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Is there a closed-form sum for the hypergeometric series $_3F_2(a+c, c+d, 1; c+1, a+b+c+d \mid 1)$ where $a, b, c, d$ are all positive and not necessarily integers?

Update: The motivation for this question comes from equation 4 of this tech report on random inequalities. In some special cases, such as when one of $a, b, c$ or $d$ is an integer, the report gives a closed form sum.

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I tried to find a simple special case that Maple did not do directly, I chose $a=0,c=1,d=0,b=2/3$. The Inverse Symbolic Calculator identified it, though: $$ {}_3F_2\Biggl([1,1,1],\biggl[\frac{5}{3},2\biggr],1\Biggr) = \frac{\Psi^{(1)} \Bigl(\frac{1}{3}\Bigr)}{6} + \frac{\Psi^{(1)} \Bigl(\frac{5}{6}\Bigr)}{6} $$

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  • $\begingroup$ It is also $\dfrac{2}{3} \Psi^{(1)}(2/3)$. According to Wolfram Alpha, ${}_3F_2([1,1,1],[2,1+b],1) = b \Psi^{(1)}(b)$. $\endgroup$ Commented Oct 7, 2012 at 21:21
  • $\begingroup$ So ... if John does not think $\Psi^{(1)}$ is "closed form", then there is no hope for his general question. $\endgroup$ Commented Oct 8, 2012 at 12:23
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According to Maple,

$${{\frac{\Gamma \left( c+1 \right) \Gamma \left( a+b+c+d \right) G^{1, 3}_{3, 3}\left(-1\, {\Big\vert}^{0, 1-a-c, 1-c-d}_{0, 1-a-b-c-d, -c}\right) }{\Gamma \left( a+c \right) \Gamma \left( c+d \right) }}} $$ where $G$ is the Meijer G function. But I don't know if you'd call that more "closed-form" than the original hypergeometric.

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  • $\begingroup$ Thanks, though that's not what I'm looking for. A practical definition of "closed form" is "can be computed with commonly available software." That would include some special functions -- Gamma, Bessel, etc. -- but probably not Meijer G. $\endgroup$ Commented Oct 7, 2012 at 10:46
  • $\begingroup$ When you tell Maple to simplify that Meijer G expression, the result is the ${}_3F_2$ that you started with. So (according to Maple) the Meijer G formula is less simple. $\endgroup$ Commented Oct 7, 2012 at 14:47
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    $\begingroup$ "Commonly available software" including Maple and Mathematica can compute the hypergeometric function. $\endgroup$ Commented Oct 7, 2012 at 18:59

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