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$$\int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .

Attempt

$[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right? Though the answer given is $\dfrac{17}{2}$.

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You must be careful about how you define your integral. I am guessing that you are defining $$ \int_a^b f(x) dg(x) = \lim \sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$ where $c_i \in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.

Then in your case, $d \lfloor x \rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at $$ \int_{1/2}^{3/2} (x^2 + 1) d\lfloor x \rfloor = \lim \sum (x_i^2 + 1)\Big( \lfloor x_{i+1} \rfloor - \lfloor x_i \rfloor\Big),$$ and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} \geq 1$. For that one term, $\lfloor x_{i+1} \rfloor - \lfloor x_i \rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus $$ \int_{1/2}^{3/2} (x^2 + 1) d\lfloor x \rfloor = \lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$

In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally $$ \int_a^b f(x) d \lfloor x \rfloor = \sum_{a < n \leq b} f(n)$$ for a continuous function $f$.


Having described this, I hope it is now not so hard to compute the entire integral.

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  • $\begingroup$ So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct? $\endgroup$ – jayant98 Nov 28 '18 at 18:17
  • $\begingroup$ @jayant98 Yes, it should be $17$. $\endgroup$ – davidlowryduda Nov 28 '18 at 18:23
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In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that $$ \int_a^b f(x) dy = \int_a^b f(x) \frac{dy}{dx} dx $$ and apply it to your integral, yielding $$ \int_0^3 (x^2+1) \frac{d[x] }{dx} dx. $$ The value of $\frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.

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