# Interesting Integration with respect to $[x]$ where $[\cdot]$ is greatest integer function

$$\int_{0}^{3} {(x^2+1)}d[x]$$ 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 $$\dfrac{17}{2}$$.

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.

• 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? – jayant98 Nov 28 '18 at 18:17
• @jayant98 Yes, it should be $17$. – davidlowryduda Nov 28 '18 at 18:23

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.