# Integral $\int_0^\pi \cos (( n-m) x )\frac{\sin^2 (nmx)}{\sin(nx)\sin(mx)} \rm dx$

I am trying to calculate the value of the integral $$\frac{1}{\pi}\int_0^\pi \cos \left(( n-m) x \right) \frac{\sin^2 (nmx)}{\sin(nx)\sin(mx)} \textrm{d}x$$ with $$n, m$$ integers greater than $$1$$.

When I give particular values ​​of $$n$$ and $$m$$, wolfram alpha uses magic simplifications and finds the value of the integral. For example if $$n=7$$ and $$m=3$$ the integral is reduced to $${\int} (\cos\left(36x\right)+\cos\left(30x\right)+\cos\left(28x\right)+\cos\left(24x\right)+2\cos\left(22x\right)+\cos\left(18x\right)+2\cos\left(16x\right)+$$$$\cos\left(14x\right)+\cos\left(12x\right)+2\cos\left(10x\right)+2\cos\left(8x\right)+\cos\left(6x\right)+2\cos\left(4x\right)+2\cos\left(2x\right)+1)\mathrm{d}x$$

It also gives a simplified form, but I do not know how to exploit it.

• After several tries with wolfram, Can someone give me an example of a couple (n, m) that returns a value of the integral different from the gcd (n, m) – cerise Jul 9 '19 at 21:24

Maybe this is an elementary solution.

Via the formula $$\begin{equation*} a^p-b^p = (a-b)\sum_{k=0}^{p-1}a^{p-1-k}b^{k} \end{equation*}$$ and Euler's formula we get $$\begin{gather*} \sin^2(nmx) = \left(\dfrac{e^{inmx}-e^{-inmx}}{2i}\right)^{2} =\dfrac{\left(e^{inx}\right)^{m}-\left(e^{-inx}\right)^{m}}{2i}\cdot \dfrac{\left(e^{imx}\right)^{n}-\left(e^{-imx}\right)^{n}}{2i}=\\[2ex] \dfrac{e^{inx}-e^{-inx}}{2i}\dfrac{e^{imx}-e^{-imx}}{2i}\cdot\sum_{k=0}^{m-1}e^{inx(m-1-k)}e^{-inxk}\cdot \sum_{j=0}^{n-1}e^{imx(n-1-j)}e^{-imxj}=\\[2ex] \sin(nx)\sin(mx)\cdot \sum_{k=0}^{m-1}\sum_{j=0}^{n-1}e^{ix(2nm-n-m-2kn-2jm)} \end{gather*}$$ Thus $$\begin{gather*} \cos((n-m)x)\dfrac{\sin^2(nmx)}{\sin(nx)\sin(mx)}=\\[2ex]\dfrac{e^{i(n-m)x}+e^{-i(n-m)x}}{2}\cdot\sum_{k=0}^{m-1}\sum_{j=0}^{n-1}e^{ix(2nm-n-m-2kn-2jm)}=\\[2ex] \dfrac{1}{2}\cdot\sum_{k=0}^{m-1} \sum_{j=0}^{n-1}e^{ix(2nm-2m-2kn-2jm)}+ \dfrac{1}{2}\cdot\sum_{k=0}^{m-1} \sum_{j=0}^{n-1}e^{ix(2nm-2n-2kn-2jm)}.\tag{1} \end{gather*}$$ However, $$\begin{equation*} \dfrac{1}{\pi}\int_{0}^{\pi}e^{i2px}\, \mathrm{d}x = \begin{cases} 0\text{ if } p\neq 0 \text{ and integer },\\ 1 \text{ if } p=0. \end{cases} \end{equation*}$$ Now we integrate the two double sums in(1). By symmetry the two integrals will have the same value. The integral $$\begin{equation*} \dfrac{1}{\pi}\int_{0}^{\pi}e^{ix(2nm-2m-2kn-2jm)}\, \mathrm{d}x = 1 \end{equation*}$$ if and only if $$\begin{gather*} nm-m-kn-jm=0 \tag{2} \end{gather*}$$ which is a linear diophantine equaion.

Put $$n= pd, m=qd$$ where $$d=\rm{gcd}(n,m)$$. Then we can write (2) as $$\begin{equation*} kp+jq=pqd-q. \end{equation*}$$ All solutions are $$\begin{equation*} \begin{cases} k=qd-rq\\ j=-1+rp \end{cases} \end{equation*}$$ where $$r$$ is an integer. But $$0 \le k \le m-1$$ and $$0\le j \le n-1$$. Thus $$\begin{equation*} \begin{cases} 0\le qd-rq \le qd-1\\ 0 \le -1+rp \le pd -1 \end{cases} \Longleftrightarrow \begin{cases} 1\le rq \text{ and } r\le d\\ 1 \le rp \text{ and } r\le d. \end{cases} \end{equation*}$$ Consequently we find $$d$$ solutions to (2). If we integrate (1) we get $$\begin{equation*} \dfrac{1}{2}d+\dfrac{1}{2}d = \rm{gcd(n,m).} \end{equation*}$$

• I like this way Thank you – cerise Jul 10 '19 at 9:27
• You are welcome! – JanG Jul 10 '19 at 9:58
• nice work and +1 – logo Jul 10 '19 at 19:53
• Thank you very much. – JanG Jul 10 '19 at 20:14

The value of the integral equals $$\gcd (m,n)$$ as stated in this article and the solution is not elemantary.

article citation: Sedjelmaci, Sidi Mohamed, Some related functions to integer GCD and coprimality, Bonomo, Flavia (ed.) et al., LAGOS’11 — VI Latin-American algorithms, graphs, and optimization symposium. Extended abstracts from the symposium, Bariloche, Argentina, March 28—April 1, 2011. Amsterdam: Elsevier. Electronic Notes in Discrete Mathematics 37, 135-140 (2011). ZBL1268.11162.

• The question was asked by a research professor Thank you for this article – cerise Jul 10 '19 at 6:57
• you are welcome – logo Jul 10 '19 at 7:00
• this result is very interesting simply because it extends the definition of the GCD to non-integral values, for example $$\gcd \left( \frac{1}{2},\frac{1}{2} \right)=\frac{1}{\pi }$$ – logo Jul 10 '19 at 7:13
• Perhaps the professor would hope that we can find a direct way to calculate the integral – cerise Jul 10 '19 at 7:20
• he hopes to much!!!! – logo Jul 10 '19 at 7:22