Expectation of Logistic Distribution/Lemma 3

Lemma for Expectation of Logistic Distribution

$\forall k \in \N$:

$\ds \int_{\to 0}^{\to 1} \map {\ln^{2k + 1} } {\dfrac {1 - u} u} \rd u = 0$

let:

$\ds x$

$=$

$\ds \paren {\dfrac 1 2 - u}$

$\ds \leadsto \ \ $

$\ds \frac {\d x} {\d u}$

$=$

$\ds -1$

Then:

$\ds \int_{\to 0}^{\to 1} \map {\ln^{2k + 1} } {\dfrac {1 - u} u} \rd u$

$=$

$\ds -\int_{\to \frac 1 2}^{\to -\frac 1 2} \map {\ln^{2k + 1} } {\dfrac {1 - \paren {\dfrac 1 2 - x} } {\paren {\dfrac 1 2 - x} } } \rd x$

$\ds $

$=$

$\ds \int_{\to -\frac 1 2}^{\to \frac 1 2} \map {\ln^{2k + 1} } {\dfrac {\paren {\dfrac 1 2 + x} } {\paren {\dfrac 1 2 - x} } } \rd x$

$\ds $

$=$

$\ds 0$

$\blacksquare$