$0 <\alpha <1$,prove $x^{-\alpha}\in L([0,1])$,and calculate $\int_{[0,1]}x^{-\alpha}dx$ Here is my method. We only need to show that $\int_{[0,1]}x^{-\alpha}dx \leq \infty$. $\int_{[0,1]}x^{-\alpha}dx=\frac{x^{1-\alpha}}{1-\alpha}\bigg|_0^1=\frac{1}{1-\alpha}-\frac{1}{1-\alpha}\lim_{x\rightarrow 0}x^{1-\alpha}\\ =\frac{1}{1-\alpha}-\frac{1}{1-\alpha}\lim_{n\rightarrow \infty}\frac{1}{n^{1-\alpha}}=\frac{1}{1-\alpha}-\frac{1}{1-\alpha}*0=\frac{1}{1-\alpha}< \infty$
So,$x^{-\alpha}\in L([0,1])$
Is my method wrong? Thx!