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I managed to derive the following integral:

$$\zeta \left( s \right) ={\frac { \left( s-2 \right)}{\Gamma \left( s \right) } \int_{0}^{\infty }\!{u}^{s-3} \left( \zeta(2)-{\it Li_2} \left(1-{{\rm e}^{-u}} \right) \right) \,{\rm d}u}\qquad \Re(s) \gt 2 \tag{1}$$

where $Li_2(z)$ is the dilogarithm.

For $s=3$ this reduces to:

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

or after the variable change $u=\ln(x)$:

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

or with $u=-\ln(x)$:

$$\zeta \left( 3 \right) =\frac12{\int_{0}^{1} \frac{1}{x}\big(\zeta(2)-{\it Li_2} \left(1-x\right)\big) \,{\rm d}x} \tag{4}$$

Searched the web for similar expressions, but haven't found anything related yet.

Could this be simplified any further into know expressions?

ADDED 1: A surprise outcome is that:

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

ADDED 2: Found one more:

$$\zeta \left( 4 \right) =\frac{4}{5}{\int_{0}^{1} \frac{1}{x}\big(\zeta(3)-{\it Li_3} \left(1-x\right)\big) \,{\rm d}x} \tag{6}$$

Agno
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1 Answers1

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One route to a connection with a simplified integral, is to integrate (4) by parts to give $$\zeta(3)=\frac{1}{2} \int_0^1 \frac{\log ^2(x)}{1-x} \, dx \tag{A1}$$ (the first term being equal to zero).

Result (4) can then be derived from $(A1)$, by integrating by parts again in the right way.

The required constant of integration in regard to the indefinite integral $$\int \frac{\log (x)}{1-x} \, dx=\text{Li}_2(1-x)+C$$ is found by equating the first term generated in this 'reversed' integration by parts to zero.

James Arathoon
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  • That makes sense for $\zeta(3)$! Do you also see a similar connection for $\zeta(5)$ (i.e. equation (5))? – Agno May 11 '20 at 10:55