91Ó°ÊÓ

Since \(f\) is a periodic function, there exists some \(p > 0\) such that \(f(x + p) = f(x)\) for all \(x \in \mathbb{R}\). Let's find the integral of \(f(t)\) over one period: $$ I = \int_{0}^{p} f(t) \mathrm{d} t $$ Now, differentiate \(F(x)\) with respect to \(x\): $$ F'(x) = \frac{d}{dx}\int_{a}^{x} f(t) \mathrm{d} t = f(x) $$ Let's consider the function \(G(x) = F(x + p) - F(x)\). Its derivative is: $$ G'(x) = F'(x + p) - F'(x) = f(x + p) - f(x) $$ Since \(f\) is periodic with period \(p\), \(f(x + p) = f(x)\). Therefore, \(G'(x) = 0\). This implies that \(G(x)\) is a constant function, say \(G(x) = C\). Then, we have: $$ F(x + p) - F(x) = C \quad \Rightarrow \quad F(x + p) = F(x) + C $$

Integrate both sides of the equation \(F(x + p) = F(x) + C\) with respect to \(x\): $$ \int_{a}^{x+p} f(t) \mathrm{d} t = \int_{a}^{x} f(t) \mathrm{d} t + C(x-a) \quad \Rightarrow \quad \int_{x}^{x+p} f(t) \mathrm{d} t = C(x-a) $$ By the Fundamental Theorem of Calculus, we have \(\int_{x}^{x+p} f(t) \mathrm{d} t = F(x+p) - F(x)\): $$ F(x+p) - F(x) = C(x-a) $$ Replace \(I = \int_{0}^{p} f(t) \mathrm{d} t\): $$ F(x+p) - F(x) = \frac{I}{p}(x-a) $$ Now, define a new function \(H(x) = F(x) - \frac{I}{p}(x-a)\). Then \(H(x)\) has period \(p\), since: $$ H(x+p) = F(x+p) - \frac{I}{p}(x+p-a) = F(x) + \frac{I}{p}(x-a) - \frac{I}{p}(x+p-a) = H(x) $$ Thus, we can represent \(F(x)\) as the sum of a linear function and a periodic function: $$ F(x) = \frac{I}{p}(x-a) + H(x) $$ We have shown that if \(f: \mathbb{R} \rightarrow \mathbb{R}\) is a periodic function that is integrable on every closed interval \([a, b] \subset \mathbb{R}\), then the function \(F(x) = \int_{a}^{x} f(t) \mathrm{d} t\) can be represented as the sum of a linear function (\(\frac{I}{p}(x-a)\)) and a periodic function (\(H(x)\)).