For which t is $x^t$ in $\mathscr{L}^1$

Question

Given is a function $f_t: ]0, \infty[ \rightarrow \mathbb{R}, \, \, x \mapsto x^t$.
The question is, for which $t \in \mathbb{R}$ is
i) $f_t \in \mathscr{L}^1(]0, 1])$
ii) $f_t \in \mathscr{L}^1([1, \infty[)$
iii) $f_t \in \mathscr{L}^1(]0, \infty[)$

My idea is to calculate the integral $\int_{0}^{1} |x^t| \, dx$ = $\int_{0}^{1} x^t \, dx$ = $\dfrac{1}{t + 1}$ which means that $f_t \in \mathscr{L}^1(]0, 1])$ for $t \in \mathbb{R}$ \ $\{-1\}$. As $\int |x^t| \, dx$ = $\int x^t \, dx$ = $ \dfrac{x^{t+1}}{t + 1}$, we get for ii) and iii) that this integral is $\infty$ and as a result, there is no t for ii) and iii).
Is my idea correct? Thanks.

Answer 1

$\int_0^{1}x^{t}dx=\frac {x^{t+1}} {t+1}|_0^{1}=\infty$ if $ t <-1$ and $\frac 1 {t+1}$ if $t >-1$. It is $\infty $ for $t=-1$. So the function is integrable iff $t>-1$. For ii) the condition is $t <-1$ and there is no $t$ for which $f \in \mathcal L^{1}(0,\infty)$.