5

It is well-known that $$10\int_0^{\ln \phi} t^2\coth t dt =\frac{5}{2}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3\binom{2n}{n}}=\zeta(3),$$ where $\phi=(1+\sqrt{5})/2$ (see Alfred van der Poorten, "A proof that Euler missed"). Let $$f(r,s)=\int_0^{s/2}t^{r-1}\coth t dt.$$ In particular, $$2f(3,s)=2\int_0^{s/2}t^2\coth t dt =\frac{s^3}{12}+\frac{s^2}{2}\log(1-e^{-s})-s\mbox{Li}_2(e^{-s})-\mbox{Li}_3(e^{-s}) +\zeta(3),$$ where $\mbox{Li}_k(x)$ is the $k$-th polylogarithm. Indeed, one can obtain similar expressions for all $\zeta(2n+1)$. Define $F(s)_3=2f(3,s)-\zeta(3)$. Since $F(0)_3=-\zeta(3)$ and $F(s)_3$ increases in $s$ without bound, there is a (unique) zero of $F(s)_3$.

Question 1: Is there a simple representation of $F(s)_3$ in terms of elementary, perhaps hyperbolic, functions?

Question 2: Can anyone see how to solve $F(s)_3=0$?

rpembroke
  • 115

1 Answers1

1

My answer to the first question is : No

For the solution of $F(s)_3=0$, notice that $F(2)_3=-0.0423076$, so the solution is just above $2$. The numerical solution, given by Newton method, is $$s=2.0318521264849950152299208376128996767397718727843\cdots$$ which is extremely close to $\pi$ times the largest root of the quartic $$2728 x^4+4144 x^3+8476 x^2-8326 x+241=0$$ (the relative error is $1.12\times 10^{-22}$).

Another good approximation is $$s \sim 2+\frac{-765-1220 e+576 e^2}{2 \left(-19-215 e+453 e^2\right)}$$

Now, if you want an expression, use a series expansion around $s=2$ and then series reversion to obtain $$s=2+(e^2-1)t-\frac{\left(1+e^2-3 e^4+e^6\right) }{2 \left(1+e^2\right)}t^2+O(t^3)$$ with $$t=\frac{3 \Re\left(\text{Li}_3\left(e^2\right)-2 \text{Li}_2\left(e^2\right)\right)+2-6 \log \left(e^2-1\right)}{3 \left(1+e^2\right)}$$

Claude Leibovici
  • 260,315