First slide

Evaluation of definite integrals

Question

The function $F (x) = \int_{0}^{x} \log \frac{1 - t}{1 + t} d t$ is

Easy

A

an even function
B

an odd function
C

a periodic function
D

none of these

Solution

$\begin{array}{l} F (- x) = \int_{0}^{- x} \log \frac{1 - t}{1 + t} d t = - \int_{0}^{x} \log \frac{1 + u}{1 - u} d u (u = - t) \\ = \int_{0}^{x} \log \frac{1 - u}{1 + u} d u = F (x) \end{array}$

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