How to calculate the Kampé de Fériet function?

Question

This is a continuation of this post.

The following is my original question in that post.

Question: Is it possible to express $$\sum_{l=0}^\infty \sum_{r=0}^\infty\frac{\Gamma(L+r-2q)}{\Gamma(L+r-1+2q)} \frac{\Gamma(L+r+l-1+2q)}{\Gamma(L+r+l+2)}\frac{r+1}{r+l+2}$$ in closed form independent of summations (possibly in terms of Gamma function)?

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@G Cab managed to reduce the double summations above using Kampé de Fériet function

$$ \bbox[lightyellow] { \eqalign{ & S(L,q) = A\;\sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{3^{\,\overline {\,r + l\,} } \left( {c - d} \right)^{\,\overline {\,r + l\,} } } \over {4^{\,\overline {\,r + l\,} } c^{\,\overline {\,r + l\,} } }}{{3^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } \left( {a - b} \right)^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } 1^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } } \over {2^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } a^{\,\overline {\,r\,} } 1^{\,\overline {\,l\,} } }}{{x^{\,r} } \over {r!}}{{y^{\,l} } \over {l!}}} } = \cr & = F\left( {\matrix{ 2 \cr 3 \cr 2 \cr 2 \cr } \,\left| {\,\matrix{ {3,\left( {c - d} \right)} \cr {3,\,1\;;\;\left( {a - b} \right),1\;;\;\;1,1} \cr {4,c} \cr {2,\,1\;;\;a,1} \cr } \,} \right|x,y} \right)\quad \quad \left| {\;x = y = 1} \right. \cr} }$$

where $r,l,L\geq 1$ are integers and $q\in [0,1]$ is a real number.

The final answer should be similar to the form

$$1 - \frac{2q^2}{(1-2q)^2} - \frac{2\pi q(1-q)}{(1-2q)(3-4q)} \cot(2\pi q).$$

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Here comes my question for this post.

Question: How to calculate the Kampé de Fériet function to get answer above?

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Based on @Nikos Bagis in that post, Mathematica 10 gives something involving gamma functions, cotangent and some generalized hypergeometric series.

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UPDATED: 08/11/18 @Nikos Bagis compuated a closed form for the double summations above.

$$ F_1(L,q,x):=\sum^{\infty}_{l=0}\sum^{\infty}_{r=0}\frac{\Gamma(L+r-2q)\Gamma(L+r+l-1+2q)}{\Gamma(L+r-1+2q)\Gamma(L+r+l+2)}x^lx^r. $$ Then $$ F_1(L,q,x)=-\frac{\Gamma(L-2q)(L-2q-1)}{4q-2}{}\frac{1}{\Gamma(L+2)} {}_2F_1(1,L-2q;L+2;x)+ $$ $$ \frac{(L+2q-2)\Gamma(L-2q)}{4q-2} \frac{1}{\Gamma(L+2)} {}_2F_1\left(1,L+2q-1;L+2;x\right)- $$ $$ -\frac{\Gamma(L-2q)}{4q-2} \frac{1}{\Gamma(L-+)} {}_2F_1\left(2,L-2q;L+2;x\right), $$ where $$ pF_q\left(a_1,a_2,\ldots,a_p;b_1,b_2,\ldots,b_q;z\right):= $$ $$ \sum_{n=0}^\infty \frac{(a_1)_n (a_2)_n...(a_p)_n}{(b_1)_n (b_2)_n...(b_p)_n}\frac{x^n}{n!}$$ is the generalized hypergeometric function.

Now my question is

How to show that $$\sum^{\infty}_{l=0}\sum^{\infty}_{r=0}\frac{\Gamma(L+r-2q)\Gamma(L+r+l-1+2q)}{\Gamma(L+r-1+2q)\Gamma(L+r+l+2)}x^lx^r = $$ $$ F_1(L,q,x)=-\frac{\Gamma(L-2q)(L-2q-1)}{4q-2}{}\frac{1}{\Gamma(L+2)} {}_2F_1(1,L-2q;L+2;x)+ $$ $$ \frac{(L+2q-2)\Gamma(L-2q)}{4q-2} \frac{1}{\Gamma(L+2)} {}_2F_1\left(1,L+2q-1;L+2;x\right)- $$ $$ -\frac{\Gamma(L-2q)}{4q-2} \frac{1}{\Gamma(L-+)} {}_2F_1\left(2,L-2q;L+2;x\right) $$ and $$ F_2(L,q,x)= $$ $$ =\frac{x\Gamma(L-2q+1)(2+L-6q-4Lq+8q^2)}{2(2q-1)(4q-3)\Gamma(L+3)}{}_2F_1\left(2,L-2q+1;L+3;x\right)- $$ $$ -\frac{\Gamma(L-2q+1)(L+2q-2)}{2(2q-1)(4q-3)\Gamma(L+2)}{}_2F_1(1,L-2q;L+2;x)+ $$ $$ +\frac{\Gamma(L-2q+1)(L+2q-2)}{2(2q-1)(4q-3)\Gamma(L+2)}{}_2F_1(1,L+2q-1;L+2;x)- $$ $$ -x\frac{\Gamma(L-2q+1)}{(4q-3)\Gamma(L+3)} {}_3F_2\left(2,2,L-2q+1;1,L+3;x\right). $$ where $$ F_2(L,q,x):=\sum^{\infty}_{l=0}\sum^{\infty}_{r=0}\frac{\Gamma(L+r-2q) \Gamma(L+r+l-1+2q)}{\Gamma(L+r-1+2q) \Gamma(L+r+l+2)}rx^lx^r? $$

Answer 1

This is another approach where I first introduce \begin{align} \Gamma(L+r-2q) &= \int_0^\infty t_1^{L+r-2q-1} \, e^{-t_1} \, {\rm d}t_1 \\ \Gamma(L+r+l+2q-1) &= \int_0^\infty t_2^{L+r+l+2q-2} \, e^{-t_2} \, {\rm d}t_2 \\ r+1 &= \frac{\rm d}{{\rm d}t_3} \, t_3^{r+1} \bigg|_{t_3=1^{-}} \\ \frac{1}{r+l+2} &= \int_0^1 t_4^{r+l+1} \, {\rm d}t_4 \\ \frac{1}{\Gamma(L+r+l+2)} &= \frac{1}{2\pi i}\int_{-\infty}^{0^+} t_5^{-(L+r+l+2)} \, e^{t_5} \, {\rm d}t_5 \\ \frac{1}{\Gamma(L+r+2q-1)} &= \frac{1}{2\pi i}\int_{-\infty}^{0^+} t_6^{-(L+r+2q-1)} \, e^{t_6} \, {\rm d}t_6 \end{align} to avoid going into the details of hypergeometric functions $_pF_q$. In the last two expressions the bounds refer to Hankel's contour starting at $-\infty$ encircling $0$ once in positive direction and going back to $-\infty$. These contours can be deformed to encircle the entire complex plane in a big circle as there is no other singularity than $0$.

Then without writing the integrals and derivatives and interchanging integration and summation order at will the summation essentially becomes two geometric series $$ \frac{1}{(2\pi i)^2} \, {\frac {{{t_1}}^{L-2q-1}\,{{ t_2}}^{L+2q-2}\, t_3 \, t_4 \,{{t_5}}^{-L}\,{{t_6}}^{-L-2q+2}\,{{\rm e}^{-{t_1}-{t_2}+{t_5}+{t_6}}}}{ \left( {t_2}{t_4}-{t_5} \right) \left( {t_1}{t_2}{t_3}{t_4}-{t_5}{t_6} \right) }} \, . $$

The procedure now is clear: We carry out the integrals in an appropriate order starting with $t_5$ since $L$ is a positive integer and the integrand is thus holomorph in $t_5$ so we can apply the residue theorem which yields 3 terms \begin{align} I_1 &= \frac{ {{t_1}}^{L-2q-1}\,{{t_2}}^{L+2q-2} \, t_3 \, t_4 \, {{t_6}}^{-L-2q+1}\,{\rm e}^{-{t_1}-{t_2}+{t_6}} }{2\pi i} \, \frac{1}{(L-1)!} \, \frac{{\rm d}^{L-1}}{{\rm d}t_5^{L-1}} \frac{{\rm e}^{t_5}}{\left( {t_2}{t_4} - {t_5} \right) \left( \frac{{t_1}{t_2}{t_3}{t_4}}{{t_6}} - {t_5} \right) } \Bigg|_{t_5=0} \\ &= \frac{ {{t_1}}^{L-2q-2}\,{{t_2}}^{L+2q-3} \, {{t_6}}^{-L-2q+2}\,{\rm e}^{-{t_1}-{t_2}+{t_6}} }{2\pi i} \, \sum_{k=0}^{L-1} \frac{1}{\Gamma(L-k)} \frac{1}{(t_2t_4)^{k+1}} \sum_{l=0}^k \left( \frac{t_6}{t_1t_3} \right)^l \\ &= \frac{ {{t_1}}^{L-2q-2}\,{{t_2}}^{L+2q-3} \, {{t_6}}^{-L-2q+2}\,{\rm e}^{-{t_1}-{t_2}+{t_6}} }{2\pi i} \, \sum_{k=0}^{L-1} \frac{1}{\Gamma(L-k)} \frac{1}{(t_2t_4)^{k+1}} \frac{\left( \frac{t_6}{t_1t_3} \right)^{k+1}-1}{\frac{t_6}{t_1t_3}-1} \\ I_2 &= \frac{1}{2\pi i} \, {\frac {{{t_1}}^{L-2q-1} \, {{t_2}}^{2q-3} \, t_3 \, {{t_4}}^{-L} \, {{t_6}}^{-L-2q+2}\,{{\rm e}^{-{t_2}(1-{t_4})-{t_1}+{t_6}}}}{{t_6} - {t_1}{t_3}}} \\ I_3 &= \frac{1}{2\pi i} \, { \frac {{{t_1}}^{-2q-1} \, {{t_2}}^{2q-3} \, t_3^{-L+1} \, t_4^{-L} \, {{t_6}}^{-2q+2}}{{t_1}{t_3}-{t_6}} \, {{\rm e}^{{-t_2\left(1-\frac{{t_1}{t_3}{t_4}}{t_6}\right)-{t_1}+t_6}}} } \, . \end{align} The best next step is to do the $t_2$ integral which leads to a set of $\Gamma$-functions after we substitute $$ \tau_2=t_2(1-t_4) \\ \tau_3=t_2\left(1-\frac{t_1t_3t_4}{t_6}\right) $$ in $I_2$ and $I_3$ respectively and so \begin{align} I_1 &= \frac{ {{t_1}}^{L-2q-2}\, {{t_6}}^{-L-2q+2}\,{\rm e}^{-{t_1}+{t_6}} }{2\pi i} \, \sum_{k=0}^{L-1} \sum_{l=0}^k \frac{\Gamma(L+2q-3-k)}{\Gamma(L-k)} \frac{1}{t_4^{k+1}} \left( \frac{t_6}{t_1t_3} \right)^l \\ I_2 &= \frac{1}{2\pi i} \, \frac {t_1^{L-2q-1} \,{t_3}\,\left( 1-{t_4} \right)^{2-2q} \, {t_4}^{-L} \,t_6^{-L-2q+2} \, {\rm e}^{-{t_1}+{t_6}} \, \Gamma \left( 2q-2 \right) }{{t_6}-{t_1}{t_3}} \\ I_3 &= -\frac{1}{2\pi i} \, \frac {{{t_1}}^{-2q-1} \, {{t_3}}^{-L+1} \, {{t_4}}^{-L} \, \left( {t_6}-{t_1}{t_3}{t_4} \right)^{2-2q} \, {\rm e}^{-t_1+t_6} \Gamma \left(2q-2 \right) }{{t_6}-{t_1}{t_3}} \, . \end{align} Now $t_6 = R\,{\rm e}^{i\phi}$ for $R\rightarrow \infty$ and $\phi \in (0,2\pi)$ and then using for $I_1,I_2$ and $I_3$ \begin{align} \lim_{R\rightarrow \infty} \frac{1}{2\pi i}\oint_{|z|=R} \frac{(z-c)^{-b} \, {\rm e}^z}{z-a} \, {\rm d}z &= \frac{(a-c)^{-b} \, {\rm e}^a \, \gamma(b,a-c)}{\Gamma(b)} \\ &= \frac{(a-c)^{-b} \, {\rm e}^a \left\{\Gamma(b)-\Gamma(b,a-c)\right\}}{\Gamma(b)} \\ \Rightarrow \qquad \lim_{R\rightarrow \infty} \frac{1}{2\pi i}\oint_{|z|=R} z^{-b-1} \, {\rm e}^z \, {\rm d}z &= \lim_{a\rightarrow 0} \lim_{c\rightarrow 0} \frac{(a-c)^{-b} \, {\rm e}^a \, \gamma(b,a-c)}{\Gamma(b)} = \frac{1}{\Gamma(b+1)} \end{align} which yields \begin{align} I_1 &= { {\rm e}^{-{t_1}} } \, \sum_{k=0}^{L-1} \sum_{l=0}^k \frac{\Gamma(L+2q-3-k)}{\Gamma(L-k)} \frac{ t_1^{L-2q-2-l} \, t_3^{-l} \, t_4^{-k-1} }{\Gamma(L+2q-2-l)} \\ I_2 &= { t_1^{-4q+1} \,{t_3}^{-L-2q+3} \, {t_4}^{-L} \, \left( 1-{t_4} \right)^{2-2q} \, {\rm e}^{-{t_1}(1-{t_3})} \, \Gamma \left( 2q-2 \right) } \left\{1 - \frac{\Gamma(L+2q-2,t_1t_3)}{\Gamma(L+2q-2)}\right\} \\ I_3 &= - {{{t_1}}^{-4q+1} \, {{t_3}}^{-L-2q+3} \, {{t_4}}^{-L} \, \left( 1-{t_4} \right)^{2-2q} \, {\rm e}^{-t_1(1-t_3)} } \left\{ \Gamma\left(2q-2\right) - \Gamma\left(2q-2,t_1t_3(1-t_4)\right) \right\} \, . \end{align} The first terms of $I_2$ and $I_3$ cancel when adding up, so we are not going to write them down anymore. Doing the $t_3$ derivative and setting $t_3=1$ the exponential functions occurring together with the incomplete Gamma functions vanish \begin{align} I_1 &= -{ {\rm e}^{-{t_1}} } \, \sum_{k=0}^{L-1} \sum_{l=0}^k \frac{l \, \Gamma(L+2q-3-k)}{\Gamma(L-k)} \frac{ t_1^{L-2q-2-l} \, t_4^{-k-1} }{\Gamma(L+2q-2-l)} \\ I_2 &= - \frac{ t_1^{-4q+1} \, {t_4}^{-L} \, \left( 1-{t_4} \right)^{2-2q} \, \Gamma \left( 2q-2 \right) }{\Gamma(L+2q-2)} \Big\{ (-L-2q+3) \, {\Gamma(L+2q-2,t_1)} \\ &\quad + t_1 \, {\Gamma(L+2q-2,t_1)} - t_1^{L+2q-2} \, {\rm e}^{-t_1} \Big\} \\ I_3 &= { {{t_1}}^{-4q+1} \, {{t_4}}^{-L} \, \left( 1-{t_4} \right)^{2-2q} } \Big\{ (-L-2q+3) \, \Gamma\left(2q-2,t_1(1-t_4)\right) \\ &\quad + t_1 \, \Gamma\left(2q-2,t_1(1-t_4)\right) - t_1^{2q-2} \left(1-t_4\right)^{2q-2} \, {\rm e}^{-t_1(1-t_4)} \Big\} \, . \end{align} We now use $$ \int_0^\infty t^a \, \Gamma(s,t) \, {\rm d}t = \frac{\Gamma(s+a+1)}{a+1} $$ which all $t_1$ integrals can be brought to after a possible substitution $t=t_1(1-t_4)$ ending up at \begin{align} I_1 &= - \sum_{k=0}^{L-1} \sum_{l=0}^k \frac{l \, \Gamma(L+2q-3-k)}{\Gamma(L-k)} \frac{ \Gamma(L-2q-1-l) \, t_4^{-k-1} }{\Gamma(L+2q-2-l)} \\ I_2 &= - \frac{ {t_4}^{-L} \, \left( 1-{t_4} \right)^{2-2q} \, \Gamma \left( 2q-2 \right) }{\Gamma(L+2q-2)} \left\{ (-L-2q+3) \, \frac{\Gamma(L-2q)}{-4q+2} + \frac{\Gamma(L-2q+1)}{-4q+3} - \Gamma(L-2q) \right\} \\ &= {\frac { {{t_4}}^{-L} \left( 1-{t_4} \right)^{2-2q} \, \Gamma \left( 2q-2 \right) \Gamma \left( L-2q \right) }{2 \left( 2q-1 \right) \left( 4q-3 \right) \Gamma \left( L+2q-3 \right) }} \\ I_3 &= { {{t_4}}^{-L} } \left\{ (-L-2q+3) \, \left( 1-{t_4} \right)^{2q} \, \frac{\Gamma\left(-2q\right)}{-4q+2} + \left( 1-{t_4} \right)^{2q-1} \, \frac{\Gamma\left(-2q+1\right)}{-4q+3} - \left(1-t_4\right)^{2q} \, \Gamma(-2q) \right\} \\ &= -{\frac {4\,{{t_4}}^{-L} \left( 1-{t_4} \right)^{2q-1}\Gamma\left(-2q \right) \left\{ \left( q-\frac{3}{4} \right) \left( L-2q-1 \right) {t_4} + \left( -L+\frac{1}{2} \right) q+\frac{3(L-1)}{4} \right\} }{2(2q-1)(4q-3)}} \, . \end{align} The remaining $t_4$ integral is divergent, but can be made finite by analytic continuation using Beta-function regularization $$ \int_0^1 t^{a-1}(1-t)^{b-1} \, {\rm d} t = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)} $$ giving \begin{align} I_1 &= \sum_{k=0}^{L-1} \sum_{l=0}^k \frac{\Gamma(L+2q-3-k)}{\Gamma(L-k)} \frac{ \Gamma(L-2q-1-l) }{\Gamma(L+2q-2-l)} \, \frac{l}{k} \\ I_2 &= -{\frac {\sin \left(\pi L + 2\pi q\right) \Gamma\left( -L+1 \right) \Gamma \left( L-2q \right) }{ 2 \left( 2q-1 \right) \left( 4q-3 \right) \sin \left( 2\pi q \right) }} \\ I_3 &= -{\frac {\pi\,\Gamma \left( -L+1 \right) }{ 2 \left( 2q-1 \right) \left( 4q-3 \right) \Gamma \left( 2q-L+1 \right) \sin \left( 2\pi q \right) }} \, . \end{align} There however remains an ambiguity if $k=0$ when $l/k=0/0$ is undefined.

For $L$ integer these expressions are not defined as well, but only the sum of $I_2$ and $I_3$ which is \begin{align} I_2 + I_3 &= -{\frac {\sin \left( \pi L \right) \Gamma \left( -L+1 \right) \Gamma \left( L-2q \right) \cot\left(2\pi q\right) }{ \left( 2q-1 \right) \left( 4q-3 \right) }} \\ &= -{\frac {\pi \, \cot\left(2\pi q\right) \, \Gamma \left( L-2q \right) }{ \left( 2q-1 \right) \left( 4q-3 \right) \Gamma\left(L\right) }} \, . \end{align} The final result would then be $$ I_1 + I_2 + I_3 \, . $$

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edit: The $k=0$ term can actually be obtained by analytic continuation. We calculate $$\sum_{l=0}^k \frac{\Gamma(\alpha-l)}{\Gamma(\beta-l)} \, x^l$$ in terms of hypergeometric functions $_2F_1$. Then we derive with respect to $x$ (the result being $_2F_1$ again) and evaluate at $x=1$. Using the connection formula we can express this in terms of $\Gamma$-functions. The result is $$ {\frac {\Gamma \left( 1+\alpha \right) \Gamma \left( \beta-k-1 \right) +\Gamma \left( \beta-1 \right) \Gamma \left( \alpha-k \right) \left\{ \left( -1-k \right) \alpha+k \left( \beta-1 \right) \right\} }{ \left( -2-\alpha+\beta \right) \left( \beta-1-\alpha \right) \Gamma \left( \beta-1 \right) \Gamma \left( \beta-k-1 \right) }} $$ or with $\alpha=L-2q-1$, $\beta=L+2q-2$ and the remaining three factors of the $k$ sum $$ {\frac { \Gamma \left( L-2\,q-1-k \right) }{\Gamma \left( L-k \right) \left( 4\,q-3 \right) }} \\ + {\frac {\Gamma \left( L-2\,q \right) \Gamma \left( L+ 2\,q-3-k \right) }{2 \, \Gamma \left( L-k \right) k \left( 2\,q-1 \right) \left( 4\,q-3 \right) \Gamma \left( L+2\,q-3 \right) }} \\ + {\frac { \left( -L+2\,q+1 \right) \Gamma \left( L-2\,q-1-k \right) }{2 \, \Gamma \left( L-k \right) k \left( 2\,q-1 \right) \left( 4\,q-3 \right) }} \tag{1} $$ which is finite for $k=0$ and any $q\in(0,1)$. The lengthy analysis of the limit $k=0$ yields $$ \frac{\Gamma(L-2q)}{2 \, \Gamma(L) (2q-1) (4q-3)} \left\{ \Psi(L-2q) - \Psi(L+2q) \right\} \\ + \frac{\Gamma(L-2q-1)}{2 \, \Gamma(L) (2q-1) (4q-3) (L+2q-1) (L+2q-2) (L+2q-3)} \Big\{ 32q^4 + (48L-144)q^3 \\ + (24L^2-144L+196)q^2 + (4L^3-36L^2+104L - 88)q + 3L^2 - 10L + 7 \Big\} \, . $$

We can use the above result (1) to symbolically carry out the $k$ sum from $1$ to $L-1$ the result being $$ -{\frac {\Gamma \left( L-2\,q-1 \right) }{2 \, q \left( 4\,q-3 \right) \Gamma \left( L-1 \right) }}+{\frac { \left\{ \Psi \left( L+2\,q-3 \right) -\Psi \left( 2\,q-2 \right) \right\} \Gamma \left( L-2\,q \right) }{ 2 \left( 2\,q-1 \right) \left( 4\,q-3 \right) \Gamma \left( L \right) }}-{\frac { \left\{ \Psi \left( L -2\,q-1 \right) - \Psi \left( -2\,q \right) \right\} \Gamma \left( L-2 \,q \right) }{ 2 \left( 2\,q-1 \right) \left( 4\,q-3 \right) \Gamma \left( L \right) }} \, . $$

Finally adding up the previous two expressions and $I_2+I_3$ after simplifying we get

\begin{align} \bbox[lightyellow] { \sum_{l=0}^\infty \sum_{r=0}^\infty\frac{\Gamma(L+r-2q)}{\Gamma(L+r-1+2q)} \frac{\Gamma(L+r+l-1+2q)}{\Gamma(L+r+l+2)}\frac{r+1}{r+l+2} \\ = -{ \frac {\pi\,\cot \left( 2\,\pi\,q \right) \Gamma \left( L-2\,q \right) }{ 2 \left( 2\,q - 1 \right) \left( 4\,q-3 \right) \Gamma \left( L \right) } } - { \frac { \left( 2\,{q}^{2}-4\,q+1 \right) \Gamma \left( L-2\,q \right) }{ 4 \left( 2\,q - 1\right) ^{2} q \left( q-1 \right) \Gamma \left( L \right) } } \, . } \end{align}

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Concerning your follow up question:

I don't have a solution, but wanted to make aware of the following procedure.

Assuming we know how to evaluate the $l$ sum, then there is the series \begin{align} &\sum_{r=0}^\infty \sum_{l=0}^\infty {\frac {\Gamma \left( L+r-2\,q \right) \Gamma \left( L+r+l-1+2\,q \right)}{\Gamma \left( L+r-1+2\,q \right) \Gamma \left( L+r+l+2 \right) } \, {x}^{r}{y}^{l}} \\ =&\sum_{r=0}^\infty \frac {\Gamma \left( L+r-2\,q \right)}{\Gamma \left( L+r+2 \right)} \, {\mbox{$_2$F$_1$}(1,L+r-1+2\,q;\,L+r+2;\,y)} \, {x}^{r} \, . \tag{2} \end{align} You can either use the conventional integral representation to carry out the sum or use the following two identities (cf. https://dlmf.nist.gov/15.6) \begin{align} {\mbox{$_2$F$_1$}(a,b;\,c;\,z)} &= \frac{\Gamma(c)}{\Gamma(d)\Gamma(c-d)} \int_0^1 {\mbox{$_2$F$_1$}(a,b;\,d;\,zt)} \, t^{d-1} (1-t)^{c-d-1} \, {\rm d}t \tag{3} \\ {\mbox{$_2$F$_1$}(a,b;\,c;\,z)} &= \frac{\Gamma(c)}{\Gamma(d)\Gamma(c-d)} \int_0^1 \frac{t^{d-1}(1-t)^{c-d-1}}{(1-zt)^{a+b-\lambda}} \, {\mbox{$_2$F$_1$}(\lambda-a,\lambda-b;\,d;\,zt)} \, {\mbox{$_2$F$_1$}\left(a+b-\lambda,\lambda-d;\,c-d;\,\frac{(1-t)z}{1-zt}\right)} \, {\rm d}t \tag{4} \end{align} where in each expression $d$ and $d,\lambda$ are arbitrary constants at our free disposal.

For example if we plug (3) in (2) and choose $d=1$ the hypergeometric function can be evaluated by $$ {\mbox{$_2$F$_1$}(1,b;\,1;\,z)} = (1-z)^{-b} $$ and (2) becomes $$ \int_0^1 (1-yt)^{1-L-2q} \, (1-t)^L \sum_{r=0}^\infty \frac{\Gamma(L+r-2q)}{\Gamma(L+r+1)} \, \left( \frac{(1-t)x}{1-yt} \right)^r \, {\rm d}t \\ = \frac{\Gamma(L-2q)}{\Gamma(L+1)} \int_0^1 \frac{(1-t)^L}{(1-yt)^{L+2q-1}} \, {\mbox{$_2$F$_1$}\left(1,L-2q;\,L+1;\,\frac{(1-t)x}{1-yt}\right)} \, {\rm d}t \, . \tag{4.1} $$ Substituting $s=\frac{1-t}{1-yt}$ reveals $$ (1-y)^{2-2q} \, \frac{\Gamma(L-2q)}{\Gamma(L+1)} \int_0^1 \frac{s^L}{(1-ys)^{3-2q}} \, {\mbox{$_2$F$_1$}\left(1,L-2q;\,L+1;\,xs\right)} \, {\rm d}s \tag{4.2} $$ so for $x=y$ only in the special case $q=3/4$ the integrand can be brought to (4) for a special choice of the parameters i.e. we first need to choosse $d=L+1$ and $c=L+2$ and $a+b-\lambda=3-2q$. Since $\lambda-a=1$ and $\lambda-b=L-2q$ we have $a=3-4q+L$ and $b=4-2q$ with $\lambda=4-4q+L$. Unfortunately the second hypergeometric function is unity only in the case $\lambda=d$ so for $3-4q=0$. Maybe this can be generalized by considering certain linear combinations; just saying.

Answer 2

Another approach comes from expressing the sums in $(s,r)$ and then managing the resulting partial sum.

1) the sum $F_1$

Starting from your first question, we write it as $$ \bbox[lightyellow] { \eqalign{ & F_{\,1} (L,q,x) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{\Gamma (L + r - 2q)} \over {\Gamma (L + r - 1 + 2q)}}{{\Gamma (L + r + l - 1 + 2q)} \over {\Gamma (L + r + l + 2)}}x^{\,l} x^{\,r} } } = \cr & = \sum\limits_{s = 0}^\infty {{{\Gamma (L + s - 1 + 2q)} \over {\Gamma (L + s + 2)}}x^{\,s} } \sum\limits_{r = 0}^s {{{\Gamma (L + r - 2q)} \over {\Gamma (L + r - 1 + 2q)}}} = \cr & = \sum\limits_{s = 0}^\infty {{{\Gamma (L - 1 + 2q)} \over {\Gamma (L + 2)}}{{\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } } \over {\left( {L + 2} \right)^{\,\overline {\,s\,} } }}x^{\,s} } {{\Gamma \left( {L - 2q} \right)} \over {\Gamma \left( {L - 1 + 2q} \right)}} \sum\limits_{r = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,r\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,r\,} } }}} = \cr & = {{\Gamma \left( {L - 2q} \right)} \over {\Gamma (L + 2)}}\sum\limits_{s = 0}^\infty {{{\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } } \over {\left( {L + 2} \right)^{\,\overline {\,s\,} } }}x^{\,s} } \sum\limits_{r = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,r\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,r\,} } }}} \cr} } \tag {1.0}$$

Let us consider the inner partial sum.
It is a sum of hypergeometric terms and to it we can apply the Gosper's algorithm, which I find to be well described and analyzed in the renowned book Concrete Mathematics.
In the following I am concisely developing the steps recommended in the relevant "Ch. 5.7 - Partial Hypergeometric Sums": please refer to that for further details.

So
- we write the sum as $$ S_P (L,q,s) = \sum\limits_{k = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}} = \sum\limits_{k = 0}^s {t_{\,k} } $$ - express the ratio of the summand as $$ {{t_{\,k + 1} } \over {t_{\,k} }} = {{\left( {L - 2q + k} \right)} \over {\left( {L - 2 + 2q + k + 1} \right)}} = {{p(k + 1)} \over {p(k)}}{{q(k)} \over {r(k + 1)}} $$ - choose the three polynomials to be $$ \left\{ \matrix{ p(k) = 1 \hfill \cr q(k) = L - 2q + k \hfill \cr r(k) = L - 2 + 2q + k \hfill \cr} \right. $$ - determine the polynomial $s(k)$ by solving $$ p(k) = q(k)s(k + 1) - r(k)s(k)\quad \Rightarrow \quad s(k) = {1 \over {4q - 2}} $$ - arrive to find the antidifference $T(k)$ of $t_k$ to be $$ \bbox[lightyellow] { \left\{ \matrix{ T(k + 1) - T(k) = t_{\,k} \hfill \cr T(k) = {{r(k)s(k)t(k)} \over {p(k)}} = {{L - 2 + 2q} \over {4q - 2}}{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 2 + 2q} \right)^{\,\overline {\,k\,} } }} \hfill \cr} \right. } \tag {1.1}$$ - and conclude that $$ \bbox[lightyellow] { \eqalign{ & S_P (L,q,s) = \sum\limits_{k = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}} = T(s + 1) - T(0) = \cr & = {{L - 2 + 2q} \over {2 - 4q}}\left( {{{\left( {L - 2q} \right)^{\,\overline {\,s + 1\,} } } \over {\left( {L - 2 + 2q} \right)^{\,\overline {\,s + 1\,} } }} - 1} \right) = \cr & = {{L - 2 + 2q} \over {2 - 4q}}\left( {{{\left( {L - 2q} \right)\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } } \over {\left( {L - 2 + 2q} \right)\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} - 1} \right) = \cr & = \left( {{{L - 2q} \over {2 - 4q}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} - {{L - 2 + 2q} \over {2 - 4q}} \cr} } \tag {1.2}$$

Note that, although $T(k)$ has a pole at $q=1/2$, it cancels out in the difference $T(k+1)-T(k)=t_k$.

Upon inserting (1.2) back into (1.0), then it is just a matter of algebraic manipulation.

2) the sum $F_2$

Since $F_2$ is same as $F_1$ with the inner sum being multiplicated by $r$, then this is rewritten as $$ \bbox[lightyellow] { \eqalign{ & S_P (L,q,s) = \sum\limits_{k = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}k} = \cr & = \sum\limits_{k = 1}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}k} = \sum\limits_{k = 1}^s {{{\left( {L-2q} \right)^{\,\overline {\,k + 1\,} } } \over {\left( {L-1+2q} \right)^{\,\overline {\,k+1\,} } }}\left( {k+1} \right)} = \cr & = {{\left( {L - 2q} \right)} \over {\left( {L - 1 + 2q} \right)}}\sum\limits_{k = 0}^{s - 1} {{{\left( {L - 2q + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {L + 2q} \right)^{\,\overline {\,k\,} } }}\left( {k + 1} \right)} \cr} } \tag {2.0}$$

Proceeding similarly to the above, the steps in summary being $$ \eqalign{ & \left\{ \matrix{ p(k) = \left( {k + 1} \right) \hfill \cr q(k) = \left( {L - \left( {2q - 1} \right) + k} \right) \hfill \cr r(k) = \left( {L + \left( {2q - 1} \right) + k} \right) \hfill \cr} \right. \cr & s(k) = {1 \over {3 - 4q}}\left( {k + {{L - 2 + 2q} \over {2\left( {2q - 1} \right)}}} \right) \cr} $$ we reach to $$ \bbox[lightyellow] { \eqalign{ & T(k) = {{r(k)s(k)t(k)} \over {p(k)}} = \cr & = {{L - 1 + 2q} \over {3 - 4q}}\left( {k + {{L - 2 + 2q} \over {4q - 2}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }} \cr} } \tag {2.1}$$

Thus $$ \bbox[lightyellow] { \eqalign{ & S_P (L,q,s) = \sum\limits_{k = 0}^s {{{\left( {L - 2q} \right)^{\,\overline {\,k\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,k\,} } }}k} = \cr & = {{\left( {L - 2q} \right)} \over {\left( {L - 1 + 2q} \right)}}\sum\limits_{k = 0}^{s - 1} {{{\left( {L - 2q + 1} \right)^{\,\overline {\,k\,} } } \over {\left( {L + 2q} \right)^{\,\overline {\,k\,} } }}\left( {k + 1} \right)} = \cr & = {{\left( {L - 2q} \right)} \over {3 - 4q}}\left( {\left( {s + {{L - 2 + 2q} \over {4q - 2}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} - \left( {{{L - 2 + 2q} \over {4q - 2}}} \right)} \right) = \cr & = {{\left( {L - 2q} \right)} \over {3 - 4q}}\left( {\left( {s + 1 + {{L - 2q} \over {4q - 2}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} - \left( {{{L - 2 + 2q} \over {4q - 2}}} \right)} \right) = \cr & = {{\left( {L - 2q} \right)} \over {3 - 4q}}\left( {{{2^{\,\overline {\,s\,} } \left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } } \over {1^{\,\overline {\,s\,} } \left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} + \left( {{{L - 2q} \over {4q - 2}}} \right){{\left( {L + 1 - 2q} \right)^{\,\overline {\,s\,} } } \over {\left( {L - 1 + 2q} \right)^{\,\overline {\,s\,} } }} - \left( {{{L - 2 + 2q} \over {4q - 2}}} \right)} \right) \cr} } \tag {2.2}$$

3) Gosper's algorithm applied directly to the original sum

The Gosper's algorithm can be applied as well to your original sum given in the precedent post, in reply to which it was already expressed in closed form as a

Kampé de Fériet function computed at $x=y=1$.

Following the same steps as above, we can now express it also as a linear combination of hypergeometric functions.

In fact, putting $$ \bbox[lightyellow] { \eqalign{ & \left\{ \matrix{ \matrix{ {a = L - 2 + 2q} \hfill & {b = \,4q} \hfill \cr {c = L} \hfill & {d = 3 - 2q\,} \hfill \cr } \hfill \cr A = {1 \over 2}\left( {c + 2} \right)^{\,\overline {\, - d\,} } \left( {a + 1} \right)^{\,\overline {\,1 - b\,} } = {1 \over 2}\left( {L + 2} \right)^{\,\overline {\, - 2 - 2q\,} } \hfill \cr} \right. \cr & S(L,q) = \sum\limits_{r = 0}^\infty {\sum\limits_{l = 0}^\infty {{{\Gamma (L + r - 2q)} \over {\Gamma (L + r - 1 + 2q)}} {{\Gamma (L + r + l - 1 + 2q)} \over {\Gamma (L + r + l + 2)}}{{r + 1} \over {r + l + 2}}} } \cr & = A\;\sum\limits_{s = 0}^\infty {{{2^{\,\overline {\,s\,} } \left( {2 + c - d} \right)^{\,\overline {\,s\,} } } \over {3^{\,\overline {\,s\,} } \left( {2 + c} \right)^{\,\overline {\,s\,} } }} \sum\limits_{r = 0}^s {{{2^{\,\overline {\,r\,} } \left( {a + 2 - b} \right)^{\,\overline {\,r\,} } } \over {1^{\,\overline {\,r\,} } \left( {a + 1} \right)^{\,\overline {\,r\,} } }}} } \cr} } \tag {3.0}$$

for the inner sum $$ S_{\,p} (a,b,s) = \sum\limits_{k = 0}^s {{{2^{\,\overline {\,k\,} } \left( {a + 2 - b} \right)^{\,\overline {\,k\,} } } \over {1^{\,\overline {\,k\,} } \left( {a + 1} \right)^{\,\overline {\,k\,} } }}} $$ we go through steps similar to the above and reach to $$ \bbox[lightyellow] { \eqalign{ & S_{\,p} (a,b,s) = \sum\limits_{k = 0}^s {{{2^{\,\overline {\,k\,} } \left( {a + 2 - b} \right)^{\,\overline {\,k\,} } } \over {1^{\,\overline {\,k\,} } \left( {a + 1} \right)^{\,\overline {\,k\,} } }}} = \cr & = {a \over {3 - b}}\left( {\left( {s + 1 - {{a - 1} \over {2 - b}}} \right){{\left( {a + 2 - b} \right)^{\,\overline {\,s + 1\,} } } \over {a^{\,\overline {\,s + 1\,} } }} + {{a - 1} \over {2 - b}}} \right) = \cr & = {a \over {3 - b}}\left( {{{\left( {a + 2 - b} \right)2^{\,\overline {\,s\,} } \left( {a + 3 - b} \right)^{\,\overline {\,s\,} } } \over {a\;1^{\,\overline {\,s\,} } \left( {a + 1} \right)^{\,\overline {\,s\,} } }} - \left( {{{a - 1} \over {2 - b}}} \right){{\left( {a + 2 - b} \right) \left( {a + 3 - b} \right)^{\,\overline {\,s\,} } } \over {a\left( {a + 1} \right)^{\,\overline {\,s\,} } }} + {{a - 1} \over {2 - b}}} \right) = \cr & = {{\left( {a + 2 - b} \right)} \over {3 - b}}\left( {{{2^{\,\overline {\,s\,} } \left( {a + 3 - b} \right)^{\,\overline {\,s\,} } } \over {1^{\,\overline {\,s\,} } \left( {a + 1} \right)^{\,\overline {\,s\,} } }} - \left( {{{a - 1} \over {2 - b}}} \right){{\left( {a + 3 - b} \right)^{\,\overline {\,s\,} } } \over {\left( {a + 1} \right)^{\,\overline {\,s\,} } }} + {{a\left( {a - 1} \right)} \over {\left( {a + 2 - b} \right)\left( {2 - b} \right)}}} \right) \cr} } \tag {3.1}$$