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

Question

This is a follow-up to the first comment (by Nemo) to the posting Compute the two-fold partial integral, where the three-fold full integral is known . (I also just asked this as a comment to that question itself--but suspected that it would remain rather obscure in that form.)

After a simplifying reparameterization (transforming $\beta$ into $\frac{b-1}{3}$) suggested by Matt F., the originally-posed problem took the form, \begin{equation} \int_{p=0}^1 \int_{q=0}^{1-p} (\mu p q (1-p-q))^b (\mu^2 q +p)^{-b-1} dq dp, \end{equation} which has been employed since.

Answers to that question have, in fact, been posted by Martin Rubey and by me, but the possible one that was apparently raised by Nemo--in terms of the original $\beta, p_{11,22}$ (rather than $b,p,q$) parameterization--has not so far been fully detailed. (In his comment, Nemo wrote "Did you try calculating Mellin transform of 𝑓(𝜇,𝛽) wrt to 𝜇 (the double integral over $p_{11,22}$ can be calculated using Dirichlet's Beta integral) and then recover 𝑓(𝜇,𝛽) by inverse Mellin transform (which reduces to Barnes integral in this case? I did some calculations and if not mistaken this will lead to representation of 𝑓(𝜇,𝛽) as Gauss hypergeometric function in variable 𝜇".)

Nemo also later commented "MartinRubey--my observation above about this function having a Gauss hypergeometric form is consistent with this logarithmic terms because 2 parameters of this hypergeometric function coincide", but this has not yet been expanded into an answer after a request of Rubey to do so.

So, to reiterate, I would like to explicitly know the presumed representation of 𝑓(𝜇,𝛽) as Gauss hypergeometric function in variable 𝜇, as this might be helpful in the pursuit of the program to construct "separability functions" presented in https://arxiv.org/abs/1701.01973.

In fact, we have been able to find, as already detailed in the earlier answers, a Gauss hypergeometric function expression for the linear (in $\log{\mu}$) term, \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} but not yet for the ``constant term'' $v(b,\mu)$ of the complete functional expression \begin{equation} v(b,\mu) + w(b,\mu) \log(\mu)= \end{equation} \begin{equation} \frac{1}{\Gamma \left(b+\frac{3}{2}\right)} \sqrt{\pi } 4^{-b} \mu^b \left(\mu^2-1\right)^{-2 b-1} \Gamma (b+1) \left(\log (\mu) \sum _{k=0}^b \mu ^{2 k} \binom{b}{k}^2-\left(\mu^2-1\right) \sum _{k=1}^b \mu^{2 k-2} \sum _{i=0}^{k-1} \binom{b}{i}^2 (\psi ^{(0)}(b-i+1)-\psi ^{(0)}(i+1))\right), \end{equation} where, \begin{equation} \sum _{k=0}^b \mu ^{2 k} \binom{b}{k}^2=\, _2F_1\left(-b,-b;1;\mu ^2\right). \end{equation}

Answer 1

In trying to implement the strategy of Nemo, I performed the integration of \begin{equation} \int_{p=0}^1 \int_{q=0}^{1-p} (\mu p q (1-p-q))^b (\mu^2 q +p)^{-b-1} dq dp, \end{equation} for $b =1,\ldots,5$, then requested Mathematica to take the Mellin transform with respect to $\mu$ of the results.

For $b=1$, I obtained \begin{equation} \frac{1}{3} \left(\mathcal{M}_\mu\left[\frac{\mu}{\left(\mu^2-1\right)^3}\right](s)+\mathcal{M}_\mu\left[\frac{\mu \log (\mu)}{\left(\mu^2-1\right)^3}\right](s)-\mathcal{M}_\mu\left[\frac{\mu^3}{\left(\mu^2-1\right)^ 3}\right](s)+\mathcal{M}_\mu\left[\frac{\mu^3 \log (\mu)}{\left(\mu^2-1\right)^3}\right](s)\right), \end{equation} for $b=2$, \begin{equation} \frac{1}{30} \left(3 \mathcal{M}_\mu\left[\frac{\mu^2}{\left(\mu^2-1\right)^5}\right](s)-3 \mathcal{M}_\mu\left[\frac{\mu^6}{\left(\mu^2-1\right)^5}\right](s)-\frac{\pi (s-2) s \left(e^{i \pi s} \left(\pi s^2-2 (\pi -2 i) s-4 i\right)-4 i (s-1)\right)}{32 \left(-1+e^{i \pi s}\right)^2}\right), \end{equation} for $b=3, \begin{equation} \frac{\pi (s-3) (s-1) (s+1) \left(2 i (3 (s-2) s-1)+e^{i \pi s} (6 i (s-2) s+\pi (s-3) (s-1) (s+1)-2 i)\right)}{161280 \left(1+e^{i \pi s}\right)^2} + \end{equation} \begin{equation} -\frac {11} {420} \mathcal {M} _\mu [\frac {\mu^9} { (\mu^2 - 1 )^7} ] (s) - \frac {9} {140} \mathcal {M} _\mu [\frac {\mu^7} { (\mu^2 - 1 )^7} ] (s) + \frac {9} {140} \mathcal {M} _\mu [\frac {\mu^5} { (\mu^2 - 1 )^7} ] (s) + \frac {11} {420} \mathcal {M} _\mu [\frac {\mu^3} { (\mu^2 - 1 )^7} ] (s), \end{equation} ....

So, since the required Mellin transforms do not all seem explicitly implementable, it appears to me that the Nemo goal of obtaining a (complete) representation of $f(\mu,b)$ (using the Matt. F. parameters) does not seem realizable.

Answer 2

I hope this is still of interest.

The integral can be solved using the Mellin representation of a sum, which is $$(A+B)^{-s} = \int_C \frac{\Gamma(s-t) \Gamma(t)}{\Gamma(s)} A^{-t} B^{t-s} \frac{\mathrm{d} t}{2 \pi \mathrm{i}} \,,$$ valid (at least) for $\Re A, \Re B, \Re s > 0$, and where the integration contour $C$ is the usual one.

We write $$(\mu^2 q + p)^{-(b+1)} = \int_C \frac{\Gamma(b+1-t) \Gamma(t)}{\Gamma(b+1)} (\mu^2 q)^{-t} p^{t-(b+1)} \frac{\mathrm{d} t}{2 \pi \mathrm{i}} \,,$$ and then interchange integrations (justified in the usual way) to obtain $$\int_0^1 \int_0^{1-p} (\mu p q (1-p-q))^b (\mu^2 q + p)^{-b-1} \mathrm{d}q \, \mathrm{d}p = \int_C \mu^{b-2t} \frac{\Gamma(b+1-t) \Gamma(t)}{\Gamma(b+1)} \int_0^1 \int_0^{1-p} (1-p-q)^b q^{b-t} p^{t-1} \mathrm{d}q \, \mathrm{d}p \, \frac{\mathrm{d} t}{2 \pi \mathrm{i}} \,.$$ The integral over $q$ gives $$\int_0^{1-p} (1-p-q)^b q^{b-t} \mathrm{d}q = (1-p)^{2b+1-t} \frac{\Gamma(1+b) \Gamma(1+b-t)}{\Gamma(2+2b-t)} \,,$$ and the remaining integral over $p$ results in $$\int_0^1 p^{t-1} (1-p)^{2b+1-t} \mathrm{d}p = \frac{\Gamma(2+2b-t) \Gamma(t)}{\Gamma(2+2b)} \,,$$ such that $$\int_0^1 \int_0^{1-p} (\mu p q (1-p-q))^b (\mu^2 q + p)^{-b-1} \mathrm{d}q \, \mathrm{d}p = \int_C \mu^{b-2t} \frac{\Gamma^2(b+1-t) \Gamma^2(t)}{\Gamma(2+2b)} \frac{\mathrm{d} t}{2 \pi \mathrm{i}} \,.$$ The remaining inverse Mellin transform finally gives a hypergeometric function, and we obtain the result $$\int_0^1 \int_0^{1-p} (\mu p q (1-p-q))^b (\mu^2 q + p)^{-b-1} \mathrm{d}q \, \mathrm{d}p = \frac{\pi \mu^b \, \Gamma^2(b+1)}{4^{1+2b} \, \Gamma^2\left( \frac{3}{2} + b \right)} {}_2\mathrm{F}{}_1\left( b+1, b+1; 2 (b+1); 1-\mu^2 \right) \,.$$ The confluent hypergeometric identity https://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/17/02/08/0003/ with $n = m = 0$ finally turns this into $$\int_0^1 \int_0^{1-p} (\mu p q (1-p-q))^b (\mu^2 q + p)^{-b-1} \mathrm{d}q \, \mathrm{d}p = - \frac{\pi \mu^b \, \Gamma(2b+2)}{2^{1+4b} \, \Gamma^2\left( \frac{3}{2} + b \right)} \left[ {}_2\mathrm{F}{}_1\left( b+1, b+1; 1; \mu^2 \right) \ln \mu - \sum_{k=0}^\infty \frac{\Gamma^2(b+1+k)}{\Gamma^2(b+1) \Gamma^2(k+1)} \mu^{2k} \left( \psi(k+1) - \psi(b+k+1) \right) \right] \,,$$ which shows explicitly the logarithmic terms. For each integer $b$, the hypergeometric function becomes a rational function of $\mu^2$, and the summand becomes a polynomial in $k$, which after summation gives another rational function of $\mu^2$. I don't think that there is a representation of the sum using Gauß hypergeometric functions, but possibly of generalized ones.