TITLE: Function in $L^p$ but not in $L^{\infty}$
QUESTION [7 upvotes]: Show that if
$$f(x) = \ln\left({1\over x}\right),\quad 0<x\le 1$$
then $f\in L^p((0, 1])$ for all $1 \le p < \infty$ but $f\not\in L^{\infty}((0, 1])$.
Intuitively I know I have to show that $\int_{[0,1]}|f|^p <\infty $ but $f$ is not essentialy bounded, that means $\exists M >0$, s.t. $|f(x)|>M$ but I forget how to compute the integral and I am not sure how to find $M$.
Any help, Thanks.

REPLY [4 votes]: You can do this change of variables : u = $\log{1/x} \implies x = e^{-u}$
So, to calculate the p-norm, consider the following integral
$\int_0^{1}\log(\frac{1}{x})^pdx = \int_{0}^{\infty} u^p e^{-u} du$
The integral in the r.h.s is the well-known gamma function $\Gamma(p+1)$, which converges for the values of p that we are interested. 
For the $L^{\infty}$ part, consider $M \in [0,\infty)$ 
So, we have that $\log(\frac{1}{x})> M \iff x < e^{-M} $
Now, consider the interval $A = (0, e^{-M}) $. We have that, for all $x \in A, f(x) > M$ because the function is strictly decreasing. Also, f  is positive in $(0,1)$, so we can throw away the modulus.
So we found a measurable set where the function is greater than M, for all M, and have positive measure.This implies that $ f \notin L^{\infty} $ and we are done.