Closed form of $\int{\lfloor{x}\rfloor}dx$ 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
$$\forall x\in\left[0,1\right] \forall \alpha\in\mathbb{Z},\ \ \ \ \ \Re\left[ \mathrm{Li_2}\left(e^{i2\pi x}\right)\right] =\pi^2\left((x-\alpha)^2-(x-\alpha)+\frac{1}{6}\right)$$
 A: 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 $c=0$.
A: 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
$$\begin{align}\frac12\{x\}(1 - \{x\}) &= 
a_0 + 2\sum_{k=1}^\infty (a_k \cos(2\pi k x) + b_k\sin(2\pi kx))\\
&= \frac{1}{2\pi^2}\sum_{k=1}^\infty \frac{1 - \cos(2\pi k x)}{k^2}\\
&= \sum_{k=1}^\infty \left(\frac{\sin(\pi k x)}{k \pi}\right)^2\end{align}$$
A: 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} $,
$$-\frac1{2\pi^2} \Re \sum_{k=1}^\infty \frac{\exp (i2\pi x)^k}{k^2}  = -\frac1{2\pi^2} \Re \operatorname{Li}_2(e^{2\pi i x}) $$
A: 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.

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