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I calculated $\int{\lfloor{x}\rfloor}dx$ and i got this result: $$\int{\lfloor{x}\rfloor}dx = \frac{x^2-x}{2}+\sum_{k=1}^{\infty}\left(\frac{\sin(k\pi x)}{k\pi}\right)^2+c$$ Do you know if this series have a closed form?

We found a nice identity!

$$\int{\lfloor{x}\rfloor}dx = \frac{\pi^2\left(x^2-x\right)+\Re\left[ \mathrm{Li_2}\left(e^{i2\pi x}\right)\right]}{2\pi^2}+c$$

So

$$\int_0^x{\lfloor{t}\rfloor}dt = \frac{\pi^2\left(x^2-x\right)+\Re\left[ \mathrm{Li_2}\left(e^{i2\pi x}\right)\right]}{2\pi^2}+\frac{1}{12}$$

Also, using $$\Re\left[ \mathrm{Li_2}\left(e^{ix}\right)\right] =\sum_{k\ge1}\frac{\cos(kx)}{k}=\frac{x^2}{4}+\frac{\pi x}{2}+\frac{\pi^2}{6}, \forall x\in\left[0,2\pi\right]$$

We get

$$\forall x\in\left[0,1\right] \forall \alpha\in\mathbb{Z},\ \ \ \ \ \Re\left[ \mathrm{Li_2}\left(e^{i2\pi x}\right)\right] =\pi^2\left((x-\alpha)^2-(x-\alpha)+\frac{1}{6}\right)$$

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I don't know if your expansion is valid. But this infinite sum can be worked with,$$\sum_{k=1}^\infty \left(\frac{\sin(k\pi x)}{k\pi}\right)^2 = \sum_{k=1}^\infty \frac{1-\cos (2k\pi x)}{2k^2\pi^2} = \frac1{2\pi^2}\sum_{k=1}^\infty \frac1{k^2} -\frac1{2\pi^2} \Re \sum_{k=1}^\infty \frac{\exp (i2\pi x)^k}{k^2} $$ The first term is $\frac1{12}$ from the Basel problem. The second term involves the dilogarithm $\operatorname{Li}_2(z) = \sum_{k=1}^\infty \frac{z^k}{k^2} $, $$-\frac1{2\pi^2} \Re \sum_{k=1}^\infty \frac{\exp (i2\pi x)^k}{k^2} = -\frac1{2\pi^2} \Re \operatorname{Li}_2(e^{2\pi i x}) $$

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Interpret the integral as a Riemann-Stieltjes integral, you can integrate it by part.

For $y > 0$, you get something like

$$\begin{align}\int_0^y \lfloor x \rfloor dx &= \int_{0-}^{y+} \lfloor x \rfloor dx = y \lfloor y \rfloor - \int_{0-}^{y+} x d\lfloor x \rfloor\\ &= y \lfloor y \rfloor - \sum_{k=0}^{\lfloor y\rfloor} k = y \lfloor y \rfloor - \frac12 \lfloor y \rfloor(\lfloor y \rfloor + 1)\\ &= \frac12y(y-1) + \frac12 \{y\}(1-\{y\}) \end{align} $$ where $\{y\} = y - \lfloor y \rfloor$ is fractional part of $y$. This suggests $$\sum_{k=1}^{\infty}\left(\frac{\sin(k\pi x)}{k\pi}\right)^2 = \frac12 \{x\} (1-\{x\})$$

One can verify this by computing the Fourier series of the periodic function on RHS.

Notice $$\begin{align} a_0 &= \frac12\int_0^1 x(1-x) dx = \frac{1}{12} = \frac{1}{2\pi^2}\frac{\pi^2}{6}\\&= \frac{1}{2\pi^2}\sum_{k=1}^\infty \frac{1}{k^2}\\ a_k &= \frac12\int_0^1 x(1-x)\cos(2\pi k x)dx = \frac{1}{4\pi k}\int_0^1 x(1-x) d\sin(2\pi k x)\\ &= \frac{1}{4\pi k}\left\{ \left[x(1-x) \sin(2\pi k x)\right]_0^1 + \int_0^1 (2x-1)\sin(2\pi k x) dx \right\}\\ &= -\frac{1}{8\pi^2 k^2}\int_0^1 (2x-1)d\cos(2\pi k x)\\ &= -\frac{1}{8\pi^2 k^2} \left\{ \left[(2x-1)\cos(2\pi k x)\right]_0^1 - 2\int_0^1 \cos(2\pi k x) dx \right\}\\ &= -\frac{1}{4\pi^2k^2} \end{align}$$

and by symmetry, $\displaystyle\;b_k = \frac12\int_0^1 x (1-x) \sin(2\pi k x)dx = 0$ for all $k$. This leads to

$$\begin{align}\frac12\{x\}(1 - \{x\}) &= a_0 + 2\sum_{k=1}^\infty (a_k \cos(2\pi k x) + b_k\sin(2\pi kx))\\ &= \frac{1}{2\pi^2}\sum_{k=1}^\infty \frac{1 - \cos(2\pi k x)}{k^2}\\ &= \sum_{k=1}^\infty \left(\frac{\sin(\pi k x)}{k \pi}\right)^2\end{align}$$

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    $\begingroup$ @TheGreatDuck It is integration by part. When one interpret an integral as RS-integral, integration by part continue to work for integrand containing jump discontinuity. It is possible the integration limit hit the jump discontinuity. So I change the limit from $[0,y]$ to $[0-,y+]$ to make sure which discontinuity we are going to include in the finite sum. The first is justified because the integrand is $0$ at $y = 0$, the second is justified because the floor function is right continuous. $\endgroup$ – achille hui Nov 26 '18 at 0:22
  • $\begingroup$ @TheGreatDuck The end result is probably the same. The advantage of this is I remember one less thing. I only need to remember integration by part works for this sort of integrand. $\endgroup$ – achille hui Nov 26 '18 at 0:31
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You don't need a series (but if you really want this series, the following can also be used to find the value of the series up to a constant term you can determine).


Let $f(x)=\int_0^x \lfloor{u}\rfloor\mathrm du$.

For a given $x\ge0$, let $n=\lfloor{x}\rfloor\ge0$. Then

$$f(x)=\int_0^n \lfloor{u}\rfloor\mathrm du+\int_n^x \lfloor{u}\rfloor\mathrm du$$

With the convention that the sum is zero if $n=0$,

$$f(x)=\sum_{k=0}^{n-1}\int_k^{k+1} \lfloor{u}\rfloor\mathrm du+\int_n^x \lfloor{u}\rfloor\mathrm du$$

On $[k,k+1[$, $\lfloor{u}\rfloor=k$, and on $[n,x[$, $\lfloor{u}\rfloor=n$, so

$$f(x)=\sum_{k=0}^{n-1}\int_k^{k+1} k\mathrm du+\int_n^x n\mathrm du$$

$$f(x)=\sum_{k=0}^{n-1}k+n(x-n)=\frac{n(n-1)}{2}+n(x-n)=\frac n2(n-1+2x-2n)=\frac n2(2x-n-1)$$

$$f(x)=\frac{\lfloor{x}\rfloor}{2}(2x-\lfloor{x}\rfloor-1)$$


Now let $x<0$, then

$$f(x)=\int_0^x \lfloor{u}\rfloor\mathrm du=-\int_x^0 \lfloor{u}\rfloor\mathrm du$$

Let $n=\lfloor{x}\rfloor<0$:

$$f(x)=-\int_x^n \lfloor{u}\rfloor\mathrm du-\int_n^0 \lfloor{u}\rfloor\mathrm du$$

$$f(x)=-\sum_{k=n}^{-1}\int_k^{k+1} \lfloor{u}\rfloor\mathrm du+\int_n^x \lfloor{u}\rfloor\mathrm du$$

$$f(x)=-\sum_{k=n}^{-1}\int_k^{k+1} k\mathrm du+\int_n^x n\mathrm du$$

$$f(x)=-\sum_{k=n}^{-1}k+ n(x-n)=\frac{-n(1-n)}2+n(x-n)$$

$$f(x)=\frac n2(2x-n-1)$$

$$f(x)=\frac{\lfloor{x}\rfloor}{2}(2x-\lfloor{x}\rfloor-1)$$


Therefore, a primitive on $\Bbb R$ is given by

$$\int \lfloor{x}\rfloor\mathrm dx=\frac{\lfloor{x}\rfloor}{2}(2x-\lfloor{x}\rfloor-1)+C$$


Now, the series, assuming your primitive is correct.

$$f(x)-\frac{x^2-x}2=\frac{2x\lfloor{x}\rfloor}2-\frac{\lfloor{x}\rfloor^2}2-\frac{x^2}{2}+\frac x2-\frac{\lfloor{x}\rfloor}{2}$$

$$f(x)-\frac{x^2-x}2=-\frac12(x-\lfloor{x}\rfloor)^2+\frac{x-\lfloor{x}\rfloor}{2}=\frac12\{x\}-\frac12\{x\}^2=\frac12\{x\}(1-\{x\})$$

That is

$$f(x)=\frac{x^2-x}{2}+\frac12\{x\}(1-\{x\})=\frac12\{x\}(1-\{x\})-\frac12x(1-x)$$

[Note that since the integrand is piecewise constant, $f(x)$ is piecewise linear and continuous, and the previous equality means that it interpolates the parabola $\frac{x^2-x}{2}$ at integer points. Furthermore, the difference between the parabola and the interpolation is never more than $\frac18$.]

The previous equality also means that there is a constant $c$ such that the following holds for all $x$:

$$\sum_{k=1}^{\infty}\left(\frac{\sin(k\pi x)}{k\pi}\right)^2+c=\frac12\{x\}(1-\{x\})$$

Now let $x=0$, then $c=0$.

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The antiderivative of a step function is a piecewise linear, continuous function, made of segments of increasing slope. If we define it as the definite integral from zero to $x$, the integer part of $x$ yields a triangular number (sum of integers), while the fractional part brings a linear contribution.

So

$$\int_0^x\lfloor x\rfloor dx=\frac{(\lfloor x\rfloor-1)\lfloor x\rfloor}2+\lfloor x\rfloor\{x\}=\frac{(2x-\lfloor x\rfloor-1)\lfloor x\rfloor}2.$$

As it turns out, the formula is also valid in the negatives.

enter image description here

It is interpolated at integer points by the parabola $\dfrac{x(x-1)}2$.

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