Reference request. A prototype case:

In $$ {}_2F_1\left(\frac{1}{12},\frac{5}{12};\frac{1}{2};x\right) = A\log\left(\frac{1}{1-x}\right) + B + o(1), \qquad x \to 1^- $$ what can we say about the connection coefficient $B \approx 0.995$? Of course already Gauss knew $$ A = \frac{\Gamma\left(\frac{7}{12}\right)\Gamma\left(\frac{11}{12}\right)}{4\pi^{3/2}} . $$ Also well-known are non-log cases of ${}_2F_1(a,b;c;x)$, such as those where $a+b-c \notin \mathbb Z$.


1 Answer 1


The result for ${}_3F_2$ was suspected by Ramanujan, and proved by Evans and Stanton (Thm 3). http://epubs.siam.org/doi/abs/10.1137/0515078 http://www.math.ucsd.edu/~revans/Stanton.pdf

The general ${}_3F_2$ result is: $${\Gamma(a)\Gamma(b)\Gamma(c)\over\Gamma(d)\Gamma(e)}{}_3F_2({a,b,c\atop d,e};z)\sim$$ $$\sim-\log(1-z)-2\gamma-{\Gamma'(a)\over\Gamma(a)}-{\Gamma'(b)\over\Gamma(b)}+\sum_{k=1}^\infty {(d-c)_k(e-c)_k\over k(a)_k(b)_k}$$ when $a+b+c=d+e$ and $\Re(c)>0$.

Transferring to ${}_2F_1$, in the specified example one has that $B/A$ is $$-{\Gamma'\over\Gamma}(1/12)-{\Gamma'\over\Gamma}(5/12)-2\gamma.$$

The above paper mentions its equation (1.10) being a "well-known" asymptotic formula for the $m$th partial sum of a zero-balanced ${}_2F_1$ (citing Luke p109 #34), and taking the limit as $m\rightarrow\infty$ should already give the desired result for ${}_2F_1$.

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    $\begingroup$ This is a good answer. Evans & Stanton point out that the ${}_2F_1$ case in Luke's book is easier than the ${}_3F_2$ they are doing. $\endgroup$ Oct 23, 2014 at 15:37
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    $\begingroup$ Now that I look at Evans/Stanton again, in $\S$6 they mention the ${}_2F_1$ case, and quote "Higher Transcendental Functions" (Erdelyi) apps.nrbook.com/bateman/Vol1.pdf $\endgroup$ Oct 23, 2014 at 17:52

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