Improper Integral of a periodic function converges

Define $$ F(x) = \int_{0}^x f(t) \, dt. $$ The $\int_0^p f = 0$ condition gives us that $F$ is also periodic, and we also have $$ \lvert F(x)\rvert \leqslant \int_0^x \lvert f(x)\rvert \, dx \leqslant \int_0^p \lvert f(x)\rvert \, dx = A, $$ say, for all $x$.

Now consider the limit used in the definition of the improper integral, and integrate by parts: $$ \int_1^R \frac{f(x)}{x} \, dx = \left[ \frac{F(x)}{x} \right]_1^R + \int_1^R \frac{F(x)}{x^2} \, dx = \frac{F(R)}{R}-F(1) + \int_1^R \frac{F(x)}{x^2} \, dx $$ The first term tends to zero since $\lvert F(R)\rvert $ is bounded by $A$, and for the last term, $$ \left\lvert\int_R^{\infty} \frac{F(x)}{x^2} \, dx\right\rvert \leqslant \int_R^{\infty} \frac{\lvert F(x)\rvert}{x^2} \, dx \leqslant A\int_R^{\infty} \frac{dx}{x^2} = \frac{A}{R} \to 0, $$ so this improper integral exists as a well-defined limit. Hence the original integral exists as an improper integral.

I will assume that $f$ is continuous. Let $F(x)=\int_1^xf(t)\,dt$. Then $F(1)=0$, $F'(x)=f(x)$ and, since $\int_{1+np}^{1+(n+1)p}f(t)\,dt=0$ for all $n\in\mathbb{N}$, $F$ is bounded. Then for any $R>1$, integrating by parts we have $$ \int_1^R\frac{f(x)}{x}\,dx=\frac{F(x)}{x}\Bigr|_1^R+\int_1^R\frac{F(x)}{x^2}\,dx=\frac{F(R)}{R}+\int_1^R\frac{F(x)}{x^2}\,dx. $$ Since $F$ is bounded $F(R)/R\to0$ as $R\to\infty$, and $$ \int_1^\infty\frac{F(x)}{x^2}\,dx. $$ is absolutely convergent.