Prove formulas for the integral from -b to b of even and odd functions

by RoRi

July 25, 2015

We define an even function to be a function $f$ such that whenever $x$ is in the domain of $f$ so is $-x$ , and for all $x$ in the domain, we have

$f(-x) = f(x).$

We define an odd function to be a function $f$ such that whenever $x$ is in the domain of $f$ so is $-x$ , and for all $x$ in the domain, we have

$f(-x) = -f(x).$

Then, for $f$ integrable on $[0,b]$ prove the following.

If $f$ is an even function then
$\int_{-b}^b f(x) \, dx = 2 \int_0^b f(x) \, dx.$
If $f$ is an odd function then
$\int_{-b}^b f(x) \, dx = 0.$

Proof. First, since the integral is additive with respect to the interval of integration (Theorem 1.17 of Apostol), we have
$\int_{-b}^b f(x) \, dx = \int_{-b}^0 f(x) \, dx + \int_0^b f(x) \, dx .$

Then, for the first integral we use the expansion/contraction of the interval of integration with $k=-1$ to get

$\int_{-b}^0 f(x) \, dx = - \int_b^0 f(-x) \, dx.$

Since $f(x)$ is an even function by assumption, we have $f(-x) = f(x)$ for all $x \in [0,b]$ . Since $-\int_b^0 = \int_0^b$ we then have,

$-\int_b^0 f(-x) \, dx = \int_0^b f(x) \, dx.$

So, putting this all together we have,

$\int_{-b}^b f(x) \, dx = \int_0^b f(x) \, dx + \int_0^b f(x) \, dx = 2 \int_0^b f(x) \, dx. \qquad \blacksquare$
Proof. Similar to part (a) we have,
$\int_{-b}^b f(x) \, dx = \int_{-b}^0 f(x) \, dx + \int_0^b f(x) \, dx = \int_0^b f(-x) \, dx + \int_0^b f(x) \, dx.$

But in this case, since $f$ is an odd function, we have $f(-x) = -f(x)$ for all $x \in [0,b]$ . Thus,

$\int_0^b f(-x) \, dx = - \int_0^b f(x) \, dx \quad \implies \quad \int_{-b}^b f(x) \, dx = 0. \qquad \blacksquare$

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One comment

September 7, 2020 at 9:47 am

Anonymous says:

I think before you come up with the additive property with respect to the interval of integration, you need to point out the integral of f from -b to 0 exists. That means you should write down the contraction / expansion with respect to the interval of integration with k = -1 first, and you can proceed as the rest of the solution.

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