Definite Integral of Odd Function

Theorem

Let $f$ be an odd function with a primitive on the open interval $\closedint {-a} a$, where $a > 0$.

Then:

$\displaystyle \int_{-a}^a \map f x \rd x = 0$

Corollary

Let $f$ be an odd function with a primitive on the open interval $\openint {-a} a$, where $a > 0$.

Then the improper integral of $f$ on $\openint {-a} a$ is:

$\displaystyle \int_{\mathop \to -a}^{\mathop \to a} \map f x \rd x = 0$

Proof

Let $F$ be a primitive for $f$ on the interval $\closedint {-a} a$.

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

 $\displaystyle \int_{-a}^a \map f x \rd x$ $=$ $\displaystyle \int_{-a}^0 \map f x \rd x + \int_0^a \map f x \rd x$

Therefore, it suffices to prove that:

$\displaystyle \int_{-a}^0 \map f x \rd x = -\int_0^a \map f x \rd x$

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 $\map {\phi'} x = -1$.

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

 $\displaystyle \int_{\map \phi a}^{\map \phi 0} \map f x \rd x$ $=$ $\displaystyle \int_a^0 \map f {-u} \paren {-1} \rd u$ $\displaystyle$ $=$ $\displaystyle \int_0^a \map f {-u} \rd u$ Definition of Definite Integral $\displaystyle$ $=$ $\displaystyle \int_0^a \map f {-x} \rd x$ Renaming integration variable $\displaystyle$ $=$ $\displaystyle -\int_0^a \map f x \rd x$ $f$ is an odd function

This concludes the proof.

$\blacksquare$