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

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