Integral $ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ Question
$ \int _0^6 \lfloor x \rfloor \sin( \frac {6x}{\pi}) \ \mathrm dx $ = ?
we tried to bound it from both sides using $x$ and $(x-1)$, which yield nice estimation ($\frac {24}\pi$) - ($\frac {36}\pi$) but not a precise one. 
we also tried to split it to 6 integrals using $\lfloor x\rfloor  $ as a different constant each time. $(0\cdot (\cos(...)-\cos(...))+1(\cos(\frac \pi3)-\cos(\frac\pi6))+2\cdot...$ etc
it produced the right result ($\frac {30}\pi$), but the number is not important as achieving the general function.
 A: x is real number, m is integer, and $\mathbb{Z}$ is the set of integers (positive, negative, and zero).
$$\lfloor x \rfloor=\max\, \{m\in\mathbb{Z}\mid m\le x\}$$
So
$$\begin{alignat*}{1}
\int_{0}^{6}\lfloor x\rfloor\sin(\frac{6x}{\pi})\ \mathrm{d}x= & \int_{0}^{1}0.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{1}^{2}1.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{2}^{3}2.\sin(\frac{6x}{\pi})\ \mathrm{d}x\\
 & +\int_{3}^{4}3.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{4}^{5}4.\sin(\frac{6x}{\pi})\ \mathrm{d}x+\int_{5}^{6}5.\sin(\frac{6x}{\pi})\ \mathrm{d}x\\
= & \frac{1}{6}\pi(\cos(\frac{6}{\pi})-\cos(\frac{12}{\pi}))+\frac{1}{3}\pi(\cos(\frac{12}{\pi})-\cos(\frac{18}{\pi}))\\
 & +\frac{1}{2}\pi(\cos(\frac{18}{\pi})-\cos(\frac{24}{\pi}))+\frac{2}{3}\pi(\cos(\frac{24}{\pi})-\cos(\frac{30}{\pi}))\\
 & +\frac{5}{6}\pi(\cos(\frac{30}{\pi})-\cos(\frac{36}{\pi}))\\
= & \frac{1}{6}\pi\left(\cos(\frac{6}{\pi})+\cos(\frac{12}{\pi})+\cos(\frac{18}{\pi})+\cos(\frac{24}{\pi})+\cos(\frac{30}{\pi})-5\cos(\frac{36}{\pi})\right)
\end{alignat*}$$
A: Here is another solution using Riemann-Stieltjes integral. Integrating by parts,
\begin{align*}
\int_{0}^{6} \lfloor x \rfloor \sin \left(\frac{6x}{\pi}\right) \, dx
&= \int_{0^{+}}^{6} \lfloor x \rfloor \sin \left(\frac{6x}{\pi}\right) \, dx \\
&= \left[ - \lfloor x \rfloor \frac{\pi}{6} \cos \left(\frac{6x}{\pi}\right) \right]_{0^{+}}^{6} + \int_{0^{+}}^{6} \frac{\pi}{6} \cos \left(\frac{6x}{\pi}\right) \, d\lfloor x \rfloor \\
&= - \pi \cos \left(\frac{36}{\pi}\right) + \sum_{k=1}^{6} \frac{\pi}{6} \cos \left(\frac{6k}{\pi}\right).
\end{align*}
More generally, let $f$ be continuous on $[a, b]$ and $F$ be an anti-derivative of $f$. Then
\begin{align*}
\int_{a}^{b} f(x)\lfloor x \rfloor \, dx
&= \int_{a^{+}}^{b} f(x)\lfloor x \rfloor \, dx \\
&= \left[ F(x)\lfloor x \rfloor \right]_{a^{+}}^{b} - \int_{a^{+}}^{b} F(x) \, d\lfloor x \rfloor \\
&= \lfloor b \rfloor F(b) - \lfloor a \rfloor F(a) - \sum_{a < k \leq b} F(k).
\end{align*}
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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\begin{align}
\int_{0}^{6}\!\!\!\!\left\lfloor x\right\rfloor\sin\pars{6x \over \pi}\,\dd x
&=
\sum_{\ell = 1}^{5}\ell\int_{\ell}^{\ell + 1}\!\!\!\!\!\!\!\sin\pars{6x \over \pi}
\,\dd x
=
\sum_{\ell = 1}^{5}\ell\braces{%
-\,{\cos\pars{6\bracks{\ell + 1}/\pi} \over 6/\pi}
+
{\cos\pars{6\ell/\pi} \over 6/\pi}}
\\[3mm]&=
-\,{\pi \over 6}\sum_{\ell = 2}^{6}\pars{\ell - 1}\cos\pars{6\ell/\pi}
+
{\pi \over 6}\sum_{\ell = 1}^{5}\ell\cos\pars{6\ell/\pi}
\\[3mm]&=
-\,{5\pi \over 6}\cos\pars{30/\pi}
-
{\pi \over 6}\sum_{\ell = 2}^{5}\pars{\ell - 1}\cos\pars{6\ell/\pi}
+
{\pi \over 6}\cos\pars{6/\pi}
\\[3mm]&\phantom{=}\mbox{}\,+
{\pi \over 6}\sum_{\ell = 2}^{5}\ell\cos\pars{6\ell/\pi}
\\[3mm]&=
{\pi \over 6}\cos\pars{6 \over \pi}
-
{5\pi \over 6}\cos\pars{30 \over \pi}
+
{\pi \over 6}\Re\sum_{\ell = 2}^{5}\pars{\expo{6\ic/\pi}}^{\ell}
\end{align}

\begin{align}
\sum_{\ell = 2}^{5}\pars{\expo{6\ic/\pi}}^{\ell}
&=
{\expo{12\ic/\pi}\pars{\expo{24\ic/\pi} - 1}
 \over
\expo{6\ic/\pi} - 1}
=
{\expo{12\ic/\pi}\pars{\expo{12\ic/\pi} - \expo{-12\ic/\pi}}
 \over
\expo{3\ic/\pi} - \expo{-3\ic/\pi}}\,
{\expo{12\ic/pi} \over \expo{3\ic/\pi}}
=
\expo{21\ic/\pi}\,
{\sin\pars{12/\pi} \over \sin\pars{3/\pi}}
\end{align}

$$\color{#0000ff}{%
\int_{0}^{6}\left\lfloor x\right\rfloor\sin\pars{6x \over \pi}\,\dd x
\color{#000000}{\LARGE\ =\ }
{1 \over 6}\,\pi\bracks{%
\cos\pars{6 \over \pi}
-
5\cos\pars{30 \over \pi}
+
{\cos\pars{21/\pi}\sin\pars{12/\pi} \over \sin\pars{3/\pi}}}}
$$

A: Its about integrals of the form
$$J:=\int_0^N\lfloor x\rfloor\sin(\lambda x)\ dx=\sum_{k=1}^{N-1} k\int_k^{k+1}\sin(\lambda x)\ dx={1\over\lambda}\sum_{k=1}^{N-1}k\bigl(\cos(\lambda k)-\cos(\lambda(k+1)\bigr)$$
for given $\lambda>0$.The sum on the right can be simplified, and one obtains
$$J={1\over\lambda}\left(\sum_{k=1}^{N-1}\cos(\lambda k)\ -\ (N-1)\cos(\lambda N)\right)\ .$$
Now, according to Mathematica,
$$\sum_{k=1}^{N-1}\cos(\lambda k)={\cot{\lambda\over2}\sin(\lambda N)-1-\cos(\lambda N)\over2}\ .$$
Therefore we obtain $J$ in closed form as follows:
$$J={1\over2\lambda}\left(\cot{\lambda\over2}\sin(\lambda N)-(2N-1)\cos(\lambda N)-1\right)\ .$$
