# Thread: does 1/x in Lp

1. ## does 1/x in Lp

I need to check whether or not $\frac{1}{x} \in {L}^p$

$\int_{0}^{1} \frac{1}{|x|^p} ~dx = \frac{1}{-p + 1} x^{-p + 1}|_{0}^1$

because this integral divergent, so $\frac{1}{x} \notin {L}^p$

is it correct??

thanks for any comment and suggestion

2. $\vert x \vert ^p$ is integrable in $(0,1)$ iff $1+p>0$ iff $p>-1$. So yeah, for any $1\leq p$ we have that $\frac{1}{\vert x \vert } \notin L^p$

3. Originally Posted by Jose27
$\vert x \vert ^p$ is integrable in $(0,1)$ iff $1+p>0$ iff $p>-1$. So yeah, for any $1\leq p$ we have that $\frac{1}{\vert x \vert } \notin L^p$
sure??

but i need to check $\frac{1}{|x|^p}$, not $\vert x \vert ^p$

$\int_{0}^{1} \frac{1}{|x|^p} ~dx = \frac{1}{-p + 1} x^{-p + 1}|_{0}^1$ convergent for $0 < p < 1$

so the right answer is $\frac{1}{x} \in {L}^p$ for $0 < p < 1$, and $\frac{1}{x} \notin {L}^p$ for $1 \leq p \leq \infty$

is it correct?need verification

thanks for any comment and suggestion

4. Originally Posted by dedust
sure??

but i need to check $\frac{1}{|x|^p}$, not $\vert x \vert ^p$

$\int_{0}^{1} \frac{1}{|x|^p} ~dx = \frac{1}{-p + 1} x^{-p + 1}|_{0}^1$ convergent for $0 < p < 1$

so the right answer is $\frac{1}{x} \in {L}^p$ for $0 < p < 1$, and $\frac{1}{x} \notin {L}^p$ for $1 \leq p \leq \infty$

is it correct?need verification

thanks for any comment and suggestion
Notice that my statement is equivalent to yours (maybe it's confusing since I named all the exponents $p$). As for your last statement, how do you define $L^p$ for $0?

5. Originally Posted by Jose27
Notice that my statement is equivalent to yours (maybe it's confusing since I named all the exponents $p$). As for your last statement, how do you define $L^p$ for $0?
ohh,..i see,..

thanks