# Hypergeometric value

Is their a closed form for the following

$${}_2F_1 \left(a,b;c;\frac{1}{2} \right)$$

I would use the following

$${}_2F_1 \left(a,b;c;x \right)= \frac{\Gamma(c)}{\Gamma(c-b)\Gamma(b)} \int^1_0 t^{b-1}(1-t)^{c-b-1} (1-xt)^{-a} \, dt$$

But it wasn't a success !

Edit: Corrected integral representation (swapped arguments in $\Gamma$ fraction)

• Since, they didn't have a closed form, they invented the cool notation. Maybe you have to look for some hypergeometric identities. You can find a closed form anytime for special $a,b,c$. – Torsten Hĕrculĕ Cärlemän Jul 26 '13 at 22:29
• @TorstenHĕrculĕCärlemän there are closed form especially for ${}_2 F_1(a,b;c;1)$ using the beta function . – Zaid Alyafeai Jul 26 '13 at 22:30
• I was talking about this function. I am not completely sure about the non-existence. – Torsten Hĕrculĕ Cärlemän Jul 26 '13 at 22:40
• The general case seems uncovered, but there are identities such as these for certain dependencies between $a,b,c$ and some transformation to main argument $-1$. – ccorn Jul 26 '13 at 22:48
• @ccorn Are the proofs difficult ? – Zaid Alyafeai Jul 26 '13 at 22:51

I would write this as a comment but I don't have the privilege yet. This is the best documentation on special functions, first book in these series gives a through Hypergeometric functions.

Higher Transcendental Functions ,H. Bateman and A. Erdelyi,

Hope this helps.

There are closed forms for the 2F1 but their exact form depend on the values and the relations between the parameters.

• The trivial case is if $a$ or $b$ is a negative integer: the function becomes a finite sum of rational expressions in the parameters. if you choose $a=-4$ and if you name $d$ the value you have fixed at $1/2$, you got

$1 - \frac{4 b d}{c} +\frac{6 b (b+1) d^2}{c (c+1)} - \frac{4 b (b+1) (b+2) d^3}{c (c+1) (c+2)} - \frac{(-b-3) b (b+1) (b+2) d^4}{c (c+1) (c+2) (c+3)}$

and setting $d = 1/2$:

$1 -\frac{2 b}{c} + \frac{3 (b+1) b}{2 c (c+1)} - \frac{(b+1) (b+2) b}{2 c (c+1) (c+2)} - \frac{(-b-3) (b+1) (b+2) b}{16 c (c+1) (c+2) (c+3)}$

• if $a$ or $b$ is equal to $c$, they cancel each other, giving a simpler $_1F_0$.

$_2F_1( a, b ; b ; d) = _1F_0(a; d) = (1-d)^{-a}$

• for small rationals, there are more interesting closed forms involving elementary functions and elliptic functions and their inverses. Among classical cases

$_1F_0(1/2 ; d) = \frac{1}{\sqrt{1- d}}$

$_2F_1(1/2, 1/2 ; 3/2; d) = \frac{1}{\sqrt d} \sin^{-1} (\sqrt d )$

$_2F_1(1/2, 1 ; 3/2; d) = \frac{1}{\sqrt d} \tanh^{-1} (\sqrt d )$

I have found a few solutions for $z=1/2$ in these references:

K. Oldham, J. Myland, & J. Spanier, An Atlas of Functions, Ch. 60, Springer.

A. Erdelyi, Higher Transcendental Functions, Vol. 1 (and particularly, Sec. 2.8), Krieger Publishing.

Bear in mind, however, that these are for specific values of $a, b, \text{and } c$. The are no general solutions for $z=1/2$. I hope you find something there that meets your needs