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This Question was asked by my teacher and the solution he presented is this:

$$\int_0^5 (x^2+1) d \lfloor x \rfloor.$$

Before integrating the above, lets see this:

$\int_a^b f'(x)g(x) + g'(x)f(x)dx = \int_a^b \frac {d(f(x)g(x))}{dx} dx = f(b)g(b)-f(a)g(a)$

Let $f(x)=x^2 +1$ and $g(x)=\lfloor x \rfloor.\ f'(x)= 2x.$ $$\int_a^b f'(x)g(x) + g'(x)f(x)dx=f(b)g(b)-f(a)g(a) \\\int_0^5 2x\lfloor x \rfloor + \frac{d\lfloor x \rfloor}{dx} (x^2 +1)dx=f(5)g(5)-f(0)g(0) \\\int_0^5 2x\lfloor x \rfloor dx + \int_0^5 (x^2 +1)d\lfloor x \rfloor=130$$

Now, $\int_0^5 2x\lfloor x \rfloor dx = 70\ .$ So $\int_0^5 (x^2 +1)d\lfloor x \rfloor$ comes out to be 60.

I dont understand how integration on $\lfloor x \rfloor$ is even possible and hence the solution seems pretty strange to me. Would love to know how this integration is possible in the first place?

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    Note \begin{align}\int_0^5(x^2+1)\lfloor x\rfloor,dx&=\sum_{i=0}^4\int_i^{i+1}(x^2+1)\cdot i,dx\end{align} However, the answer should be $430/3$ and not $60$. – TheSimpliFire Jul 27 '19 at 08:22
  • @giobrach $\lfloor \cdot \rfloor$ is not discontinuous at every integer. But it is not differentiable at every integer. – Hussain-Alqatari Jul 27 '19 at 08:23
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    @Simplifire, the answer is 60 and you must find $\int_0^5 2x\lfloor x \rfloor dx = \sum_{1}^{4} \int_{i}^{i+1}2x.idx = 70$ – Satish Ramanathan Jul 27 '19 at 10:49

3 Answers3

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As a Riemann–Stieltjes integral, we have that $$d \lfloor x \rfloor=\sum_{k\in\mathbb{Z}}\delta(x-k)dx$$ where $\delta(x)$ is the Dirac delta function. Hence $$\int_a^b f(x)d \lfloor x \rfloor=\sum_{k\in\mathbb{Z}\cap (a,b]}f(k).$$

Robert Z
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It's "possible" because what you're computing is a Riemann–Stieltjes integral, i.e. $\int_a^b fdg$ is identified with $f(b)g(b)-f(a)g(a)-\int_a^b gdf$ if the latter is definable by narrower notions of integrability. There's nothing illegitimate or unfamiliar about this; but, as with summing infinite series, integration admits multiple definitions that vary in how many answers they define. (This definition achieves the same thing.) You'll see a very similar problem here, which you can solve the same way. Here are others.

J.G.
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Also the integral can be understood as a Lebesgue-Stieltjes integral, where $dg$ is a notation to represent the Lebesgue-Stieltjes measure $\mu_g$ generated by $g$.

For this case we have that $\mu_{\lfloor \cdot\rfloor}((a,b])=\lfloor b\rfloor-\lfloor a\rfloor$, what imply that $\mu_{\lfloor \cdot\rfloor}(\{a\})=\chi_{\Bbb Z}(a)$, and so

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