# Definite Integral of Odd Function

## Theorem

Let $f$ be an odd function with a primitive on the closed interval $\left[{-a \,.\,.\, a}\right]$, where $a > 0$.

Then:

$\displaystyle \int_{-a}^a f \left({x}\right) \ \mathrm d x = 0$

## Proof

Let $F$ be a primitive for $f$ on the interval $\left[{-a \,.\,.\, a}\right]$.

Then, by Sum of Integrals on Adjacent Intervals for Integrable Functions, we have:

 $\displaystyle \int_{-a}^a f \left({x}\right) \ \mathrm d x$ $=$ $\displaystyle \int_{-a}^0 f \left({x}\right) \ \mathrm d x + \int_0^a f \left({x}\right) \ \mathrm d x$

Therefore, it suffices to prove that:

$\displaystyle \int_{-a}^0 f \left({x}\right) \ \mathrm d x = - \int_0^a f \left({x}\right) \ \mathrm d x$

To this end, let $\phi: \R \to \R$ be defined by $x \mapsto -x$.

From Derivative of Identity Function and Derivative of Constant Multiple, for all $x \in \R$, we have $\phi' \left({x}\right) = -1$.

Then, by means of Integration by Substitution, we compute:

 $\displaystyle \int_{\phi \left({a}\right) }^{\phi \left({0}\right)} f \left({x}\right) \ \mathrm d x$ $=$ $\displaystyle \int_a^0 f \left({-u}\right) \left({-1}\right) \ \mathrm d u$ $\displaystyle$ $=$ $\displaystyle \int_0^a f \left({-u}\right) \ \mathrm d u$ Definition of integral $\displaystyle$ $=$ $\displaystyle \int_0^a f \left({-x}\right) \ \mathrm d x$ Renaming integration variable $\displaystyle$ $=$ $\displaystyle -\int_0^a f \left({x}\right) \ \mathrm d x$ $f$ is an odd function

This concludes the proof.

$\blacksquare$