Given that $f(x)$ is Riemann integrable and periodic with period $1$ and $\int_{0}^{1}f\left(x\right)dx=0$, prove that $$\int_{1}^{\infty}\frac{f\left(x\right)}{x^{s}}dx$$ exists for $s>0$
Here's my attempt to prove it for $s>1$, Define a sequence {$a_n$} by $$ {a_n}=\int_{1}^{n}\left|\frac{f\left(x\right)}{x^{s}}\right|dx$$
Using the fact that $f(x)$ is bounded, we have $$\int_{1}^{n}\left|\frac{f\left(x\right)}{x^{s}}\right|dx\ \le\ M\ \int_{1}^{n}\frac{1}{x^{s}}dx=M\left(\frac{1}{s-1}-\frac{1}{\left(s-1\right)n^{s-1}}\right)<\frac{M}{s-1}$$ for all $n$. So {$a_n$} is an increasing sequence bounded above and therefore converges which implies that {$b_n$}= $\int_{1}^{n}\frac{f\left(x\right)}{x^{s}}dx$ also converges. Is this correct? and even if it is, this argument can't be applied when $0<s\le1$.
Let $g(x)=\int_1^{x} f(t) dt$. Then is also periodic with period $1$. Now $\int_1^{n} \frac {f(x)} {x^{s}} dx=\frac {g(x)} {x^{s}}|_1^{n}+s\int_1^{n} \frac {g(x)} {x^{s+1}} dx$. Note that $g(1)=g(n)=0$ so the first term vanishes. Now note that $g$ is bounded and repeat your argument with $g$ in place of $f$. Since we have $s+1$ in place of $f$ now your argument works fine now.