Is this infinite product formula for Apéry's constant correct $\zeta(3)=\frac{\pi^3}{28}\prod_{n=1}^\infty \ln(1+1/n)(n+1/2)$?

Question

Hello fellow integral enthusiasts,

today, by accident, I derived the following formula for Apéry's constant (which is $\zeta(3)$ where $\zeta$ is the Riemann zeta function) $$\zeta(3)=\frac{\pi^3}{28}\prod_{n=1}^\infty \ln(1+1/n)(n+1/2)$$

Numerically, I obtain $\zeta(3)=1.1998$ using a million product terms, although the correct result should be $\zeta(3)=1.2021$. I am not 100% sure if the deviation is just due to numerical precision issues.

My derivation starts with the integral $$\int_0^\pi dt \frac{t(\pi-t)}{\sin(t)}$$ I use the substitution $t=\pi/2+x$, then symmetry of the integrand, and then the infinite product formula for the cosine. Then I interchange product and integral (which I hope is okay). On the other hand, a solution to the above integral which contains $\zeta(3)$ is given here (after substitution $\pi x=t$). Equating both evaluations of the integral yields the product formula for $\zeta(3)$.

Questions:

a) Is the formula correct?

b) Do you have another derivation for it?

Answer 1

I think your problem lies in the interchanging of the product and the integral. Even if it wasn't an infinite product, the interchanging would still be invalid. Here's a simple example:

$$\int_0^1x\sin x\ dx\approx 0.30116867894$$ while $$\left(\int_0^1xdx\right)\left(\int_0^1 \sin x\ dx\right)\approx 0.229848847066,$$ so those are obviously not the same.

Although $$\int \sum_{i}f_i(x)\ dx=\sum_{i}\int f_i(x)dx$$ holds under special circumstances, very rarely (if ever) does $$\int\prod_{i}f_i(x)\ dx=\prod_i \int f_i(x)dx$$ hold.

Answer 2

If you want an infinite product involving $\zeta(3)$, I suggest the starting formula $$\frac1{\sin\pi x}=\frac1\pi\sum_{k\in\Bbb Z}\frac{(-1)^k}{k+x},$$ from which you can show that $$\frac{\pi}{\sin\pi x}=\frac1x+\sum_{k\ge1}(-1)^k\frac{2x}{x^2-k^2}.$$ Hence, multiplying both sides by $x-x^2$ and integrating over $[0,1]$, $$\begin{align} \frac{7\zeta(3)}{\pi^2}&=\int_0^1\frac{x(1-x)}{x}dx+2\sum_{k\ge1}(-1)^k\int_0^1\frac{x^2(1-x)}{x^2-k^2}dx\\ &=\frac12+2\int_0^1\frac{x^2(1-x)}{1-x^2}dx+2\sum_{k\ge2}(-1)^k\int_0^1\frac{x^2(1-x)}{x^2-k^2}dx\\ &=-\frac12+2\ln2+2\sum_{k\ge2}(-1)^k\int_0^1\frac{x^2(1-x)}{x^2-k^2}dx. \end{align}$$ The final integral is a little tedious, but its value (after a lot of simplification) ends up being $$\int_0^1\frac{x^2(1-x)}{x^2-k^2}dx=\ln\left[\sqrt{e}\left(\frac{j(k)}{\sqrt{j(k+1)j(k-1)}}\right)^k\right]$$ where $j(x)=x^x$. With this and the definition $e_1(k)=\sqrt{e}\left(\tfrac{j(k)}{\sqrt{j(k+1)j(k-1)}}\right)^k$, we see that $$\begin{align} \frac{7\zeta(3)}{2\pi^2}+\frac14-\ln2&=\sum_{k\ge2}(-1)^k\ln\left[e_1(k)\right]\\ &=\sum_{k\ge2}\ln\left[e_1(k)^{(-1)^k}\right]\\ &=\sum_{k\ge2}\ln\left[e_2(k)\right]\\ &=\ln\left[\prod_{k\ge2}e_2(k)\right]. \end{align}$$ Thus the marvelous product $$\prod_{k\ge2}e^{(-1)^k/2}\left(\frac{j(k)}{\sqrt{j(k+1)j(k-1)}}\right)^{(-1)^kk}=\frac12\exp\left[\frac{7\zeta(3)}{2\pi^2}+\frac14\right].$$