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This is a follow-up question to the one posed in Compute the two-fold partial integral, where the three-fold full integral is known . (I hope that doing so is viewed as a legitimate step. If not so, I can withdraw the question, and fall back on the original.)

The problem is to integrate over $p \in [0,1]$ the integrand, \begin{equation} -\frac{(p-1)^{2 b+1} \mu^b \Gamma (b+1)^2 \, _2F_1\left(b+1,b+1;2 (b+1);\frac{(p-1) \mu^2}{p}\right)}{p \Gamma (2 (b+1))}, \end{equation} where $\mu \in [0,1]$ and $b$ is a nonnegative integer.

The answer takes the form $v(b,\mu) + w(b,\mu) \log(\mu)$, where it is now known that \begin{equation} w(b,\mu)=\frac{\sqrt{\pi } 4^{-b} \mu^b \left(\mu^2-1\right)^{-2 b-1} \Gamma (b+1) \, _2F_1\left(-b,-b;1;\mu^2\right)}{\Gamma \left(b+\frac{3}{2}\right)}. \end{equation}

Additionally, \begin{equation} v(b,1)=\frac{\pi 4^{-2 b-1} \Gamma (b+1)^2}{\Gamma \left(b+\frac{3}{2}\right)^2}. \end{equation} So, a general formula for $v(b,\mu)$ is sought.

So, we would like the counterpart for $v(b,\mu)$ of the Rubey formula for $w(b,\mu)$, that is, \begin{equation} 4 u^b \left(u^2-1\right)^{-2 b-1} \frac{1}{4 \left(4 b^2-1\right)} \frac{b} {\binom{2 (b-1)}{b-1}} \Sigma_{k=0}^b u^{2 k} \binom{b}{k}^2, \end{equation} which he apparently obtained using the general purpose computer algebra system, FriCAS

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    $\begingroup$ Having open questions in three places (your original post, your answer to it, and now this post) is a bit much. $\endgroup$
    – user44143
    Commented Feb 14, 2019 at 22:05
  • $\begingroup$ OK--sorry! ............... $\endgroup$ Commented Feb 15, 2019 at 2:19

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The answer to the indicated integration over $p \in [0,1]$ is \begin{equation} \frac{\pi (-1)^{2 b} 4^{-2 b-1} \mu^{-b-2} \Gamma (b+1)^2 \, _2F_1\left(b+1,b+1;2 (b+1);1-\frac{1}{\mu^2}\right)}{\Gamma \left(b+\frac{3}{2}\right)^2}. \end{equation}

This was obtained by transforming the hypergeometric argument $\frac{(p-1) \mu^2}{p}$ to $v$, and integrating the result over $v \in [-\infty,0]$.

This, in effect, also provides the answer to the related questions

Compute the two-fold partial integral, where the three-fold full integral is known

(validating the intuition of @Nemo, in his comments there)

and

"Find a representation [using Mellin transform] of 𝑓(𝜇,𝛽) as Gauss hypergeometric function in variable 𝜇"

(which asked for an implementation of the Mellin transform proposal of @Nemo).

Even though this specific question had been down-voted and commented on as "a bit much" by Matt F., I think, in retrospect, it was useful to pose, since it changed the focus from a double integral to an (eventually doable) single integral.

As a bit of background, I should credit Charles Dunkl, whom I quite recently informed of this mathoverflow thread. He wrote--in terms of the two-fold integration--"I could only think of changing variables (so $\mu ^2 p +q$) is one of the variables, but then the limits of integration come in several cases". This led me to investigate similar changes-of-variables in the single integration.

Of some further interest, Dunkl wrote: "I don't recall ever seeing a worked out triple integral where the intermediate double isn't known (here is the answer - what is the question?)". This led me to inform him that the triple integration result was not strictly in the form of a proof (as I should have made clear at the outset)--but rather simply based on an application of the Mathematica FindSequenceFunction command to a finite series $(b=1,2\ldots)$ of $b$-specific calculations.

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