^{\infty}\frac{(a)_n}{(c)_n}\,\frac{z^n}{n!}={}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right).\]

This integral representation can be used to prove Kummer's transformation formula:

Theorem: \(\displaystyle{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)=e^z\,{}_1F_1\left(\genfrac{}{}{0pt}{}{c-a}{c}\,;\,-z\right)\).

Proof: We use the substitution \(t=1-u\) to obtain

\[\int_0^1t^{a-1}(1-t)^{c-a-1}e^{zt}\,dt=\int_0^1(1-u)^{a-1}u^{c-a-1}e^{z(1-u)}\,du=e^z\int_0^1u^{c-a-1}(1-u)^{a-1}e^{-zu}\,du.\]

This implies that

\[{}_1F_1\left(\genfrac{}{}{0pt}{}{a}{c}\,;\,z\right)=e^z\,{}_1F_1\left(\genfrac{}{}{0pt}{}{c-a}{c}\,;\,-z\right).\]

Note that this also follows from Pfaff's transformation formula for the \({}_2F_1\):

\[{}_2F_1\left(\genfrac{}{}{0pt}{}{a,b}{c}\,;\,z\right)=(1-z)^{-b}\,{}_2F_1\left(\genfrac{}{}{0pt}{}{b,c-a}{c}\,;\,\frac{z}{z-1}\right).\]

Replace \(z\) by \(z/b\) and take the limit \(b\to\infty\).

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Last modified on May 15, 2021