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

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Theorem

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


$\ds \int_0^1 \map {\ln^{2n} } {\dfrac {1 - x} x} \rd x = \paren {-1}^{n + 1} B_{2 n} \paren {2^{2 n} - 2} \pi^{2 n}$


where $B_{2 n}$ is the $2 n$th Bernoulli number.


Corollary

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

let:

\(\ds \map \ln {\dfrac {1 - x} x}\) \(=\) \(\ds -u\) Integration by Substitution
\(\ds \dfrac {1 - x} x\) \(=\) \(\ds e^{-u}\)
\(\ds \dfrac 1 x - 1\) \(=\) \(\ds e^{-u}\)
\(\ds \dfrac 1 x\) \(=\) \(\ds 1 + e^{-u}\)
\(\ds x\) \(=\) \(\ds \dfrac 1 {1 + e^{-u} }\)
\(\ds \frac {\rd x} {\rd u}\) \(=\) \(\ds \dfrac { {e^{-u} } } {\paren { 1 + e^{-u} }^2 }\) Power Rule for Derivatives, Chain Rule for Derivatives and Derivative of Exponential Function: Corollary 1

and also:

\(\ds \lim_{x \mathop \to 0} \map \ln {\dfrac {1 - x} x}\) \(=\) \(\ds \infty\)
\(\ds \lim_{x \mathop \to 1} \map \ln {\dfrac {1 - x} x}\) \(=\) \(\ds -\infty\)

Then:

\(\ds \int_0^1 \map {\ln^{2n} } {\dfrac {1 - x} x} \rd x\) \(=\) \(\ds -\int_{\infty}^{-\infty} \dfrac { u^{2n} e^{-u} \rd u } {\paren { 1 + e^{-u} }^2 }\)
\(\ds \) \(=\) \(\ds \int_{-\infty}^{\infty} \dfrac { u^{2n} e^{-u} \rd u } {\paren { 1 + e^{-u} }^2 }\)
\(\ds \) \(=\) \(\ds \int_{-\infty}^0 \dfrac { u^{2n} e^{-u} \rd u } {\paren { 1 + e^{-u} }^2 } + \int_0^{\infty} \dfrac { u^{2n} e^{-u} \rd u } {\paren { 1 + e^{-u} }^2 }\) Sum of Integrals on Adjacent Intervals for Integrable Functions
\(\ds \) \(=\) \(\ds \int_0^{\infty} \dfrac { u^{2n} e^{u} \rd u } {\paren { 1 + e^{u} }^2 } + \int_0^{\infty} \dfrac { u^{2n} e^{-u} \rd u } {\paren { 1 + e^{-u} }^2 }\) Rewriting the first integral
\(\ds \) \(=\) \(\ds \int_0^{\infty} \dfrac { u^{2n} e^{u} \rd u } {\paren { 1 + e^{u} }^2 } \paren {\dfrac{e^{-2u} }{e^{-2u} } } + \int_0^{\infty} \dfrac { u^{2n} e^{-u} \rd u } {\paren { 1 + e^{-u} }^2 }\) Multiplying numerator and denominator by $e^{-2u}$
\(\ds \) \(=\) \(\ds 2\int_0^{\infty} \dfrac { u^{2n} e^{-u} \rd u } {\paren { 1 + e^{-u} }^2 }\)


From Sum of Infinite Geometric Sequence, we have:

$\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {e^{-u} }^k = \dfrac 1 {1 + e^{-u} }$

Taking the derivative of both sides, we have:

\(\ds \sum_{k \mathop = 0}^\infty \paren {-k} \paren {-1}^k e^{-ku} \rd u\) \(=\) \(\ds \dfrac { e^{-u} \rd u} {\paren {1 + e^{-u} }^2}\)
\(\ds \sum_{k \mathop = 1}^\infty k \paren {-1}^{k + 1} e^{-ku} \rd u\) \(=\) \(\ds \dfrac { e^{-u} \rd u} {\paren {1 + e^{-u} }^2}\) simplifying


Therefore:

\(\ds \int_0^1 \map {\ln^{2n} } {\dfrac {1 - x} x} \rd x\) \(=\) \(\ds 2\int_0^{\infty} u^{2n} \paren { \sum_{k \mathop = 1}^\infty k \paren {-1}^{k + 1} e^{-ku} \rd u}\)
\(\ds \) \(=\) \(\ds 2\sum_{k \mathop = 1}^\infty k \paren {-1}^{k + 1} \int_0^{\infty} u^{2n} e^{-ku} \rd u\) Fubini's Theorem
\(\ds \) \(=\) \(\ds 2\sum_{k \mathop = 1}^\infty k \paren {-1}^{k + 1} \bigintlimits {\frac {e^{-k u} } {-k} \paren {u^{2n} - \dfrac {2n u^{2n - 1} } {\paren {-k} } + \dfrac {2n \paren {2n - 1} u^{2n - 2} } { \paren {-k}^2 } - \dfrac {2n \paren {2n - 1} \paren {2n - 2} u^{2n - 3} } { \paren {-k}^3 } + \cdots + \dfrac {\paren {-1}^{2n} \paren {2n}!} { \paren {-k}^{2n} } } } 0 \infty\) Primitive of Power of x by Exponential of a x
\(\ds \) \(=\) \(\ds 2\sum_{k \mathop = 1}^\infty k \paren {-1}^{k + 1} \paren{0 + \frac 1 k \paren{0 - 0 + 0 - 0 + \cdots + \dfrac {\paren {-1}^{2n} \paren {2n}!} { \paren {-k}^{2n} } } }\)
\(\ds \) \(=\) \(\ds 2 \paren {2n}! \sum_{k \mathop = 1}^\infty {-1}^{k + 1} \paren{\dfrac 1 { k^{2n} } }\)
\(\ds \) \(=\) \(\ds 2 \paren {2n}! \paren {\paren {-1}^{n + 1} \dfrac {B_{2 n} \paren {2^{2 n - 1} - 1} \pi^{2 n} } {\paren {2 n}!} }\) Sum of Reciprocals of Even Powers of Integers Alternating in Sign
\(\ds \) \(=\) \(\ds \paren {-1}^{n + 1} B_{2 n} \paren {2^{2 n} - 2} \pi^{2 n}\)


$\blacksquare$

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