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I was experimenting with Frullani integral again, and obtained a very curious series:

$$\sum_{k=0}^\infty \frac{{_2 F_1} (2k+1,2k+1;4k+2;s)}{(2k+1)^2 \binom{4k+2}{2k+1} r^{2k+1}}= \frac{1}{4} \log (a) \log (b)$$

Here:

$$r= \frac{1}{2} \frac{ab+1+a+b}{ab+1-a-b} \left(1+\sqrt{1-\frac{16 ab}{(ab+1+a+b)^2}} \right)$$ $$s= \frac{2\sqrt{1-\frac{16 ab}{(ab+1+a+b)^2}}}{1+\sqrt{1-\frac{16 ab}{(ab+1+a+b)^2}}} $$

For example:

$$\sum_{k=0}^\infty \frac{{_2 F_1} (2k+1,2k+1;4k+2;\sqrt{3}-1)}{(2k+1)^2 \binom{4k+2}{2k+1} (3+\sqrt{3})^{2k+1}}= \frac{1}{4} \log (2) \log (3)$$

What's really amazing is that $7$ terms of the series already give $16$ correct digits for the right hand side: $0.1903750026047022 \ldots$. On the other hand $48$ terms give $100$ correct digits.

The result might be pretty useless for computations, because the terms feature hypergeometric functions, but they are a very special case (${_2 F_1} (n,n;2n;x)$) and probably have some special properties which could make them easier to evaluate.

Have you seen any series like that? Is there a list of series with ${_2 F_1}$ terms which have elementary closed forms? How would you prove this result? Can it lead to any useful or interesting identities? As a more practical question, can we express $a(r,s)$ and $b(r,s)$ in radicals?

The way I obtained the series is too long to fully provide here, but I started with a double Frullani integral:

$$\int_0^\infty \int_0^\infty \frac{d x dy}{x y} (e^{-x}-e^{-a x})(e^{-y}-e^{-b y})=\log (a) \log (b)$$

Then used polar substitution $x= \rho \cos \phi$, $y= \rho \sin \phi$, integrated w.r.t. $\rho$, used half-angle tangent substitution, expanded the logarithm and then integrated each term using Appell function which then reduced to hypergeometric function.

Update:

Using a known transformation, we can write:

$${_2 F_1} (2k+1,2k+1;4k+2;x)= \frac{1}{(1-x/2)^{2k+1}} {_2 F_1} \left(k+\frac12,k+1;2k+\frac32;\frac{x^2}{(2-x)^2}\right)$$

Which makes the particular case above more beautiful since both the parameters become rational:

$$\color{blue}{\sum_{k=0}^\infty \frac{{_2 F_1} \left(k+\frac12,k+1;2k+\frac32;\frac{1}{3}\right)}{(2k+1)^2 \binom{4k+2}{2k+1} 3^{2k+1}}= \frac{1}{4} \log (2) \log (3)}$$

In the general case the parameters also become rational:

$$\sum_{k=0}^\infty \frac{{_2 F_1} \left(k+\frac12,k+1;2k+\frac32;u\right)}{(2k+1)^2 \binom{4k+2}{2k+1} v^{2k+1}}= \frac{1}{4} \log (a) \log (b)$$

Where:

$$u= 1-\frac{16 ab}{(ab+1+a+b)^2} $$ $$v= \frac{1}{2} \frac{ab+1+a+b}{ab+1-a-b} $$

It seems that for $a,b>0$ we have $0<u<1$ and $v>1/2$ which is good for convergence.

Update 2:

Using Euler integral for the hypergeometric function, and summing the series we obtain another, more simple identity:

$$\int_0^1 \text{arctanh} \left(\frac{1}{2v} \sqrt{\frac{x(1-x)}{1-u x}} \right) \frac{dx}{x \sqrt{(1-x)(1-u x)}}=\frac{1}{2} \log (a) \log (b)$$

While the general solution for $a(u,v)$ and $b(u,v)$ eludes me, there's a single parameter case that's easy to express:

$$\sum_{k=0}^\infty \frac{{_2 F_1} \left(k+\frac12,k+1;2k+\frac32;\frac{1}{p}\right)}{(2k+1)^2 \binom{4k+2}{2k+1} p^{2k+1}}= \frac{1}{4} \log \left(\frac{2 p+\sqrt{8 p+1}+1}{2 (p-1)} \right) \log \left(\frac{2 p+\sqrt{8 p+1}+1}{2 p} \right)$$

$$p>1$$