Definite Integral from 0 to 1 of Logarithm of x by Logarithm of One minus x over One minus x
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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 1
| \(\ds \int_0^1 \dfrac {\ln x \map \ln {1 - x} } {\paren {1 - x} } \rd x\) | \(=\) | \(\ds \int_1^0 \dfrac {\map \ln {1 - u} \ln u } u \paren {-\rd u}\) | $x \to \paren {1 - u}$ and $\rd x \to -\rd u$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \int_0^1 \dfrac {\map \ln {1 - u} \ln u } u \rd u\) | reversing limits of integration | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\int_0^1 \frac {\ln u} u \paren {\sum_{n \mathop = 1}^\infty \frac {u^n} n}\) | Power Series Expansion for $\map \ln {1 + x}$: Corollary | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\int_0^1 \ln u \paren {\sum_{n \mathop = 1}^\infty \frac {u^{n - 1} } n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\sum_{n \mathop = 1}^\infty \frac 1 n \paren {\int_0^1 u^{n - 1} \ln u \rd u}\) | Fubini's Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {-1}^2 \sum_{n \mathop = 1}^\infty \frac 1 n \paren {\frac {\map \Gamma 2} {\paren {\paren {n - 1} + 1}^2} }\) | Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1! \sum_{n \mathop = 1}^\infty \frac 1 {n^3}\) | Gamma Function Extends Factorial | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \zeta {3}\) | Definition of Apéry's Constant |
$\blacksquare$
Proof 2
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 |
$\blacksquare$