Definite Integral from 0 to 1 of Logarithm of One minus x over x
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Theorem
- $\ds \int_0^1 \frac {\map \ln {1 - x} } x \rd x = -\frac {\pi^2} 6$
Proof
\(\ds \int_0^1 \frac {\map \ln {1 - x} } x \rd x\) | \(=\) | \(\ds -\int_0^1 \frac 1 x \paren {\sum_{n \mathop = 1}^\infty \frac {x^n} n} \rd x\) | Power Series Expansion for $\map \ln {1 + x}$: Corollary | |||||||||||
\(\ds \) | \(=\) | \(\ds -\sum_{n \mathop = 1}^\infty \paren {\frac 1 n \int_0^1 x^{n - 1} \rd x}\) | Power Series is Termwise Integrable within Radius of Convergence | |||||||||||
\(\ds \) | \(=\) | \(\ds -\sum_{n \mathop = 1}^\infty \frac 1 {n^2}\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac {\pi^2} 6\) | Basel Problem |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Logarithmic Functions: $15.94$