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I have an expression I want to evaluate

$$\int_a^bf'(x)\lfloor x\rfloor dx $$

where $f'(x)$ is continuous over the interval. I am looking for a solution without using integration by parts.

My attempt to find the value of this integral is to divide it into parts:

$$\int_a^{\lfloor a+1\rfloor}f'(x)\lfloor x\rfloor dx+\int_{\lfloor a+1\rfloor}^{\lfloor a+2\rfloor}f'(x)\lfloor x\rfloor dx +... +\int_{\lfloor b\rfloor}^{b}f'(x)\lfloor x\rfloor dx.$$

With the reasoning being that $\lfloor x\rfloor$ should be constant under these intervals. But I can't really convince myself of why this should be the case. The intervals $[a,\lfloor a+1\rfloor),[\lfloor a+1\rfloor,\lfloor a+2\rfloor),...$ do the trick but when we split up the intergrals as above my concern is that the intervals we get are actually the closed ones $[a,\lfloor a+1\rfloor],[\lfloor a+1\rfloor,\lfloor a+2\rfloor],...$ over which the floor function is not constant. So what gives?

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    This looks very general. What do you expect to obtain in the end? Another formula? – 2017-01-10
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    I think don't it is possible to use integration by part here. a good int should be to consider all integers in [a,b). which can be easily done by considering the case where $a$ is integer or not – 2017-01-10

3 Answers 3

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The point is: Is does not make any difference if you consider the half-open intervals or the closed ones, because the integral is stable against changing the value of the function at one point. That means even if you integrate over a closed intervall, you have $$ \def\f#1{\left\lfloor#1\right\rfloor}\int_{\f{a+i}}^{\f{a+i+1}}\f{x}f'(x)\,dx = \f{a+i} \Bigl(f\bigl(\f{a+i+1}\bigr) - f\bigl(\f{a+i}\bigr)\Bigr) $$ for each $i$.

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    What do you mean by "is stable against changing the value of the function at one point"? – 2017-01-10
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    @StrangeBrew: whether the bounds are included or not doesn't matter, they have a null measure. – 2017-01-10
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    @YvesDaoust What does null measure mean? Is there perhaps a way to realise that the open/closeness of the intervals is immaterial with more rudimentary means? – 2017-01-10
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    @StrangeBrew: what is $\int_a^a f(x) dx$ ? – 2017-01-10
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    @YvesDaoust Hmm okay, so I could as well evaluate the improper integrals: $$\lim_{\delta\to 0}\int_a^{\lfloor a+1\rfloor -\delta}f(x)+\int_{\lfloor a+1\rfloor -\delta}^{\lfloor a+1\rfloor }f(x)?$$ – 2017-01-10
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    It follows from the very definition of the Riemann-integral. If $f,g \colon [a,b] \to \mathbf R$ are any integrable function and $c\in [a,b]$ such that $f(x) = g(x)$ for $x \in [a,b]-\{c\}$, say $f(c) < g(c)$. Then every lower sum for $g$ is a lower sum for $f$, and any upper sum for $g$ can be approximated arbitrarlty close with a upper sum of $f$, just making the intervall of the partition containing $c$ small. – 2017-01-10
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    @StrangeBrew: no, this doesn't help because the second integral still has the discontinuity. (And these integrals are not improper.) – 2017-01-10
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    @YvesDaoust Okay, so then I don't quite understand your question. (Why are they not improper? They are of the same form as https://en.wikipedia.org/wiki/Improper_integral) – 2017-01-10
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    @StrangeBrew: last attempt, $[a,b]=[a,b)\cup[b,b]$. – 2017-01-10
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    @martini Thank you, I think I got it :)! – 2017-01-10
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I summarize the answers of others and provide more details that should clarify and drastically simplify the final result:

1) The integral is independent of whether the intervals from $a$ to $b$ considered are open or closed (including $[a,b]$, $(a,b]$, $[a,b)$, $(a,b)$) as long as the function that shall be integrated is integrable, i.e. the other comments on open vs. closed interval are correct.

2) combining the two answers provided [by martini and Guy Fsone] so far, we obtain a complete and finally simple answer:

[following Guy Fsone:]

$$\int_a^b f'(x)\lfloor x\rfloor dx = \int_{a}^{\lfloor a\rfloor+1 } f'(x)\lfloor a \rfloor dx + \sum_{j=0}^{\lfloor b\rfloor-\lfloor a\rfloor-2} \int_{\lfloor a\rfloor+j+1}^{\lfloor a\rfloor+ j+2 } f'(x)\left(\lfloor a\rfloor +j+1\right) dx\\ + \int_{\lfloor b \rfloor}^{b} f'(x)\lfloor b\rfloor dx $$ Here, as stated before, the sum appears if it makes sense, i.e. if $\lfloor b\rfloor-\lfloor a\rfloor-2\ge 0$; otherwise, the sum is not present.

Moreover, [following martini] we can further simplify the final result by noticing that the term $\left(\lfloor a\rfloor +j+1\right)$ in the integrand does not depend on $x$. So all those terms can be moved out of the integral such that

$$ \int_a^b f'(x)\lfloor x\rfloor dx = \lfloor a \rfloor \int_{a}^{\lfloor a\rfloor+1 } f'(x)dx + \sum_{j=0}^{\lfloor b\rfloor-\lfloor a\rfloor-2} \left(\lfloor a\rfloor +j+1\right) \int_{\lfloor a\rfloor+j+1}^{\lfloor a\rfloor+ j+2 } f'(x) dx \\ + \lfloor b\rfloor\int_{\lfloor b \rfloor}^{b} f'(x) dx $$ and thus

$$ ... = \lfloor a \rfloor\left[ - f(a) + f(\lfloor a\rfloor+1) \right] + \sum_{j=0}^{\lfloor b\rfloor-\lfloor a\rfloor-2} \left(\lfloor a\rfloor +j+1\right) \left[-f\left(\lfloor a\rfloor+j+1\right)+f\left(\lfloor a\rfloor+ j+2 \right)\right]\\ + \lfloor b\rfloor~\left[f(b)-f\left( \lfloor b \rfloor\right) \right] $$

Now this is similar to a telescope sum; if written out explicitly, we observe that

$$ ... \qquad = - \lfloor a \rfloor f(a) + \lfloor a \rfloor f\left(\lfloor a\rfloor+1\right)\\ + \left(\lfloor a \rfloor + 1 \right) \left[-f\left(\lfloor a\rfloor+1\right)+f\left(\lfloor a\rfloor+2 \right)\right]\\ + \left(\lfloor a \rfloor + 2 \right) \left[-f\left(\lfloor a\rfloor+2\right)+f\left(\lfloor a\rfloor+3 \right)\right]\\ + ... \\ + \left(\lfloor b \rfloor -1 \right) \left[-f\left(\lfloor b\rfloor-1\right)+f\left(\lfloor b\rfloor\right)\right]\\ + \lfloor b\rfloor~\left[f(b)-f\left( \lfloor b \rfloor\right) \right]\\[5mm] = \quad - \lfloor a \rfloor f(a) - f\left(\lfloor a\rfloor+1\right) - f\left(\lfloor a\rfloor+2 \right) ... - f\left(\lfloor b\rfloor\right) + \lfloor b\rfloor~f(b) $$ where the notation above is slightly sloppy as some terms may not appear if $\lfloor b\rfloor - \lfloor a\rfloor$ is not sufficiently large. In summary, we obtain $$\boxed{\int_a^b f'(x)\lfloor x\rfloor dx = \lfloor b\rfloor~f(b) - \lfloor a \rfloor f(a) - \sum_{j=\lfloor a+1\rfloor}^{\lfloor b\rfloor} f(j)} $$

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$$\int_a^b f'(x)\lfloor x\rfloor dx = \int_{a}^{\lfloor a\rfloor+1 } f'(x)\lfloor a \rfloor dx + \sum_{j=0}^{\lfloor b\rfloor-\lfloor a\rfloor-2}\int_{\lfloor a\rfloor+j+1}^{\lfloor a\rfloor+ j+2 } f'(x)\left(\lfloor a\rfloor +j+1\right) dx+ \int_{\lfloor b \rfloor}^{b} f'(x)\lfloor b\rfloor dx $$ Where the terms under summation hold only when it makes sens. Namely $\lfloor b\rfloor-\lfloor a\rfloor-2\ge 0$. The remaining computations are left to the OP