Find $f:(0,1] \to \mathbb{R}$ s.t. $f^s \in L^1$ for $s < 1$ and $f^s\notin L^1$ for $s \geq 1$ where $L^1$ stands for the lebesgue-integrable functions.
Attempt:
$f(x) = \frac{1}{x}$
Then $f^s$ is for $s \leq 0$ just a polynomial with positive exponent, so it is continuous and hence $f^s \in L^1$.
How do I prove it for the cases $0 < s < 1$ and $s > 1 $. I know that for $0 < s < 1$ the limit of the series $\sum f^s$ exists and for $s > 1$ it doesn't. But can I conclude from that, that it is not lebesgue-integrable?
EDIT
Lets take $f(x) = \frac{1}{x}$ and try to compute (as suggested by @TitoEliatron) $\int_{0}^1 \frac{1}{x^s} dx = \lim_{a \to 0^{+}}\int_a^1 x^{-s} dx$ = $\lim_{a \to 0^{+}}[\frac{x^{1-s}}{1-s}]|_a^1$ =$\lim_{a \to 0^{+}}$ $\frac{1^{1-s}}{1-s} - \frac{a^{1-s}}{1-s}$.
For $s=1, \int_0^1 \frac{1}{x}=ln(x)|_0^1$ and we know that $\lim_{a\to 0} ln(a) = - \infty$ and $s>1$ is not possible as well, since $a \to 0^{+}$. Hence, for $s < 1$ it is in $L^1$