# 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.

• What are you asking? – Git Gud Oct 19 '13 at 10:13
• do you want $\int \lfloor x \rfloor \sin(\frac{6x}{\pi}) \ dx$ ?? – what'sup Oct 19 '13 at 10:25
• Yes, I'm asking what is the way to solve this integral. – Splash Oct 20 '13 at 10:53


\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}

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

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*}

• Tnx, that's what I did. I thought there's a generic way to solve it. I mean, if the boundaries were 0 to 100, what would you do? – Splash Oct 20 '13 at 10:56
• For $0$ to $100$ look at @sos440 's answer – Ömer Oct 20 '13 at 11:39

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*}

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: $$J={1\over2\lambda}\left(\cot{\lambda\over2}\sin(\lambda N)-(2N-1)\cos(\lambda N)-1\right)\ .$$