Finding $p>0$ for which $\log x\in L^p$

Question

Q) Find all $p>0$ s.t. $\log x \in L^p((0,1))$.

Since $\mu(X)=1$, I know that for a function $f\in L^p((0,1))$, $||f||_r\leq ||f||_s$ for $0<r<s\leq \infty$. Thus since $\log x \in L^1((0,1))$, $\log x \in L^p((0,1))$ for all $0<p\leq 1$. But I am not sure how to address $p>1$. Someone suggested using Holder's but not sure what should be the choice of the other function? Thanks.

Answer 1

$$\int_{0}^1|\log{x}|^pdx=\int_{-\infty}^0e^{u}|u|^pdu=\int_{0}^{\infty}\frac{|u|^p}{e^u}du$$ $$=\int_0^{\infty}e^{-u}u^pdu=\Gamma(p+1)<+\infty,\forall p>0$$