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.

  • 1
    $\begingroup$ 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) $\endgroup$ – 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*}

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

  • $\begingroup$ The question was asked by a research professor Thank you for this article $\endgroup$ – cerise Jul 10 '19 at 6:57
  • $\begingroup$ you are welcome $\endgroup$ – logo Jul 10 '19 at 7:00
  • 3
    $\begingroup$ this result is very interesting simply because it extends the definition of the GCD to non-integral values, for example $$\gcd \left( \frac{1}{2},\frac{1}{2} \right)=\frac{1}{\pi }$$ $\endgroup$ – logo Jul 10 '19 at 7:13
  • $\begingroup$ Perhaps the professor would hope that we can find a direct way to calculate the integral $\endgroup$ – cerise Jul 10 '19 at 7:20
  • $\begingroup$ he hopes to much!!!! $\endgroup$ – logo Jul 10 '19 at 7:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.