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In this MSE-question I've asked about three, similarly shaped, integrals for integer vales of $\zeta(s)$ that I found numerically:

$$\zeta \left( 3 \right) =\frac12{\int_{0}^{1} \frac{1}{x}\big(\zeta(2)-{\it Li_2} \left(1-x\right)\big) \,{\rm d}x} \tag{1}$$

$$\zeta \left( 4 \right) =\frac{4}{5}{\int_{0}^{1} \frac{1}{x}\big(\zeta(3)-{\it Li_3} \left(1-x\right)\big) \,{\rm d}x} \tag{2}$$

$$\zeta \left( 5 \right) =\frac12{\int_{0}^{1} \frac{1}{x}\big(\zeta(2)-{\it Li_2} \left(1-x\right)\big)^2 \,{\rm d}x} \tag{3}$$

ADDED: found one more:

$$\zeta \left( 3 \right) = \frac32 - \frac14{\int_{0}^{1} \big(\zeta(2)-{\it Li_2} \left(1-x\right)\big)^2 \,{\rm d}x} \tag{4}$$

where ${\it Li_n}(z)$ is the PolyLogarithm.

I have not found any similar expressions at other integer values.

The answer to the MSE-question helped reducing the integral for $\zeta(3)$ to a known integral, however still curious whether the other two could be reduced to something known as well.

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  • $\begingroup$ Again indicate the series corresponding to the integral $\endgroup$ – reuns Jun 21 at 18:14
  • $\begingroup$ "a known integral" --- if Mathematica knows the integral, does that count? $\endgroup$ – Carlo Beenakker Jun 21 at 19:58
  • $\begingroup$ @Carlo Beenhakker, yes, that counts since Mathematica also yields the indefinite integral for the $\zeta(4)$-case. One down; how about the integral for $\zeta(5)$? $\endgroup$ – Agno Jun 21 at 21:24
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These identities, and many more, follow from a theorem in Integrals of polylogarithmic functions, recurrence relations, and associated Euler sums,

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For example,

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There are also variations with $\log x$ factor, such as

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  • $\begingroup$ Thanks Carlo. This will solve the $\zeta(5)$ case in equation (3)! I had actually just found (numerically) this simple equation for integers $n \ge 2$: $${\int_{0}^{1} \zeta(n)-{\it Li_n} \left(1-x\right) \,{\rm d}x} = (-1)^n \left(1-\sum_{k=2}^{n-1} (-1)^k\,\zeta(k) \right)$$ $\endgroup$ – Agno Jun 23 at 18:16

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