# Three integral expressions for integer values of $\zeta(s)$. Could these be further reduced to known integrals?

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

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

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

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

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

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

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.

• Again indicate the series corresponding to the integral – reuns Jun 21 at 18:14
• "a known integral" --- if Mathematica knows the integral, does that count? – Carlo Beenakker Jun 21 at 19:58
• @Carlo Beenhakker, yes, that counts since Mathematica also yields the indefinite integral for the $\zeta(4)$-case. One down; how about the integral for $\zeta(5)$? – Agno Jun 21 at 21:24

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