How find this integral $\int_{0}^{x}\left(\frac{1}{2}-\{t\}\right)dt$

Question

Find the value

$$F(x)=\int_{0}^{x}\left(\dfrac{1}{2}-\{t\}\right)dt$$ where$\{x\}=x-[x]$

my try: since $$F(x)=\int_{0}^{x}\left(\dfrac{1}{2}-t+[t]\right)dt$$ when

$$k-1\le t<=k,[t]=k-1,k\in Z$$ because $x$ is not integer,and maybe $x\to \infty$ so follow I can't.Thank you

A simpler way is to turn it into two integrals (one of 0.5-t and the other of [t]) then just draw [t], it is a very nice step function, the easiest thing to integrate! — Alec Teal, Nov 27 '13 at 16:25
*second easiest, the constant function would be the easiest, a step function is a sum of these :P — Alec Teal, Nov 27 '13 at 16:25

score 6 · Accepted Answer · 2013-11-27T16:47:08.610

6

Hint

Let $[x]=p$ then write $$\int_0^x f(t)dt=\sum_{k=0}^{p-1}\int_k^{k+1}f(t)dt+\int_p^x f(t)dt$$

Added

$$\int_0^x[t]dt=\sum_{k=0}^{p-1}\int_k^{k+1}[t]dt+\int_p^x [t]dt=\sum_{k=0}^{p-1}\int_k^{k+1}k\ dt+\int_p^x p\ dt\\=\sum_{k=0}^{p-1}k+p(x-p)=\frac{p(p-1)}{2}+p(x-p)$$

There's no problem to calculate the integral $$\int_0^x\left(\frac{1}{2}-t\right) dt$$ and you have $F(x)$.

edited Nov 27 '13 at 16:47

answered Nov 27 '13 at 16:05

1

I have use this Euler-Maclaurin summion,But I can't find this $F(x)$ form,can you post full your solution?Thank you – Nov 27 '13 at 16:21
1

@nanchangjian I edited my answer. – Nov 27 '13 at 16:47
Nice.Thank you very much @SamiBenRomdhane – math110 Nov 27 '13 at 16:56
You're welcome @chinamath – Nov 27 '13 at 16:58
@SamiBenRomdhane: I made this question favorited just case of your approach. :+) – Mikasa Nov 28 '13 at 04:37
Nicely done, Sami! +1 – amWhy Dec 01 '13 at 00:04

How find this integral $\int_{0}^{x}\left(\frac{1}{2}-\{t\}\right)dt$

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