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Question:

let

$$f(t)=\int_0^t\left(\{x\}-\dfrac{1}{2}\right)dx$$ where $\{t\}$ is the fractional part of $t$,

then find this integral value

$$I=\int_0^{+\infty}\dfrac{f(t)}{1+t^2}dt$$

My try: I have $$f(t)=\int_0^t\left(\{t\}-\dfrac{1}{2}\right)dt=\dfrac{1}{2}\{t\}(\{t\}-1)$$

also can see:How find this integral $\int_{0}^{x}\left(\frac{1}{2}-\{t\}\right)dt$ so $$I=\dfrac{1}{2}\int_0^{+\infty}\dfrac{\{t\}(\{t\}-1)}{1+t^2}dt$$ so $$\sum_{n=0}^{+\infty}\int_n^{n+1}\dfrac{\{t\}(\{t\}-1)}{1+t^2}dt=\sum_{n=1}^{\infty}\int_n^{n+1}\dfrac{(t-n)(t-n-1)}{t^2+1}dt$$ since $$\int_n^{n+1}\dfrac{(t-n)(t-n-1)}{t^2+1}dt=\left(1-\dfrac{2n+1}{2}\ln{\dfrac{(n+1)^2+1}{n^2+1}}+(n^2+n-1)(\arctan{(n+1)}-\arctan({n})\right)$$ so $$I=\sum_{n=0}^{\infty}\left(1-\dfrac{2n+1}{2}\ln{\dfrac{(n+1)^2+1}{n^2+1}}+(n^2+n-1)(\arctan{(n+1)}-\arctan({n}))\right)$$

and this integral $I$ is convege, because $2f(t)=\{t\}(\{t\}-1)$ is bounded so and $$\int_0^{+\infty}\dfrac{1}{1+t^2}dt=\dfrac{\pi}{2}$$ then I can't find this sum,

Thank you very much!

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  • 1
    $\begingroup$ The natural next step would be to split it into a series of $\int_{n}^{n+1}$ integrals (which makes the fractional-parts brackets go away), integrate each of them separately, and then sum the series. $\endgroup$ – Henning Makholm Dec 10 '13 at 13:35
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Dec 10 '13 at 13:36
  • $\begingroup$ @HenningMakholm,yes,That's my mean,But follow I fell ugly $\endgroup$ – math110 Dec 10 '13 at 13:41
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    $\begingroup$ @MhenniBenghorbal: I don't think it's related in any way except for the fact that a fractional part appears... $\endgroup$ – Najib Idrissi Dec 10 '13 at 16:06
  • $\begingroup$ Which integral are you asking for? The one in the title is not the same one in the body of the post. $\endgroup$ – Jon Claus Dec 11 '13 at 15:17
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The function $$f(t):={1\over2}\{t\}(\{t\}-1)$$ is periodic with period $1$ and is $\ ={1\over2}(t^2-t)$ for $0\leq t\leq1$, whence continuous. It can be developed into a Fourier series as follows: $$f(t)=-{1\over12}+\sum_{k=1}^\infty{1\over 2\pi^2 k^2}\>\cos(2k\pi t)\ .$$ Therefore we obtain $$\eqalign{I&:=\int_0^\infty{f(t)\over 1+t^2}\ dt=-{1\over12}\int_0^\infty{1\over 1+t^2}\ dt +\sum_{k=1}^\infty{1\over 2\pi^2 k^2}\int_0^\infty{\cos(2k\pi t)\over 1+t^2}\ dt\cr &\ =-{\pi\over24}+\sum_{k=1}^\infty{1\over 4\pi k^2}e^{-2\pi k}\quad .\cr}$$ Here we have used the well known integral $$\int_0^\infty{\cos(c\>t)\over 1+t^2}\ dt={\pi\over2}e^{-c}\qquad(c\geq0)\ .$$ The resulting sum cannot be expressed in elementary terms, but it is extremely well convergent. Taking the first six terms we obtain the numerical approximation $$I\doteq -0.13075101809261383373\ .$$

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  • $\begingroup$ Oh,It's nice solution,Thank you+1 $\endgroup$ – math110 Dec 13 '13 at 14:16
  • $\begingroup$ $$ \sum_{k = 1}{{\rm e}^{-2\pi k} \over 4\pi k^{2}} ={{\rm Li}_{2}\left(\,{\rm e}^{-2\pi}\,\right) \over 4\pi} \approx 1.4868\times 10^{-4} $$ $\endgroup$ – Felix Marin Sep 13 '14 at 18:51
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It's considerably easier to split the integral into parts. Let $T:=[t]$, the floor of $t$. We have:

$$\int_{0}^t\{t\}dt=T\int_0^1tdt+\int_0^{t-T}tdt=\frac{T}{2}+\frac{(t-T)^2}{2}.$$

So,

$$f(t)=\frac{T}{2}+\frac{(t-T)^2}{2}-\frac{t}{2}$$

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  • $\begingroup$ Yes,this is evry easy,and How prove find follow integral $\endgroup$ – math110 Dec 11 '13 at 11:53

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