Prove some properties of even, periodic, integrable functions

Let $f$ be an integrable, even, periodic function with period 2. Define

$g(x) = \int_0^x f(t) \, dt, \qquad \text{and} \qquad A = g(1).$

Prove $g$ is odd and $g(x+2) - g(x) = g(2)$ .
Compute $g(2)$ and $g(5)$ using that $A = g(1)$ .
Find the value of $A$ that makes $g$ periodic with period 2.

Proof. Using the expansion/contraction property (Apostol, Thm 1.19 with $k = -1$ ) we compute,
$\begin{align*} g(-x) &= \int_0^{-x} f(t) \, dt \\ & = - \int_0^x f(-t) \, dt \\ & = - \int_0^x f(t) \, dt & (f \text{ is even}) \\ & = - g(x). \end{align*}$

Thus, $g$ is odd. Next,

$\begin{align*} g(x+2) - g(x) &= \int_0^{x+2} f(t) \, dt - \int_0^x f(t) \, dt \\ &= \int_0^2 f(t) \,dt + \int_2^{x+2} f(t) \, dt - \int_0^x f(t) \, dt \\ &= g(2) + \int_0^x f(t+2) \, dt - \int_0^x f(t) \, dt \\ & = g(2) + \int_0^x f(t) \, dt - \int_0^x f(t) \, dt & (f \text{ periodic}) \\ & = g(2). \qquad \blacksquare \end{align*}$
First, we compute $g(2)$ ,
$\begin{align*} g(2) &= g(3) - g(1) &(\text{part (a) with } x = 1) \\ &= \int_0^3 f(t) \, dt - \int_0^1 f(t) \, dt \\ &= \int_1^3 f(t) \, dt \\ & = \int_{-1}^1 f(t+2) \, dt &(\text{Translation prop, Thm 1.18})\\ & = 2 \int_0^1 f(t) \, dt \\ & = 2A, \end{align*}$

where the second to last line followed from the fact that $f$ is even and this exercise, so $\int_{-b}^b f(t)\, dt = 2 \int_0^b f(t) \, dt$ .

Since $g(1) = A$ by definition, the above also gives us $g(3) = g(2) + g(1) = 3A$ . So, again using part (a) (this time with $x = 3$ ) we have

$\begin{align*} g(5) &= g(2) + g(3) \\ &= 2A + 3A \\ &= 5A. \end{align*}$
Here we want to find a value of $A$ such that $g$ is periodic with period 2. For $g$ to be periodic with period 2, we must have
$g(x+2) = g(x) \quad \implies \quad g(x+2) - g(x) = 0.$

But, since $g(0) = \int_0^0 f(t) \, dt = 0$ , and the above must be true for all $x$ , we evaluate at $x=0$ and use part (b) that gave us $g(2) = 2A$ to compute,

$g(2) = 0 \quad \implies \quad 2A = 0 \quad \implies \quad A = 0.$