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What does $\int(x^2-4)d \lfloor x \rfloor$ mean? How do I integrate wrt $\lfloor x\rfloor$? This has to be a typo. enter image description here

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    $\begingroup$ It means a Riemann-Stieltjes integral. $\endgroup$ – Jakobian Jun 28 '19 at 17:25
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    $\begingroup$ Possible duplicate of Integration of a function with respect to another function. $\endgroup$ – InterstellarProbe Jun 28 '19 at 17:33
  • $\begingroup$ @Jakobian Could you link me to a definition of RS Integrals that uses that notation? The wiki article doesn't seem to. $\endgroup$ – Akash Gaur Jun 28 '19 at 17:36
  • $\begingroup$ Where are these questions from? $\endgroup$ – Ovi Jun 28 '19 at 17:37
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    $\begingroup$ @rhaldryn it's on the wiki. $f(x) = x^2-4$, $g(x) = \lfloor x \rfloor$, $\int_0^4 f(x) dg(x)$ $\endgroup$ – Jakobian Jun 28 '19 at 17:39
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Working with Riemann-Stieltjes integrals,$$\int_0^4(x^2-4)d\lfloor x\rfloor=\left[(x^2-4)\lfloor x\rfloor\right]_0^4-\int_0^4\lfloor x\rfloor 2xdx\\=48-\int_1^22xdx-2\int_2^32xdx-3\int_3^42xdx\\=48-(2^2-1^2)-2(3^2-2^2)-3(4^2-3^2)=48-3-10-21=14.$$ More generally,$$\int_{0}^{n}f\left(x\right)d\left\lfloor x\right\rfloor =nf\left(n\right)-\sum_{k=1}^{n-1}k\left(f\left(k+1\right)-f\left(k\right)\right)=\sum_{k=1}^{n}f\left(k\right)$$for $n\in\mathbb{N}$.

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You can use integration by parts as follows. $$\int_0^4 (x^2-4)d\lfloor x\rfloor = (x^2-4)\lfloor x\rfloor |_{x=0}^{x=4} - \int_0^4 \lfloor x\rfloor d(x^2-4) = \\ 48-\int_0^4 2x\lfloor x\rfloor dx = 48-(4-1)-2(9-4)-3(16-9) = 14 $$

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