Prove some formulas for odd periodic functions

by RoRi

August 24, 2015

Let $f$ be an odd, periodic, integrable everywhere function with period 2. Then, define

$g(x) = \int_0^x f(t) \, dt.$

Prove that $g(2n) = 0$ for all $n \in \mathbb{Z}$ .
Prove that $g$ is an even, periodic function with period 2.

Proof. We can establish this by a direct computation.
$\begin{align*} g(2n) &= \int_0^{2n} f(t) \, dt \\ &= \int_{-2n}^0 f(t+2n) \, dt & (\text{translation property, Thm 1.18})\\ &= \int_{-2n}^0 f(t) \, dt & (f \text{ periodic, period 2}) \\ &= - \int_0^{-2n} f(t) \, dt \\ &= \int_0^{2n} f(-t) \, dt & (\text{Expansion/contraction property}) \\ &= - \int_0^{2n} f(t) \, dt & (f \text{ is odd so } f(-t) = -f(t)) \end{align*}$

But then we have

$\int_0^{2n} f(t) \, dt = - \int_0^{2n} f(t) \, dt \quad \implies \quad \int_0^{2n} f(t) \, dt = 0$

Since $g(2n) = \int_0^{2n} f(t) \, dt$ we then have the requested result,

$g(2n) = \int_0^{2n} f(t) \, dt = 0. \qquad blacksquare$
First, to show that $g$ is even we need to show $g(-x) = g(x)$ for all $x$ . We compute, using the fact that $f$ is odd and the expansion/contraction property of the integral (Theorem 1.19 in Apostol) with $k=-1$ .
$\begin{align*} g(-x) &= \int_0^{-x} f(t) \, dt \\ &= - \int_0^x f(-t) \, dt & (\text{Exp/Cont property}) \\ &= \int_0^x f(t) \, dt & (f \text{ is odd}) \\ &= g(x) & (\text{Def of } g). \end{align*}$

Next, to show that $g$ is periodic with period 2 we must show $g(x+2) = g(x)$ for all $x$ .

$\begin{align*} g(x+2) &= \int_0^{x+2} f(t) \, dt \\ &= \int_0^2 f(t) \, dt + \int_2^{x+2} f(t) \, dt \\ &= 0 + \int_0^x f(t+2) \, dt &(\text{part (a) and translation property}) \\ &= \int_0^x f(t) \, dt & (f \text{ is periodic, period 2}) \\ & = g(x). \qquad \blacksquare \end{align*}$

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