Definite Integral from 0 to 1 of Logarithm of x by Logarithm of One minus x over One minus x/Proof 2

Theorem

$\ds \int_0^1 \dfrac {\ln x \map \ln {1 - x} } {\paren {1 - x} } \rd x = \map \zeta 3$

where $\map \zeta 3$ 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:

\(\ds u\)

\(=\)

\(\ds \ln x\)

\(\ds \leadsto \ \ \)

\(\ds \frac {\d u} {\d x}\)

\(=\)

\(\ds \frac 1 x\)

Derivative of $\ln x$

and let:

\(\ds \frac {\d v} {\d x}\)

\(=\)

\(\ds \frac {\map \ln {1 - x} } {\paren {1 - x} }\)

\(\ds \leadsto \ \ \)

\(\ds v\)

\(=\)

\(\ds - \frac 1 2 \paren {\map \ln {1 - x} }^2\)

Primitive of Power

Then:

\(\ds \int_0^1 \frac {\map \ln {1 - x} \ln x } {\paren {1 - x} } \rd x\)

\(=\)

\(\ds \intlimits {\ln x \paren {-\frac 1 2 \paren {\map \ln {1 - x} }^2} } 0 1 + \frac 1 2 \int_0^1 \frac {\paren {\map \ln {1 - x} }^2} x \rd x\)

Integration by Parts

\(\ds \)

\(=\)

\(\ds 0 + \frac 1 2 \int_0^1 \frac {\paren {\map \ln {1 - x} }^2} x \rd x\)

\(\ds \)

\(=\)

\(\ds \frac 1 2 \int_1^0 \frac {\paren {\map \ln u}^2} {1 - u} \paren {-\d u}\)

$x \to \paren {1 - u}$ and $\rd x \to -\rd u$

\(\ds \)

\(=\)

\(\ds \frac 1 2 \int_0^1 \frac {\paren {-\map \ln u} \paren {-\map \ln u} } {1 - u} \rd u\)

reversing limits of integration

\(\ds \)

\(=\)

\(\ds \frac 1 2 \int_0^1 \frac {\paren {\map \ln {\dfrac 1 u} }^{3 - 1} } {1 - u} \rd u\)

Logarithm of Reciprocal

\(\ds \)

\(=\)

\(\ds \frac {\map \zeta 3 \map \Gamma 3} 2\)

Integral Representation of Riemann Zeta Function in terms of Gamma Function: $\ds \map \zeta s \map \Gamma s = \int_0^1 \frac {\paren {\map \ln {\frac 1 u} }^{s - 1} } {1 - u} \rd u$

\(\ds \)

\(=\)

\(\ds \map \zeta 3\)

Gamma Function of 3

$\blacksquare$