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I have been searching through some books and also this

but I have not succeed.

I wonder if there is a simple equivalent form for ${}_2F_1(a,a+\tfrac{1}{2};a+1;z)$, in terms of elementary functions or other simpler functions.

In principle, $a>0$ but a solution for integer positive values of $a$ would be useful to me as well.

Thank you in advance.

user 1
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Chopin
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  • I can give you an answer in terms of Gamma functions when $z = 1$. For $z\neq 1$, the incomplete beta function will be involved. Is that ok with you? – kobe Feb 18 '15 at 21:24
  • Yes that would be great. Can you please let me know that solution? – Chopin Feb 19 '15 at 09:24

3 Answers3

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I have found a possible solution and thought I would post it here.

Applying the Euler transformation:

http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/17/02/07/0001/

the original expression ${}_2F_1(a,a+\tfrac{1}{2};a+1;z)$ can be transformed into one where the following form appears: ${}_2F_1(1,\frac{1}{2};a+1;z)$. If $a$ is a positive integner then it can be expressed in elementary (power) functions using the following relation:

http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/03/06/07/10/0001/

Any other ideas and suggestions are still welcome.

Chopin
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For the sake of simplicity, I'll assume $z$ is real and $0 < a \le 1/2$. First, suppose $a = 1/2$ and $z\neq 0,1$. Then

\begin{align}{}_2F_1\left(a, a+\frac{1}{2}; a + 1; z\right) &= {}_2F_1\left(\frac{1}{2},1;\frac{3}{2};z \right)\\ &= \frac{1}{i\sqrt{z}}\cdot i\sqrt{z}\,{}_2F_1\left(1,\frac{1}{2};\frac{3}{2};-(i\sqrt{z})^2\right)\\ &=\frac{1}{\sqrt{z}}\arctan(i\sqrt{z})\\ &= \frac{1}{2\sqrt{z}}\ln\frac{1 + \sqrt{z}}{1 - \sqrt{z}}. \end{align}

Now suppose $a < 1/2$ and $z = 1$. Since $(a + 1) - a - (a + 1/2) = 1/2 - a > 0$, Gauss's summation formula gives

\begin{align} {}_2F_1\left(a, a + \frac{1}{2};a + 1; z\right) &= \frac{\Gamma(a+1)\Gamma\left(\frac{1}{2}-a\right)}{\Gamma(1)\Gamma\left(\frac{1}{2}\right)}\\ &= \frac{\Gamma(a+1)\Gamma\left(\frac{1}{2}-a\right)}{\sqrt{\pi}}. \end{align}

In general, since $a + 1 > a > 0$, Euler's integral representation of ${}_2F_1$ gives

\begin{align} {}_2F_1\left(a,a+\frac{1}{2};a+1;z\right) &= \frac{\Gamma(a+1)}{\Gamma(a)\Gamma(1)} \int_0^1 t^{a-1} (1 - zt)^{-a-1/2}\, dt\\ &= a\int_0^z u^{a-1}(1 - u)^{-a-1/2}\, \frac{du}{z^a}\\ &= az^{-a}B\left(a,\frac{1}{2}-a;z\right) \end{align}

for $z \neq 0$, where the $B$ function in the last step is the incomplete Beta function.

kobe
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  • In your Gauss summation formula, you probably meant to set $z=1$. – Alex R. Feb 20 '15 at 23:28
  • @AlexR. I already said to suppose $z = 1$. So in that calculation, $z$ is assumed to take the value $1$. – kobe Feb 20 '15 at 23:31
  • Thanks a lot Kobe. The incomplete beta function can also be expressed as an infinite series of elementary functions (this is what I actually need) using the relation shown for example here: http://mathworld.wolfram.com/IncompleteBetaFunction.html. I suspect that this procedure may lead to the same final result obtained by using my other suggested approach. I will analyse this. Anyway thank you for your contribution, it is helpful to me. – Chopin Feb 23 '15 at 09:58
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Maple can do some of these, $a$ integer or half-integer. But not, for example $$ {}_2F_1\left(\frac{1}{3},\frac{5}{6};\frac{4}{3};z\right) $$ In fact, $$ {}_2F_1\left(\frac{1}{3},\frac{5}{6};\frac{4}{3};\frac{1}{2}\right) \approx 1.1460295323744285$$ and if I put $1.1460295323744285$ into the ISC, the result is $$ {}_2F_1\left(\frac{1}{3},\frac{5}{6};\frac{4}{3};\frac{1}{2}\right) $$ So I am guessing that no simpler form is known.

added, based on kobe's comment

Maple says $$ {}_2F_1\left(a,a+\frac{1}{2};a+1;1\right) = \frac{\Gamma(a+1)\Gamma\left(\frac{1}{2}-a\right)}{\sqrt{\pi}} $$ Of course the series diverges when $a>1/2$, so I guess this is an analytic continuation in that case.

Also, the case $a=1/2$ is $$ {}_2F_1\left(\frac{1}{2},1;\frac{3}{2};z\right) = \frac{1}{2\sqrt{z}}\log\frac{1+\sqrt{z}}{1-\sqrt{z}} $$

GEdgar
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