After playing around with transforms of a certain parametric integral, I am inclined to think that the linear combination $$f(n):=\dfrac1{n-2}\left({\,}_2F_1(\dfrac{n-2}{4n},\dfrac12;\dfrac{5n-2}{4n};-1)\right)+\dfrac1{n+2}\left({\,}_2F_1(\dfrac{n+2}{4n},\dfrac12;\dfrac{5n+2}{4n};-1)\right)$$ has a closed form for integer $n$. I know for example that $f(3)=\dfrac{1}{12^{3/4}}\dfrac{\Gamma(\frac14)^2}{\sqrt{\pi}}$. Any ideas?

Edit: putting $a:=\frac14-\frac1{2n}$, we can define $g(a):= \frac{8}{1-4a}f(\frac{8}{1-4a})$ to get arguments closer to the "standard" notation used in formula collections. The question then becomes:

Which rational values of $a$, other than $a= \frac1{12}$, allow a closed form for$$g(a)=\dfrac1{a}\left({\,}_2F_1(a,\frac12;a+1;-1)\right)+\dfrac1{\frac{1}{2}-a}\left({\,}_2F_1(\dfrac{1}{2}-a,\frac12;\frac32-a;-1)\right)?$$

  • $\begingroup$ Can you include that "parametric integral"? I might be a clue to why the $n$ in my answer has closed-forms. $\endgroup$ – Tito Piezas III Jan 28 at 15:52
  • $\begingroup$ @TitoPiezasIII oh sorry, that was such a long time ago that I don't remember at all what that was and where it occurred! Nothing for f(24)? And it is surprising that nothing came up for f(10). $\endgroup$ – Wolfgang Jan 28 at 17:05
  • $\begingroup$ Actually, "apparently" nothing for $n<48$, though I did use a hundred decimal digits. $\endgroup$ – Tito Piezas III Jan 28 at 17:22
  • $\begingroup$ I guess you have already raised $f(n)/K(k_{whatever})$ to the 4th power to increase your chances? $\endgroup$ – Wolfgang Jan 28 at 17:56
  • $\begingroup$ Yes, and I assume that the $d$ of $K(k_d)$ divides $n$. So it was natural to check $f(5)/K(k_5)$ and $f(7)/K(k_7)$. Why the former is algebraic, while the latter apparently is not is intriguing. P.S. However, I've found that certain patterns using eta quotients may involve only $p = 2,3,5$. $\endgroup$ – Tito Piezas III Jan 28 at 18:12

If we use the original function,


then there are several other $f(n)$ with a closed form. Let $K(k_d)$ be an elliptic integral singular value. Then,

$$f(3) = \sqrt{\frac{2}{3\sqrt3}}\; K(k_1)$$ $$f(4) = \sqrt{\frac{-1+\sqrt2}{2\sqrt2}}\; K(k_2)$$ $$f(5) = \frac{\sqrt2}{(\sqrt5\phi)^{5/4}}\; K(k_5)$$ $$f(6) = \sqrt{\frac{1}{12\sqrt3}}\; K(k_3)$$ $$f(12) = \sqrt{\alpha}\; K(k_6)$$

where $\phi$ is the golden ratio and $\alpha$ is a quartic root.

P.S. Try as I might, I couldn't find anything else with $2<n<20$ which begs the question what makes these $n$ so special.


I have found two instances of $_2F_1(a,b;c;z), z=-1$ in

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

that may be of help to you.

$$ _2F_1(a,b;1+a-b;-1)=2^{-a}\frac{\Gamma(1+a-b)\Gamma(1/2)}{\Gamma(1-b+a/2)\Gamma(1/2+a/2)}, \quad 1+a-b\ne0,-1,-2,... $$ $$ (a+1)_2F_1(-a,1;b+2;-1)+(b+1)_2F_1(-b,1;a+2;-1)=2^{a+b+1}\frac{\Gamma(a+2)\Gamma(b+2)}{\Gamma(a+b+2)}, \quad a,b\ne-2,-3,-4,... $$ And this one from the NIST Handbook of Mathematical Functions, which is a variation of the first one above

$$ _2F_1(a,b;1+a-b;-1)=\frac{\Gamma(1+a-b)\Gamma(a/2+1)}{\Gamma(1-b+a/2)\Gamma(1+a)} $$

There was one additional relation for $_2F_1(a,b;c;z), z=-1$, but it was in terms of the digamma function.

  • $\begingroup$ Thank you for searching, but this helps only for $_2F_1(a,1/2;a+1/2;-1)$, What I need is $_2F_1(a,1/2;a+1;-1)$. And the digamma one (15.4.27) has 1 not 1/2 as argument. $\endgroup$ – Wolfgang Jul 24 '17 at 6:51
  • $\begingroup$ @spanferkel I haven't had a chance to see if the case for $n=3$ you presented fits any of the solutions I found. I did, however, check that solution numerically. At any rate, my point is that there are only a limited number of known solutions for $z=-1$ and if these don't do the trick it's unlikely that you'll find a general solution for arbitrary $n$. $\endgroup$ – Cye Waldman Jul 24 '17 at 21:15
  • $\begingroup$ You say "it's unlikely that you'll find a general solution", and of course I'd agree if it was just about one of the two terms. But here is a (conveniently weighted) sum. I remember having seen once a certain weighted sum of two order $n$ digamma functions with arguments $\frac{1\pm\sqrt{5}}2$ which had a closed form... $\endgroup$ – Wolfgang Jul 25 '17 at 8:38
  • $\begingroup$ @spanferkel Apologies. I didn't mean to disparage your efforts, but merely to point out that such solutions are rare and limited, as evidenced by what I turned up in my own references. $\endgroup$ – Cye Waldman Jul 25 '17 at 15:52
  • $\begingroup$ Sure, I am just searching a needle in a hay stick... :) $\endgroup$ – Wolfgang Jul 25 '17 at 16:00

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