A series for $\log (a) \log (b)$ in terms of hypergeometric function

Solution 1:

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.

• Using this hypergeometric transformation, on the second series, I was able to obtain another identity:

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

Where:

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

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

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.

Solution 2:

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.