8
$\begingroup$

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:

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

$$p>1$$

$\endgroup$
6
$\begingroup$

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.

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

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.