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A${\;}_3F_2$ identity

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Post Tue May 03, 2016 8:22 am
Shobhit Site Admin
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Location: Jaipur, India

Prove that $${\;}_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};1 \right)=\frac{4G}{\pi}$$

where $G$ is Catalan's constant.

BONUS Problem

For $0< a\leq 1$, $${\;}_3F_2\left(\frac{1}{2},\frac{1}{2},a;1,1+a;1 \right)=\frac{4a}{\pi}{\;}_3F_2\left(1,1,1-a;\frac{3}{2},\frac{3}{2};1 \right)$$

Post Tue May 03, 2016 11:48 am

Posts: 20
Location: Israel
Well, after simplifying the pochhammer symbols we get that it is is equal to the following sum:
$$S= \sum_{n=0}^{\infty} \binom{2n}{n}^2 \frac1{2^{4n} (2n+1)}.$$

It is now easy to finish off using the integral (which can be calculated using the beta function and Legendre's duplication formula):
$$\int_0^1 \frac{x^{2 n}}{\sqrt{1-x^2}}dx=\frac{\pi}{2} \binom{2n}{n}\frac1{2^{2n}}$$

and also the power series expansion of arcsin:

$$\sin^{-1} x=\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{2n+1}}{2^{2n}(2n+1)}.$$

Thus we get:
$$ S= \frac{2}{\pi}\sum_{n=0}^{\infty} \binom{2n}{n} \frac1{2^{2n}(2n+1)} \int_0^1 \frac{x^{2 n}}{\sqrt{1-x^2}}dx
\\= \frac{2}{\pi} \int_0^1 \frac1{\sqrt{1-x^2}} \frac{\sin^{-1} x}{x} dx
\\=\frac{2}{\pi} \int_0^{\large\frac{\pi}{2}} \frac{x}{\sin x} dx.$$

That last integral is very easy to calculate using the sub $\large t=\tan\frac{x}{2}$.

Post Wed May 04, 2016 2:42 am
Shobhit Site Admin
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Posts: 852
Location: Jaipur, India

Well done! :) The bonus problem is actually a special case of an identity due to Bailey. $${\;}_3F_2\left(a,b,c;d,e;1 \right)=\frac{\Gamma(d)\Gamma(e)}{\Gamma(a)}\frac{\Gamma(s)}{\Gamma(s+b)\Gamma(s+c)}{\;}_3F_2\left(d-a,e-a,s;s+b,s+c;1 \right)$$ for $s=d+e-a-b-c$, when $\text{Re}(s)>0$ and $\text{Re}(a)>0$.

ADDED

Using the bonus problem and G.N. Watson's identity, we can evaluate the case when $a=\frac{1}{4}$.
\begin{align*}
{\;}_3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{4};1,\frac{5}{4};1 \right) &=\frac{1}{\pi}{\;}_3F_2\left(1,1,\frac{3}{4};\frac{3}{2},\frac{3}{2};1 \right) \\
&= \frac{\Gamma^4\left(\frac{1}{4}\right)}{16\pi^2}
\end{align*}

This is equivalent to Ramanujan's classic series: $$\sum_{n=0}^\infty\frac{\binom{2n}{n}^2}{(4n+1)16^k}= \frac{\Gamma^4\left(\frac{1}{4}\right)}{16\pi^2}$$

Post Thu Dec 08, 2016 3:34 pm

Posts: 1
How about this generalization .......
$$
\frac{\Gamma(s+1)^2{\;}_3F_2(s+1,s+1,s+1;s+2,2s+2;1)}
{2(s+1)\Gamma(2s+2)}
=
\sum_{j=0}^\infty \frac{(-1)^j}{(s+1+j)^2}
$$
with $-\frac{1}{2} \le s \le 0$.


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