Let $[x]$ be defined as the greatest integer part of $x \in \mathbb{R}$. Let $0<t<1$ and $\alpha(x) = [1/x]$. Compute the integral:
$I(t) = \displaystyle\int_{t}^{1}x^{a}\mathrm{d\alpha(x)}$
I was trying to first define $\alpha(x)$ in terms of the linear combinations of the Heavyside Function so that by linearity of $\mathrm{d\alpha(x)}$ the integral can be easily computed (as we have that $\int_{t}^{1}x^{a}\mathrm{d(\alpha_{1}(x) + \alpha_{2}(x))} = \int_{t}^{1}x^{a}\mathrm{d\alpha_{1}(x)} + \int_{t}^{1}x^{a}\mathrm{d\alpha_{2}(x)}$. However, I'm not having much luck in this regard. What would be a way to define $\alpha(x)$ properly?
Thanks for the help.
If you mean for the differential to contain a floor term, then $d\alpha(x)=0$ when $x\notin\mathbb{Z}$, $d\alpha(x)= \delta(x-t)dt$ when $x \in \mathbb{N}$ where $\delta$ is the Dirac delta function. Therefore $$\int_{a}^{b} f(x)d\alpha(x)= \sum_{i=\lceil a \rceil}^{\lfloor b \rfloor} f(i) $$
If you meant for the floor function to be in the integrand I would use the identity $\forall$ $a \in \mathbb{N}$, $\int_0^{a} \lfloor x \rfloor dx= {a \choose 2}$. Hope that helps.