Definite Integral from 0 to 1 of Logarithm of One minus x over x

From ProofWiki
Jump to navigation Jump to search


This article has been identified as a candidate for Featured Proof status.
If you do not believe that this proof is worthy of being a Featured Proof, please state your reasons on the talk page.
To discuss this page in more detail, feel free to use the talk page.


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

Retrieved from "https://proofwiki.org/w/index.php?title=Definite_Integral_from_0_to_1_of_Logarithm_of_One_minus_x_over_x&oldid=669351"