I view this as both a mathematics and a Mathematica question -- so apologies if it is thought I should have sent it alternatively to the mathematics stack exchange.

I want to perform the two-dimensional integration (or, possibly, reduce to a one-dimensional integration)

Integrate[Y^(-1 + d)
   Hypergeometric2F1Regularized[d/2, -k, (2 + d)/
   2, ((Y^2 - e^2 Subscript[r, 14]^2) (-1 + e^2 Subscript[r, 14]^2))/(
   e^2 (Y^2 - Subscript[r, 14]^2) (-1 + Subscript[r, 14]^2))] (1/(
   e Subscript[r, 14]))^(
  1 + d) (1 + Y^2 (1 - 1/Subscript[r, 14]^2) - Subscript[r, 14]^2)^
  k (1 - e^2 Subscript[r, 14]^2)^(
  d/2) (-Y^2 + e^2 Subscript[r, 14]^2)^(d/2), {Subscript[r, 14], 0, 
  1}, {Y, e Subscript[r, 14]^2, e Subscript[r, 14]}, 
 Assumptions -> d >= 1 && k >= 0 && 0 < e <= 1]

So, $ d $ and $ k $ are parameters, and $ Y $ and $ r_{14} $ the variables of integration, with $ e $ being a free variable.

In $ \mathrm\TeX $, the integrand is the product of

\begin{equation} Y^{d-1} \left(\frac{1}{r_{14} \epsilon }\right){}^{d+1} \left(1-r_{14}^2 \epsilon ^2\right){}^{d/2} \left(\left(1-\frac{1}{r_{14}^2}\right) Y^2-r_{14}^2+1\right){}^k \left(r_{14}^2 \epsilon ^2-Y^2\right){}^{d/2} \end{equation}


\begin{equation} \, _2\tilde{F}_1\left(\frac{d}{2},-k;\frac{d+2}{2};\frac{\left(r_{14}^2 \epsilon ^2-1\right) \left(Y^2-r_{14}^2 \epsilon ^2\right)}{\left(r_{14}^2-1\right) \epsilon ^2 \left(Y^2-r_{14}^2\right)}\right) . \end{equation}

The two-dimensional domain of integration is

\begin{equation} r_{14} \in [0,1], \hspace{.25in} Y \in [\varepsilon r_{14}, \varepsilon^2 r_{14}] . \end{equation}

For $ k=0 $, the integral evaluates (as can be confirmed by setting $ d $ to a positive integer -- even integers evaluate more readily) to

1/4 e^(-1 + d)
  Gamma[d/2] Gamma[
  d] HypergeometricPFQRegularized[{-(d/2), d/2, d}, {1 + d/2, 
   1 + (3 d)/2}, e^2]

That is, \begin{equation} \frac{1}{4} \epsilon ^{d-1} \Gamma \left(\frac{d}{2}\right) \Gamma (d) \, _3\tilde{F}_2\left(-\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;\epsilon ^2\right). \end{equation}

I've been trying all possible integration-by-parts combinations with no success to this point in time.

For even, nonnegative $ d $ and nonnegative $ k $, the integrand evaluates to a polynomial in $ e $. For odd $ d $, logs and polylogs appear.

The question stated here pertains to the issue discussed in sec. IX.B of my posting, Qubit-qudit separability/PPT-probability investigations, including Lovas-Andai formula advancements, of finding an "extended Lovas-Andai master formula", denoted there by $ \tilde{\chi}_{d,k}(\varepsilon) $.


1 Answer 1


Charles F. Dunkl has provided me with essentially the answer to the bivariate ($r_{14}, r_{23}$) integration question at hand. (However, I had brought to his attention, a problem with a slightly reexpressed integrand, using a change-of-variables, now designated $x,y$. Also, I had omitted certain accompanying [gamma,...] factors, not functions of the two variables, from the stated problem for the sake of conciseness. These are now present in his write-up. The important parameters $d,k$ and (not integrated out) variable $\varepsilon$ remain. (Let me note that CFD works in Maple.)

I now provide his write-up (certain of the TeX commands do not seem to "take" in this stack-exchange setting). The central problem is the evaluation of $I(\varepsilon)$, which is given near the end.

Let us simplify \begin{align*} I\left( \varepsilon\right) & :=\frac{\Gamma\left( 1+d+k\right) ^{2}% }{\Gamma\left( \frac{d}{2}\right) ^{3}\Gamma\left( 1+\frac{d}{2}+k\right) \Gamma\left( 1+k\right) }\times\frac{2}{d}\varepsilon^{-d}\\ & \times\int_{0}^{1}dx\int_{\varepsilon^{2}x}^{\varepsilon^{2}}dy[\left\{ \left( 1-x\right) \left( 1-y\right) \right\} ^{k}\left( xy\right) ^{d/2-1}\left\{ \left( \varepsilon^{2}-y\right) \left( 1-x\varepsilon ^{2}\right) \right\} ^{d/2}\\ & \times~_{2}F_{1}\left( -k,\frac{d}{2};1+\frac{d}{2};T\right) ]\\ T & :=\frac{\left( \varepsilon^{2}-y\right) \left( 1-x\varepsilon ^{2}\right) }{\left( 1-x\right) \left( 1-y\right) \varepsilon^{2}}. \end{align*} First we apply the transformation $_{2}F_{1}\left( a,b;c;t\right) =\left( 1-t\right) ^{-a}~_{2}F_{1}(a,c-b;c;\frac{t}{t-1}$), but the series on the right side only converges if $a$ is a negative integer or $t<\frac{1}{2}$, not the case in our application, thus \textbf{henceforth assume} $k=0,1,2,3,\ldots $then% \begin{align*} 1-T & =\frac{\left( 1-\varepsilon^{2}\right) \left( y-x\varepsilon ^{2}\right) }{\left( 1-x\right) \left( 1-y\right) \varepsilon^{2}},\\ \frac{T}{T-1} & =-\frac{\left( \varepsilon^{2}-y\right) \left( 1-x\varepsilon^{2}\right) }{\left( 1-\varepsilon^{2}\right) \left( y-x\varepsilon^{2}\right) }; \end{align*} the integrand becomes% \begin{align*} & \left( xy\right) ^{d/2-1}\left( 1-\varepsilon^{2}\right) ^{k}% \varepsilon^{-2k}\left( y-x\varepsilon^{2}\right) ^{k}\left( 1-x\varepsilon ^{2}\right) ^{d/2}\left( \varepsilon^{2}-y\right) _{~}^{d/2}\\ & \times~_{2}F_{1}\left( -k,1;1+\frac{d}{2};-\frac{\left( \varepsilon ^{2}-y\right) \left( 1-x\varepsilon^{2}\right) }{\left( 1-\varepsilon ^{2}\right) \left( y-x\varepsilon^{2}\right) }\right) . \end{align*} Substitute $y=\varepsilon^{2}u$ so $dy=\varepsilon^{2}du$ and $0\leq x\leq u\leq1.$ This gives a factor of $\varepsilon^{2d}$ in front of% \begin{align*} & \int\limits_{0\leq x\leq u\leq1}\int dx~du~\left( xu\right) ^{d/2-1}\\ & \sum_{j=0}^{k}\frac{\left( -k\right) _{j}}{\left( 1+\frac{d}{2}\right) _{j}}\left( -1\right) ^{j}\left( 1-x\varepsilon^{2}\right) ^{d/2+j}\left( 1-u\right) ^{d/2+j}\left( 1-\varepsilon^{2}\right) ^{k-j}\left( u-x\right) ^{k-j}. \end{align*} Isolate the $x$-integral (use the negative binomial series for $\left( 1-x\varepsilon^{2}\right) ^{d/2+j}$)% \begin{align*} & \int_{0}^{u}x^{d/2-1}\left( u-x\right) ^{k-j}\sum_{i=0}^{\infty}% \frac{\left( -\frac{d}{2}-j\right) _{i}}{i!}x^{i}\varepsilon^{2i}dx\\ & =\sum_{i=0}^{\infty}\frac{\left( -\frac{d}{2}-j\right) _{i}}{i!}% \frac{\Gamma\left( \frac{d}{2}+i\right) \Gamma\left( k-j+1\right) }% {\Gamma\left( \frac{d}{2}+i+k-j+1\right) }\varepsilon^{2i}u^{d/2+i+k-j}, \end{align*} by use of $\int_{0}^{u}x^{\alpha-1}\left( u-x\right) ^{\beta-1}% dx=u^{\alpha+\beta-1}B\left( \alpha,\beta\right) $. The inner $u$-integral is% [ \int_{0}^{1}u^{d/2-1}u^{d/2+i+k-j}\left( 1-u\right) ^{d/2+j}du=\frac {\Gamma\left( d+i+k-j\right) \Gamma\left( \frac{d}{2}+j+1\right) }% {\Gamma\left( \frac{3d}{2}+i+k+1\right) }. ] Thus the integral is% \begin{align*} & \sum_{j=0}^{k}\frac{\left( -k\right) _{j}}{\left( 1+\frac{d}{2}\right) _{j}}\varepsilon^{2d}\left( 1-\varepsilon^{2}\right) ^{k-j}\left( -1\right) ^{j}\\ & \times\sum_{i=0}^{\infty}\frac{\left( -\frac{d}{2}-j\right) _{i}}{i!}% \frac{\Gamma\left( \frac{d}{2}+i\right) \Gamma\left( k-j+1\right) }% {\Gamma\left( \frac{d}{2}+i+k-j+1\right) }\varepsilon^{2i}\frac {\Gamma\left( d+i+k-j\right) \Gamma\left( \frac{d}{2}+j+1\right) }% {\Gamma\left( \frac{3d}{2}+i+k+1\right) }\\ & =\sum_{j=0}^{k}\frac{\left( -k\right) _{j}}{\left( 1+\frac{d}{2}\right) _{j}}\left( -1\right) ^{j}\varepsilon^{2d}\left( 1-\varepsilon^{2}\right) ^{k-j}\frac{\Gamma\left( \frac{d}{2}\right) \Gamma\left( k-j+1\right) \Gamma\left( d+k-j\right) \Gamma\left( \frac{d}{2}+j+1\right) }% {\Gamma\left( \frac{d}{2}+k-j+1\right) \Gamma\left( \frac{3d}% {2}+k+1\right) }\\ & \times\sum_{i=0}^{\infty}\frac{\left( -\frac{d}{2}-j\right) _{i}\left( \frac{d}{2}\right) _{i}\left( d+k-j\right) _{i}}{i!\left( \frac{d}% {2}+k-j+1\right) _{i}\left( \frac{3d}{2}+k+1\right) _{i}}\varepsilon^{2i}. \end{align*} The last sum is a $_{3}F_{2}$ with argument $\varepsilon^{2}.$

Simplify the Gamma terms and note $\left( -k\right) _{j}=\left( -1\right) ^{j}\frac{k!}{\left( k-j\right) !}$ and $\Gamma\left( k-j+1\right) =\left( k-j\right) !.$ Then% \begin{align*} & \frac{\left( -k\right) _{j}}{\left( 1+\frac{d}{2}\right) _{j}}\left( -1\right) ^{j}\frac{\Gamma\left( \frac{d}{2}\right) \Gamma\left( k-j+1\right) \Gamma\left( d+k-j\right) \Gamma\left( \frac{d}% {2}+j+1\right) }{\Gamma\left( \frac{d}{2}+k-j+1\right) \Gamma\left( \frac{3d}{2}+k+1\right) }\\ & =\frac{k!}{\left( 1+\frac{d}{2}\right) _{j}}\frac{\Gamma\left( \frac {d}{2}\right) \Gamma\left( \frac{d}{2}+1\right) \left( \frac{d}% {2}+1\right) _{j}\Gamma\left( d\right) \left( d\right) _{k-j}}% {\Gamma\left( \frac{d}{2}+1\right) \left( \frac{d}{2}+1\right) _{k-j}\Gamma\left( \frac{3d}{2}+1+k\right) }\\ & =\frac{k!\left( d\right) _{k-j}\Gamma\left( \frac{d}{2}\right) \Gamma\left( d\right) }{\left( \frac{d}{2}+1\right) _{k-j}\Gamma\left( \frac{3d}{2}+1+k\right) }. \end{align*} Combine the factors (correctly +/- ??)% \begin{align*} I\left( \varepsilon\right) & =\frac{2\Gamma\left( 1+d+k\right) ^{2}\Gamma\left( \frac{d}{2}\right) \Gamma\left( d\right) k!\varepsilon ^{d}}{d\Gamma\left( \frac{d}{2}\right) ^{3}\Gamma\left( 1+\frac{d}% {2}+k\right) \Gamma\left( 1+k\right) \Gamma\left( \frac{3d}{2}+1+k\right) }\\ & \times\sum_{j=0}^{k}\frac{\left( d\right) _{k-j}}{\left( \frac{d}% {2}+1\right) _{k-j}}\left( 1-\varepsilon^{2}\right) ^{k-j}~_{3}F_{2}\left( % %TCIMACRO{\QATOP{-d/2-j,d/2,d+k-j}{1+d/2+k-j,1+k+3d/2}}% %BeginExpansion \genfrac{}{}{0pt}{}{-d/2-j,d/2,d+k-j}{1+d/2+k-j,1+k+3d/2}% %EndExpansion ;\varepsilon^{2}\right) . \end{align*} The first line simplifies to% [ \frac{\Gamma\left( 1+d+k\right) ^{2}\Gamma\left( 1+d\right) }% {2\Gamma\left( 1+\frac{d}{2}+k\right) \Gamma\left( 1+\frac{d}{2}\right) ^{2}\Gamma\left( 1+\frac{3d}{2}+k\right) }\varepsilon^{d}% ] (apparently agrees with the postulated $k=0$ expression)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.