Definite Integral from 0 to 1 of Even Powers of Logarithm of 1 - x over x/Corollary

Corollary to Definite Integral from 0 to 1 of Even Powers of Logarithm of 1 - x over x

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.

$\ds \int_0^1 \map {\ln^{2n} } {\dfrac {1 - x} x} \rd x = 2 \map \zeta {2 n} \paren {2 n}! \paren {1 - \dfrac 1 {2^{2n-1} } }$

Proof

\(\ds \int_0^1 \map {\ln^{2n} } {\dfrac {1 - x} x} \rd x\)

\(=\)

\(\ds \paren {-1}^{n + 1} B_{2 n} \paren {2^{2 n} - 2} \pi^{2 n}\)

Definite Integral from 0 to 1 of Even Powers of Logarithm of 1 - x over x

\(\ds \)

\(=\)

\(\ds \paren {-1}^{n + 1} B_{2 n} \paren {2^{2 n} - 2} \pi^{2 n} \dfrac {\paren {2 n}!} {\paren {2 n}!} \dfrac {\paren {2^{2 n} } } {\paren {2^{2 n} } }\)

multiply by 1

\(\ds \)

\(=\)

\(\ds 2 \paren {\paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {2^{2 n} \paren {2 n}!} } \paren {2 n}! \paren {2^{2 n} - 2}\)

rearranging

\(\ds \)

\(=\)

\(\ds 2 \paren {\paren {-1}^{n + 1} \dfrac {B_{2 n} 2^{2 n - 1} \pi^{2 n} } {\paren {2 n}!} } \paren {2 n}! \paren {1 - \dfrac 1 {2^{2n-1} } }\)

\(\ds \)

\(=\)

\(\ds 2 \map \zeta {2 n} \paren {2 n}! \paren {1 - \dfrac 1 {2^{2n-1} } }\)

Riemann Zeta Function at Even Integers

$\blacksquare$