Question about the proof of Theorem 3.1 (Page 54) of the book Introduction to Analytic Number Theory by Apostol

Question

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.

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 06:34

score 6 · Accepted Answer · answered Dec 14 '11 at 06:13

6

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.

answered Dec 14 '11 at 06:13

N. S.

132,525

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? – Consider Non-Trivial Cases Oct 09 '20 at 13:14
@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. – N. S. Oct 09 '20 at 22:47
@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. – N. S. Oct 09 '20 at 22:49
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: https://math.stackexchange.com/questions/239324/is-integrability-equivalent-to-having-antiderivative – Consider Non-Trivial Cases Oct 10 '20 at 07:43
1

@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. – N. S. Oct 10 '20 at 15:46

score 5 · Answer 2 · answered Dec 14 '11 at 06:14

5

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

answered Dec 14 '11 at 06:14

Gerry Myerson

179,216

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? – Consider Non-Trivial Cases Oct 09 '20 at 13:15

Question about the proof of Theorem 3.1 (Page 54) of the book Introduction to Analytic Number Theory by Apostol

2 Answers2