I'm trying to calculate directly following the definition the Lebesgue integral $$ x^{-\alpha}, x \in [0,1], 0<\alpha<1$$
My attempts: To prove that the integral exists and calculate it at the same time, let
$$ f_n(x) = f(\frac{k}{2^n}), x\in(\frac{k-1}{2^n},\frac{k}{2^n}], k = 1,\ldots,2^n $$
and $f_n = 0$ at any other point. Then $f_n$ converges monotonically to $f$. And
$$ \int f_n dx = \sum_{k=1}^{2^n} \frac{1}{2^n}(\frac{k}{2^n})^{-\alpha}$$
Therefore the integral of $f$ over $[0,1]$ is the limit of the above formula as $n\rightarrow\infty$, i.e.
$$\int f dx = \lim_{n\rightarrow\infty}\sum_{k=1}^{2^n} \frac{1}{2^n}(\frac{k}{2^n})^{-\alpha}$$
So I need to compute the limit, but how?
Thanks in advance!
The sum in question is
$$\frac{1}{2^{n(1-\alpha)}}\sum_{k=1}^{2^n} k ^{-\alpha}.$$
Using the binomial expansion we have
$$\begin{align}(k-1)^{1- \alpha} &= k^{1 - \alpha}(1 - k^{-1})^{1-\alpha} \\ &= k^{1 - \alpha}(1 - (1- \alpha)k^{-1} + O(k^{-2})) \\ &= k^{1-\alpha}- (1-\alpha)k^{-\alpha} + O(k^{-\alpha-1})\end{align},$$
Hence,
$$k^{-\alpha} = (1-\alpha)^{-1}(k^{1-\alpha} - (k-1)^{1 - \alpha} + O(k^{-\alpha-1})),$$
and, taking advantage of a telescoping sum,
$$\frac{1}{2^{n(1-\alpha)}}\sum_{k=1}^{2^n} k ^{-\alpha} = \frac{(1- \alpha)^{-1}}{2^{n(1-\alpha)}}\sum_{k=1}^{2^n}(k^{1-\alpha} - (k-1)^{1 - \alpha} + O(k^{-\alpha-1})) \\ = \frac{(1- \alpha)^{-1}}{2^{n(1-\alpha)}} (2^{n(1 - \alpha)} + O(2^{-n\alpha})) \\ = (1- \alpha)^{-1} + O(2^{-n}).$$
Taking the limit as $n \to \infty$ we find the value of the integral is $(1 - \alpha)^{-1}$, as expected.