For continuous 1-periodic f. $\int_1^{\infty}\frac{f(x)}{x}dx $ converges. Let $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous and 1 - periodic. If $ \int_0^1f(x)dx = 0$. Then $$\int_1^{\infty}\frac{f(x)}{x}dx $$ converges.
My question is that does a bound on the $\int_1^{\infty}\frac{f(x)}{x}dx$ will imply convergence of the integral? Please justify as well. 
And if not then how does one show convergence.
My Attempt
Let $ G(x) = \int_0^xf(x)dx$ then $G'(x)=f(x)$ we have to show that  $\int_1^{\infty}\frac{G'(x)}{x}dx$ converges. 
We note that $f(x)$ is bounded on $[0,1]$ say by $M$ and hence on $[0, \infty)$. Then so is $G(x) =\int_0^xf(x)dx= \int_{[x]}^xf(x)dx <M(x- [x])\leq M$.
Using integration by parts we have 
$$ \left.\frac{G(x)}{x}\right|_1^{\infty} + \int_1^{\infty}\frac{G(x)}{x^2}dx < M(2)$$
 A: Dirichlet's test is a consequence of summation/integration by parts.
In our case, it is useful in its integral formulation:

If $f(x)$ has a bounded primitive and $g(x)$ is decreasing to zero on $\mathbb{R}^+$, the integral $\int_{0}^{+\infty}\frac{f(x)}{x}\,dx$ exists.

If $f(x)$ is continuous, $1$-periodic and with mean zero, surely
$$\left|\int_{0}^{M}f(x)\,dx\right| = \left|\int_{\left\lfloor M\right\rfloor}^{M}f(x)\,dx\right|\leq \max_{x\in[0,1]}|f(x)| = C $$
for any $M\in\mathbb{R}^+$, so the claim trivially follows from Dirichlet's test.
A: You are on the right track. Integrate by parts to see that
$$
\int_1^z{f(x)\over x}\,dx = {G(z)\over z}-{G(1)\over 1}+\int_1^z {G(x)\over x^2}\,dx,\quad z>1.
$$
As you observe, $G$ is bounded. This ensures that the first term on the right of the display above converges to $0$, and that the integral on the right converges. 
A: Your problem is a special case that can be solved by Abel-Dirichlet Criterion. You only need to prove that the function $F(x) = \int_1^x f(u)du$ is bounded on $[1, \infty)$ to use it.
Proof of Abel-Dirichlet Criterion :
Let $f, g : [a, \infty) \mapsto \Bbb R$ be two functions such that $f$ is continuous, $F : [a, \infty) \mapsto \Bbb R$ defined by $F(x) = \int_a^x f(u)du$ is bounded (say by a positive constant $M$), and such that $g$ is monotonic, continuously differentiable, with $\lim \limits_{x \rightarrow \infty} g(x) = 0$. Then $\int_a^{\infty} f(u)g(u)du$ converges.
Indeed, consider $\int_a^x f(u)g(u)du$. 
We integrate it by parts : \begin{equation}\tag{1} \int_a^x f(u)g(u)du = \left.F(u)g(u)\right|_a^{x} - \int_a^x F(u)g'(u)du = F(x)g(x) - F(a)g(a) - \int_a^x F(u)g'(u)du = F(x)g(x) - \int_a^x F(u)g'(u)du \end{equation} because $F(a) = 0$.
Since $F$ is bounded, $\lim \limits_{x \rightarrow \infty} g(x) = 0$, we have $\lim \limits_{x \rightarrow \infty} F(x)g(x) = 0$.
Finally, notice that the improper integral $\int_a^{\infty} F(u)g'(u)du$ converges absolutely (so it converges) :
Indeed, $\lvert F(u)g'(u) \rvert \leq \lvert M g'(u) \rvert = M \lvert g'(u) \rvert $. Since $g$ is monotonic, either $g'$ is positive, or $g'$ is negative. Without loss of generality, it is positive. So $\lvert F(u)g'(u) \rvert \leq M g'(u)$ (just add a minus sign if $g'$ is negative) 
Now, $\int_a^{\infty} Mg'(u)du = \lim \limits_{x \rightarrow \infty} \int_a^{x} Mg'(u)du = \lim \limits_{x \rightarrow \infty} (Mg(x) - Mg(a)) = -Mg(a)$. So $\int_a^{\infty} Mg'(u)du$ converges. We conclude by comparison test that $\int_a^{\infty} F(u)g'(u)du$ converges absolutely.
Thus, by taking the limit $x \rightarrow \infty$ in $(1)$, we have that $\int_a^{\infty} f(u)g(u)du$ converges.
