I have come upon the hypergeometric function $$_2F_1\left(k+\frac{1}{2},k+\frac{1}{2};\frac{3}{2},z\right)$$ where $k \geq 1$ is an integer, and I believe that this is equal to $$\frac{p(z)}{(1-z)^{(4k-1)/2}}$$ where $p$ is a polynomial of degree $k-1$ (Wolframalpha confirms the first few values). I understand that this must follow from some relationship involving contiguous hypergeometric functions, but I don't know how, and don't have a good reference (library at my uni is closed for COVID-19). I actually don't care about the coefficients in the polynomial, because I'm just trying to show an integral is finite. Is anyone able to put me on the right track?

Many thanks, Greg

  • $\begingroup$ Done, thanks. Cheers, Greg $\endgroup$
    – user387394
    Commented Aug 18, 2020 at 6:20
  • $\begingroup$ May I ask if you are this user? This MO problem on hypergeometric series attracts me very much. I transferred it to MSE and it is recently solved here (after $9$ years...). $\endgroup$ Commented Aug 31, 2020 at 3:37
  • $\begingroup$ Yes I am, and thanks! Please see the discussion there. $\endgroup$
    – user387394
    Commented Sep 1, 2020 at 4:51

1 Answer 1


It follows from the Euler's transformation

$$\displaystyle {}_{2}F_{1}(a,b;c;z)\\=(1-z)^{c-a-b}\,{}_{2}F_{1}(c-a,c-b;c;z)$$

In your case, we have

$$\displaystyle {}_{2}F_{1}\left(k+\frac 12,k+\frac 12;\frac 32;z\right)\\ =(1-z)^{\frac 12-2k}{}_{2}F_{1}\left(1-k,1-k;\frac 32;z\right)$$

Now the hypergeometric function in the RHS can be expanded as a finite series of $k$ elements. This creates the polynomial of grade $k-1$ noted in the OP. By the usual power series definition, it reduces to

$$\begin{aligned} &{k=1 \rightarrow 1}\\ &k=2 \rightarrow 1+ \frac{2z}{3}\\ &k=3 \rightarrow 1+\frac{8z}{3}+\frac{8z^2}{15}\\ &k=4 \rightarrow 1+6z+\frac{24z^2}{5}+\frac{16z^3}{35} \end{aligned} $$ and so on. Generalizing, the polynomial is

$$p(z)=\sum_{n=0}^{k-1} \frac{[(1-k)_n]^2 }{(3/2)_n}\frac{z^n}{n!}$$

where $(z)_n$ is the Pochhammer symbol for rising factorial. We conclude that

$$\displaystyle {}_{2}F_{1}\left(k+\frac 12,k+\frac 12;\frac 32;z\right)\\ =\frac{p(z)}{(1-z)^{2k-\frac{1}{2}}}$$

  • $\begingroup$ Wow, that was fast! Many thanks, that completely settles it. $\endgroup$
    – user387394
    Commented Aug 18, 2020 at 6:18

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