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Why $\displaystyle\int_{n-1}^{n}[t]f'(t)dt=\int_{n-1}^{n}(n-1)f'(t)dt$? IMO, since $t\leq n$, $[t]$ could be $n$. How to explain this? Thanks in advance!

This is used in the proof of Euler's summation formula (Theorem 3.1) in Apostol's Introduction to Analytic Number Theory, page 54.

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    $\begingroup$ Since $n-1 \le t \le n$, $[t] = n$ only when $t = n$, so you're essentially looking at two identical integrals. $\endgroup$ Dec 14 '11 at 6:34
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I assume $[t]$ denotes the integer part.

Then $[t]=n$ only when $t=n$. If two functions only differ at finitely many points, their integrals are equal.

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  • $\begingroup$ I don't get it why $[t]=(n-1)$ ? looks like we are treating $ [t]$ as constant, isn't $[t]$ is a floor function? floor function can not be integrated ? why? $\endgroup$ Oct 9 '20 at 13:14
  • $\begingroup$ @Andrew The integral is from $n-1$ to $n$. And the floor function, while not continuous, has a well defined Riemann integral over any bounded interval. It does not have an antiderivative over $\mathbb R$, but the question is about Riemann integral not antiderivative. $\endgroup$
    – N. S.
    Oct 9 '20 at 22:47
  • $\begingroup$ @Andrew For functions which are NOT continuous, having an antiderivative over the interval $[a,b]$, and the integral $\int_a^b f(t) dt$ existing are NOT equivalent. $\endgroup$
    – N. S.
    Oct 9 '20 at 22:49
  • $\begingroup$ Yes..... sounds reasonable, could you plz provide name(s) of book(s) where your explanation can be found, if possible, specifically on floor function, so I can give reference... or any document that discusses (like you did)? Thanks. I found below answer, so suggest something else, if possible: math.stackexchange.com/questions/239324/… $\endgroup$ Oct 10 '20 at 7:43
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    $\begingroup$ @Andrew Since on the interval $[ n-1,n)$ the function is constant, it is an easy exercise to show Riemann integrability, it is the type of exercise which we give as easy Homework in real analysis. If you want a reference, just cite the Lebesgue Integrability condition, but that's a big overkill for this problem. $\endgroup$
    – N. S.
    Oct 10 '20 at 15:46
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$[t]=n$ at only one point, and no one point affects an integral.

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  • $\begingroup$ I don't get it why $[t]=(n-1)$ ? looks like we are treating $ [t]$ as constant, isn't $[t]$ is a floor function? floor function can not be integrated ? why? $\endgroup$ Oct 9 '20 at 13:15

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