# Expectation of Logistic Distribution/Lemma 3

## Lemma for Expectation of Logistic Distribution

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

## Proof

let:

 $\ds x$ $=$ $\ds \paren {\dfrac 1 2 - u}$ Integration by Substitution $\ds \leadsto \ \$ $\ds \frac {\d x} {\d u}$ $=$ $\ds -1$ Power Rule for Derivatives

Then:

 $\ds \int_{\to 0}^{\to 1} \map {\ln^{2k + 1} } {\dfrac {1 - u} u} \rd u$ $=$ $\ds -\int_{\to \frac 1 2}^{\to -\frac 1 2} \map {\ln^{2k + 1} } {\dfrac {1 - \paren {\dfrac 1 2 - x} } {\paren {\dfrac 1 2 - x} } } \rd x$ $\ds$ $=$ $\ds \int_{\to -\frac 1 2}^{\to \frac 1 2} \map {\ln^{2k + 1} } {\dfrac {\paren {\dfrac 1 2 + x} } {\paren {\dfrac 1 2 - x} } } \rd x$ $\ds$ $=$ $\ds 0$ Definite Integral of Odd Function

$\blacksquare$