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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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Since $n-1 \le t \le n$, $[t] = n$ only when $t = n$, so you're essentially looking at two identical integrals. – Patrick Da Silva Dec 14 '11 at 6:34

2 Answers 2

up vote 5 down vote accepted

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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$[t]=n$ at only one point, and no one point affects an integral.

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