Definite Integral from 0 to 1 of Logarithm of x by Logarithm of One minus x
Jump to navigation
Jump to search
Theorem
- $\ds \int_0^1 \ln x \map \ln {1 - x} \rd x = 2 - \frac {\pi^2} 6$
Proof
\(\ds \int_0^1 \ln x \map \ln {1 - x} \rd x\) | \(=\) | \(\ds -\int_0^1 \ln x \paren {\sum_{n \mathop = 1}^\infty \frac {x^n} n} \rd x\) | Power Series Expansion for $\map \ln {1 + x}$: Corollary | |||||||||||
\(\ds \) | \(=\) | \(\ds -\sum_{n \mathop = 1}^\infty \frac 1 n \paren {\int_0^1 x^n \ln x \rd x}\) | Fubini's Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-1}^2 \sum_{n \mathop = 1}^\infty \frac 1 n \paren {\frac {\map \Gamma 2} {\paren {n + 1}^2} }\) | Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1! \sum_{n \mathop = 1}^\infty \frac 1 {n \paren {n + 1}^2}\) | Gamma Function Extends Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {n + 1} } - \sum_{n \mathop = 1}^\infty \paren {\frac 1 {\paren {n + 1}^2} }\) | partial fraction expansion | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 - \paren {\sum_{n \mathop = 1}^\infty \frac 1 {n^2} - 1}\) | Telescoping Series: Example 1 | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 - \frac {\pi^2} 6\) | Basel Problem |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Logarithmic Functions: $15.96$