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I was reading about Euler's approach for the closed form of $\zeta(2)$. On the same lines, if we instead consider the factorization

$x(1-\frac{x}{\pi})(1+\frac{x}{\pi})(1-\frac{x}{2\sqrt2\pi})(1+\frac{x}{2\sqrt2\pi})(1-\frac{x}{3\sqrt3\pi})(1+\frac{x}{3\sqrt3\pi})(1-\frac{x}{4\sqrt4\pi})(1+\frac{x}{4\sqrt4\pi})...$,

i.e.,

$x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{2^3\pi^2})(1-\frac{x^2}{3^3\pi^2})(1-\frac{x^2}{4^3\pi^2})...$,

will it just be a matter of finding a trigonometric function representation of this polynomial ($\sin(x)$ in case of Euler's method), that when expanded as a Taylor series, and its co-efficients compared with the polynomial, that we shall be able to find a closed form for

$1 + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + ...$ ?

Further, if it can be proved that no trigonometric function can express the above polynomial, will that mean that no closed form for $\zeta(3)$ exists?

Please forgive my limited knowledge on this topic, lack of research, and only quoting Wikipedia.

Kenta S
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Neeraj Gupta
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2 Answers2

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I've seen this closed form over internet a long time ago, wolframalpha confirm that it is correct, but I've also a very limited math knowledge, I have no idea if it can be considered as closed form since it involve the third-order derivative of Gamma or polyGamma functions

I think it's coming from Plouffe's inverter

$\zeta(3) = -{1\over2}\biggl(\gamma^3+{1\over2}(\gamma\pi^2)+\Gamma'''(1)\biggl)$

wolframalpha computation result

Proposed as alternate form by wolframalpha :

$\zeta(3) = -{{\psi^{(2)}(1)}\over2}$

wolframalpha computation result

Ch'nycos
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    Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. – Community Sep 13 '23 at 07:34
  • The idea for your formula is simply to start with this digamma expression : $$\psi(z) = \frac{\mathrm{d}}{\mathrm{d}z}\ln\Gamma(z) = \frac{\Gamma'(z)}{\Gamma(z)}$$ so that $\Gamma'(z)=\Gamma(z) \psi(z)$
    and $;\displaystyle\Gamma'''(z)=\Gamma(z) \psi(z)^3 + 3,\Gamma(z) \psi'(z) \psi(z) + \Gamma(z) \psi''(z)$ and conclude using : $\psi(1)=-\gamma,;\psi'(1)=\zeta(2),;\psi''(1)=-2,\zeta(3)$.     – Raymond Manzoni Sep 13 '23 at 12:14
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Not an answer.

I'm not sure if Ch'nycos's answer will stay, but to support its claims a bit, by rearranging the Polygamma function's recurrence relation we obtain, $$\frac{1}{n^3}=\frac{\psi^{(2)}(n+1)-\psi^{(2)}(n)}{2}$$ summing over $n\in\mathbb{N}$, $$\zeta(3)=\sum_{n=1}^\infty\frac{1}{n^3}=\frac{1}{2}\sum_{n=1}^\infty\psi^{(2)}(n+1)-\psi^{(2)}(n)=-\frac{\psi^{(2)}(1)}{2}$$ after telescoping the series.

bob
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