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I'm trying to solve the following

$$ \int_0^n f(x) d \lfloor x \rfloor $$

where $\lfloor x \rfloor$ is the floor function, $f : [0,n] \rightarrow \mathbb{R}$ is continuous, and $n \in \mathbb{N}$.

I know $\frac{d}{dx} \lfloor x \rfloor$ itself cannot be secured (i.e. not differentiable). I cannot proceed further.

Can anyone give some hints?

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    $\begingroup$ The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit. $\endgroup$ – Kavi Rama Murthy Aug 31 '18 at 8:06
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    $\begingroup$ @KaviRamaMurthy Shouldn't it be $$\int_0^n\,f(x)\,\text{d}\lfloor x\rfloor = f(1)+f(2)+\ldots+f(n)\,?$$ I think we have $$\int_0^n\,f(x)\,\text{d}\lceil x\rceil=f(0)+f(1)+\ldots+f(n-1)\,.$$ $\endgroup$ – Batominovski Aug 31 '18 at 8:08
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Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:

$$\int_{[0,n]}f(x)d\lfloor x\rfloor =\sum_{i=0}^nf(i)$$

$$\int_{[0,n)}f(x)d\lfloor x\rfloor =\sum_{i=0}^{n-1}f(i)$$

$$\int_{(0,n]}f(x)d\lfloor x\rfloor =\sum_{i=1}^nf(i)$$

$$\int_{(0,n)}f(x)d\lfloor x\rfloor =\sum_{i=1}^{n-1}f(i)$$

since the associated measure for the floor-function is

$$\mu_{\lfloor\cdot\rfloor}=\sum_{k=-\infty}^\infty \delta_k$$

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