Integration involving d[x] instead of dx

Question

Find the value of: $$\int_0^3(x^2+1)d\lfloor x\rfloor$$

Shouldn't the answer be zero since $[x]$ is always an integer? But the options given are $12,17,15,19$. ([x] denotes floor(x))

From $0$ to $1$, $[x]=0$ and $d[x]=0$. Similarly, from $1$ to $2$, $[x]=2$ and $d[x]=0$ and so on...

I got this question from my book. I did not make it. Is it wrong? — Aditya Dev, Nov 14 '15 at 07:12
Do you know what Lebesgue Stieltjes integration is? (And is $[x]$ the floor or ceiling?) — copper.hat, Nov 14 '15 at 07:18
No. But I would like to learn it. (And learn how to pronounce it). Let me Google it. — Aditya Dev, Nov 14 '15 at 07:19
I didn't understand a thing given in wikipedia. Can you suggest a site? — Aditya Dev, Nov 14 '15 at 07:23
Do you know what integration with respect to an arbitrary measure is? How did you encounter this problem? — copper.hat, Nov 14 '15 at 07:24
No. I can integrate wrt d(sinx). You can simply multiply and divide by dx to get $\frac{d(\sin x)}{dx}$. — Aditya Dev, Nov 14 '15 at 07:26
@copper.hat: what should I learn before learning lebesgue theorem? — Aditya Dev, Nov 14 '15 at 07:32
http://math.stackexchange.com/questions/1230680/integration-with-respect-to-the-floor-function — Sujith Sizon, Nov 14 '15 at 07:32
Probably just the Riemann Stiljes integral. https://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral — Alan, Nov 14 '15 at 07:33
@sujith, this question is slightly different. My question is more basic. — Aditya Dev, Nov 14 '15 at 07:34
@AdityaDev Reimann integration is something like applying chain rule and taking limit , there must be some secret manipulation of $x^2+1$ to make it in terms of $\floor{x}$ — Sujith Sizon, Nov 14 '15 at 07:36
@AdityaDev: It is a little more involved that can be conveniently explained in a comment, but it starts with defining a measure (or length) as $\mu_{\lfloor \cdot \rfloor}((a,b]) = \lfloor b \rfloor - \lfloor a \rfloor$. — copper.hat, Nov 14 '15 at 07:56
$[x]$ was for a long time the standard notation for "floor (x)". — DanielWainfleet, Nov 14 '15 at 08:58

score 1 · Accepted Answer · answered Nov 18 '15 at 00:23

Let $f(x) = x^2+1$, $\alpha(x) = \lfloor x \rfloor$, and let $P$ be a partition of $[0,3]$ such that the points $1,2,3$ are in different partitions.

Let $I_k$ be the intervals in the partition and without loss of generality, suppose $k \in I_k$ for $k = 1,2,3$. Note that we have $\alpha(\sup I_k) - \alpha(\inf I_k) = 1$ for $k=1,2,3$. Also note that on any other interval $J$ in the partition, we have $\alpha(\sup J) - \alpha(\inf J) = 0$.

Hence we have $L(P,f,\alpha) = \sum_{k=1}^3 \inf_{x \in I_k} f(x)(\alpha(\sup I_k) - \alpha(\inf I_k)) = \sum_{k=1}^3 \inf_{x \in I_k} f(x)$, and similarly we have $U(P,f,\alpha) = \sum_{k=1}^3 \sup_{x \in I_k} f(x)$.

Since $k \in I_k$ we see that $L(P,f,\alpha) \le \sum_{k=1}^3 f(k)$ and similarly $U(P,f,\alpha) \ge \sum_{k=1}^3 f(k)$.

Since $f$ is continuous, for any $\epsilon>0$ we can choose a partition so that $L(P,f,\alpha) \ge \sum_{k=1}^3 f(k)-\epsilon$, $U(P,f,\alpha) \le \sum_{k=1}^3 f(k)+\epsilon$, from which it follows that $\int_0^3 f(x) d \alpha(x) = \sum_{k=1}^3 f(k)$.

Evaluating for the specific $f$ gives $17$.

Integration involving d[x] instead of dx

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