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$0 <\alpha <1$,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!

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Should be $\alpha < 1$, otherwise the integral doesn't converge. For $\frac 1 x$ neither $\int_0^1 \frac 1 x dx$ nor $\int_1^\infty \frac 1 x dx$ exist – Stefan Dec 5 '12 at 12:34
sorry, you're right! I have edited it. – user39843 Dec 5 '12 at 12:37
Otherwise your calculation are right – Stefan Dec 5 '12 at 12:40

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