Improper Integral of a periodic function converges Given that $f(x)$ is a $p$-periodic function and $\int_0^p{f(x)}dx=0$. Show $\int_1^\infty\frac{f(x)}{x}dx$ converges.

*

*I know this integral can be broken into $\int_1^\infty\frac{f(x)}{x}dx=\int_0^\infty\frac{f(x)}{x}dx-\int_0^1\frac{f(x)}{x}dx$ for easier dealing if needed


*My first thought it so say that with $f(x)$ periodic, we can break the integral into parts such that: $\int_0^\infty\frac{f(x)}{x}dx=\int_0^p\frac{f(x)}{x}dx+\int_p^{2p}\frac{f(x)}{x}+...+\int_{(n-1)p}^{np}\frac{f(x)}{x}dx+...$ from here we can somehow use the given fact that $\int_0^pf(x)dx=0$ and that $\int_a^{a+np}f(x)dx=n\int_0^pf(x)dx$
We also can use that $\lim_{x->\infty}\frac{1}{x}=0$
But can somehow help me put these thoughts together in such a way that my mathematical proof is rigorous?
 A: 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.
A: 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.
A: Let us define 
$$F(x):=\int_1^x{\frac{f(y)}{y}dy}$$ 
In order to prove convergence of $F$ as $x\rightarrow \infty$  we will show  that $F$ is bounded with derivative that tends to zero.
Since $f$ periodic with $\int_0^pf(x)dx=0$ there is some constant $c>0$ such that $|\int_0^xf(y)dy|\leq c$ for all $x\in[0,\infty)$ (uniformly bounded). Then $F$ can be written as
$$F(x)=\int_1^x{\frac{1}{y}\frac{d}{dy}\bigg[\int_0^y{f(u)du}\bigg]dy}=\frac{1}{x}\int_0^x{f(u)du}-\int_0^1{f(u)du}+\int_1^x{\frac{1}{y^2}\bigg[\int_0^y{f(u)du}\bigg]dy}$$
Note that for $x>1$
$$\Bigg|\int_1^x{\frac{1}{y^2}\bigg[\int_0^y{f(u)du}\bigg]}dy\Bigg|\leq c\int_1^x{\frac{1}{y^2}dy}\leq c$$
and
$$\Bigg|\frac{1}{x}\int_0^x{f(u)du}\Bigg|\leq c $$
Combining the previous inequalities with the above identity boundedness of $F$ can be deduced.
Also $\lim_{x\rightarrow\infty}\frac{dF(x)}{dx}=\lim_{x\rightarrow \infty}\frac{f(x)}{x}=0$. 
Since $F$ is bounded with $\lim_{x\rightarrow\infty} F'(x)=0$ it converges.
A: Te method describe by other solutions can be used to solve  a slightly more general solution to the OP.
This is problem in T. Apostol's analysis book (2nd ed)

Problem: Suppose $f$ is $p$ periodic and integrable (either as in the sense or Riemann  or Lebesgue)  over $[0,p]$
Then $\lim_{A\rightarrow \infty}\int^A_p x^{-s}f(x)\,dx$ exists for all $s>0$.

Let $F(x)=\int^x_p f(t)\,dt$, for $x\geq p$. It is easy to check that $F$ can be extended as a continuous periodic function (period $p$) over the real line.
Integration by parts gives
\begin{aligned}
\int^A_p x^{-s}f(x)\,dx&=\int^A_p x^{-s} dF(x)= x^{-s}f(x)|^A_p +s \int^A_p \frac{F(x)}{x^{s+1}}\,dx\\
&=A^{-a}F(A)+s\int^A_p\frac{F(x)}{x^{s+1}}\,dx
\end{aligned}
Since $F$ is bounded and $\int^\infty_0\frac{1}{x^{s+1}}\,dx <\infty$ for all $s>0$, we have that $\int^\infty_A\frac{F(x)}{x^{s+1}}\,dx$ converges (both as an improper Riemann integral and also in the sense of Lebesgue) and so
$$\lim_{A\rightarrow\infty}\int^A_p x^{-s}\,f(x)\,dx=s\int^\infty_p\frac{F(x)}{x^{s+1}}\,dx
$$
