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

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


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

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

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.


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:

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


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

$$v= \frac{1}{2} \frac{ab+1+a+b}{ab+1-a-b} $$

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:

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



This is not an answer, but I ran out of space in the post, so I will be adding any new results on this topic here.

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


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

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

I think the symmetry of this is beautiful, and this leads me to believe that more identities like this one are possible.

Using Euler integral and simplifying, we obtain:

$$\sum_{k=0}^\infty \frac{\beta^{2k+1}}{2k+1} \int_0^1 \frac{t^{2k} dt}{\sqrt{(1-t)(1- \alpha t)}}= \log (a) \log (b)$$

Summation gives us:

$$\int_0^1 \frac{\tanh^{-1} (\beta t) dt}{t\sqrt{(1-t)(1- \alpha t)}}= \log (a) \log (b)$$

After working on the explicit expression for the hypergeometric function in the first series, we can now write:

$$\sum_{n=0}^\infty \frac{(r s)^{-2n-1}}{2n+1} \left( \sum _{k=0}^{2 n} (-1)^k \binom{2 n}{k} \binom{2 n+k}{k} \frac{H_{2 n}-H_k}{s^k}-\frac{\log(1-s)}{2} P_{2n} \left(\frac{2}{s}-1 \right) \right) = \\ = \frac{1}{4} \log (a) \log (b)$$

Surprisingly enough, both terms inside the series seem to converge individually, in particular:

$$\sum_{n=0}^\infty \frac{(r s)^{-2n-1}}{2n+1} P_{2n} \left(\frac{2}{s}-1 \right) = \frac{1}{2} \log (c), \qquad c = \begin{cases} a, & 1<a< b \\ b & 1<b< a \end{cases}$$

I don't know how to prove this last result, but it works numerically.


It is entirely possible to express $a$ and $b$ as functions of $r,s$.

Writing $$\sqrt{1-\frac{16 ab}{(ab+1+a+b)^2}}=\frac s{2-s}\implies ab=\frac{1-s}{4(2-s)^2}\cdot(ab+1+a+b)^2$$ and letting $$t=\frac{(2-s)r+1}{(2-s)r-1}$$ yields \begin{align}r=\frac12\cdot\frac{ab+1+a+b}{ab+1-a-b}\cdot\frac2{2-s}&\implies ab-t(a+b)+1=0\\&\implies b=\frac{ta-1}{a-t}\end{align} so $$t(a+b)-1=\frac{1-s}{4(2-s)^2}(1+t)^2(a+b)^2\implies a+b=k$$ with $$k=\frac{2(2-s)^2}{1-s}\left(t\pm\sqrt{t^2-\frac{1-s}{(2-s)^2}(1+t)^2}\right)$$ where the positive root must be taken for $s\le1$, giving $$a+\frac{ta-1}{a-t}=k\implies a=\frac{k\pm\sqrt{k^2-4(kt-1)}}2$$ where the positive root must be taken for $kt\ge1$, and therefore $a(r,s)$ and $b(r,s)$ are expressed up to radicality.

  • $\begingroup$ Thank you! This allows to treat $r,s$ as independent variables, and the rhs as a two variable generating function $\endgroup$ – Yuriy S Jul 19 '19 at 7:36

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