Definite Integral from 0 to 1 of Zeta of 2 minus Dilogarithm of x over One minus x
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Theorem
- $\ds \int_{\to 0}^{\to 1} \dfrac {\paren {\map \zeta 2 - \map {\Li_2} x } } {1 - x} \rd x = 2 \map \zeta 3$
where:
- $\map {\Li_2} x$ is the Dilogarithm function of $x$
- $\map \zeta 2$ is the Riemann $\zeta$ function of $2$
- $\map \zeta 3$ is Apéry's constant: the Riemann $\zeta$ function of $3$.
Proof
With a view to expressing the primitive in the form:
| \(\ds \int u \frac {\d v} {\d x} \rd x\) | \(=\) | \(\ds u v - \int v \frac {\d u} {\d x} \rd x\) | Integration by Parts |
let:
| \(\ds u\) | \(=\) | \(\ds \map \zeta 2 - \map {\Li_2} x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac {\map \ln {1 - x} } x\) | Definition of Spence's Function: $\ds \map {\Li_2} z = -\int_0^z \frac {\map \Ln {1 - t} } t \rd t$ |
and let:
| \(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds \frac 1 {\paren {1 - x} }\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds -\paren {\map \ln {1 - x} }\) | Primitive of $\dfrac 1 {a x + b}$ |
Then:
| \(\ds \int_{\to 0}^{\to 1} \dfrac {\paren {\map \zeta 2 - \map {\Li_2} x } } {1 - x} \rd x\) | \(=\) | \(\ds \bigintlimits {\paren {\map \zeta 2 - \map {\Li_2} x} \paren {-\paren {\map \ln {1 - x} } } } 0 1 + \int_0^1 \frac {\paren {\map \ln {1 - x} }^2} x \rd x\) | Integration by Parts | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0 + \int_0^1 \frac {\paren {\map \ln {1 - x} }^2} x \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \int_1^0 \frac {\paren {\map \ln u}^2} {1 - u} \paren {-\d u}\) | $x \to \paren {1 - u}$ and $\rd x \to -\rd u$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^1 \frac {\paren {-\map \ln u} \paren {-\map \ln u} } {1 - u} \rd u\) | reversing limits of integration | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^1 \frac {\paren {\map \ln {\dfrac 1 u} }^{3 - 1} } {1 - u} \rd u\) | Logarithm of Reciprocal | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \zeta 3 \map \Gamma 3\) | Integral Representation of Riemann Zeta Function in terms of Gamma Function: $\ds \map \zeta s \map \Gamma s = \int_0^1 \frac {\paren {\map \ln {\frac 1 u} }^{s - 1} } {1 - u} \rd u$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \map \zeta 3\) | Gamma Function of $3$ |
$\blacksquare$