Evaluating $I=\int_{0}^{\frac{\pi}{2}}\prod_{k=1}^{7}\cos kx \, dx$

Question

I am attemping to show that $$ I \equiv \int_{0}^{\pi/2}\left[\prod_{k = 1}^{7}\cos\left(kx\right)\right] {\rm d}x = \frac{\pi}{32} $$ So far I have tried expressing $\cos kx$ as $\frac{1}{2}(e^{ikx}+e^{-ikx})$, resulting in $$I=\frac{1}{2^7}\int_{0}^{\frac{\pi}{2}}e^{-28ix}\prod_{k=1}^{7}(e^{2ikx}+1)\, dx,$$ but I am not sure how to proceed futher. As this problem was sourced from the UNSW Integration Bee 2022, I assume there is a 'trick' so that the integral can be evaluated without tediously expanding the product using trig identities, but I am also interested in these answers.

EDIT:

This is not a duplicate of Closed form of $\int\limits_0^{2\pi} \prod\limits_{j=1}^n \cos(jx)dx$ and combinatorial link as the following does not apply:

As suggested by Winther in the comments, the problem can be viewed from a combinatorial standpoint. Looking at the complex exponential representation one gets $2^n$ integrals of the form $\int_{0}^{2\pi}e^{iNx}dx$ , which is only nonzero, if $N=0$ . The integral evaluates to $\frac{M\pi}{2^{n-1}}$ , where M is the number of nonzero integrals.

As $$\int_{0}^{\frac{\pi}{2}}e^{iNx}dx = \frac{1}{N}\sin\frac{N\pi}{2}+\frac{i}{N}\left(1-\cos\frac{N\pi}{2}\right) \neq 0 \text{ for } N \text{ mod } 4\neq 0, $$ I do not believe this method applies.

Answer 1

Given $n \in \mathbb{N}$ let's denote $I_n =\int_{0}^{\frac{\pi}{2}} \prod_{k=1}^{n} \cos(kx)\, \mathrm{d}x$ and $J_n = \int_{0}^{2\pi} \prod_{k=1}^{n} \cos(kx)\, \mathrm{d}x$. Using symmetry we get \begin{align*} J_n & \overset{\color{blue}{x \to\pi+x}}{=} \int_{-\pi}^{\pi} \prod_{k=1}^{n}\cos(kx)(-1)^k\, \mathrm{d}x\\ & \overset{\color{purple}{\text{Even}}}{=} \color{purple}{2}(-1)^{\left\lfloor\frac{n+1}{2} \right\rfloor} \int_{\color{purple}{0}}^{\pi} \prod_{k=1}^{n} \cos(kx)\, \mathrm{d}x\\ & = 2(-1)^{\left\lfloor\frac{n+1}{2} \right\rfloor} \left[\int_{0}^{\frac{\pi}{2}} \prod_{k=1}^{n} \cos(kx)\, \mathrm{d}x +\int_{\frac{\pi}{2}}^{\pi} \prod_{k=1}^{n} \cos(kx)\, \mathrm{d}x\right]\\ &\overset{\color{blue}{x \to \pi-x}}{=} 2(-1)^{\left\lfloor\frac{n+1}{2} \right\rfloor} \left[\int_{0}^{\frac{\pi}{2}} \prod_{k=1}^{n} \cos(kx)\, \mathrm{d}x -\int_{\color{blue}{\frac{\pi}{2}}}^{\color{blue}{0}} \prod_{k=1}^{n} \cos(kx) (-1)^k\, \mathrm{d}x\right]\\ & = 2 \left( 1 + (-1)^{\left\lfloor\frac{n+1}{2} \right\rfloor} \right) \int_{0}^{\frac{\pi}{2}} \prod_{k=1}^{n} \cos(kx)\, \mathrm{d}x\\ & =\begin{cases}4I_n , & \text{if }n\equiv 0 \pmod{4} \text{ or }n\equiv 3 \pmod{4}\\ 0, &\text{if } n\equiv 1 \pmod{4} \text{ or }n\equiv 2 \pmod{4} \end{cases} \end{align*} Since you're interested in $I_7$ then your case does reduce to $J_7$. Noticing that \begin{align*} + 1 + 2 - 3 + 4 - 5 - 6 + 7\\ + 1 + 2 - 3 - 4 + 5 + 6 - 7\\ + 1 - 2 + 3 + 4 - 5 + 6 - 7\\ + 1 - 2 - 3 - 4 - 5 + 6 + 7\\ - 1 + 2 + 3 + 4 + 5 - 6 - 7\\ - 1 + 2 - 3 - 4 + 5 - 6 + 7\\ - 1 - 2 + 3 + 4 - 5 - 6 + 7\\ - 1 - 2 + 3 - 4 + 5 + 6 - 7 \end{align*} are the only $\color{green}{8}$ ways to add up to $0$ with the numbers $\{\pm 1 \pm2, \dots, \pm 7\}$, by the duplicate questions you get $$ I_7 = \frac{J_7}{4} = \frac{2\pi \cdot\mathbin{\color{green}{8}} }{4\cdot 2^{7}} = \frac{\pi}{32} $$ as desired.

Answer 2

Expanding using product to sum identities on pairs of factors whose arguments add to 8x yields the following, where $c_n$ is short for $\cos(nx)$. \begin{align*} I&=\frac{1}{2^{3}}\int_0^{\frac{\pi}{2}}(c_8+c_6)(c_8+c_4)(c_8+c_2)c_4\,dx\\ &=\frac{1}{2^4}\int_0^{\pi}(c_4+c_3)(c_4+c_2)(c_4+c_1)c_2\,dx. \end{align*} By King's Property, we have \begin{equation} I=\frac{1}{2^4}\int_0^{\pi}(c_4-c_3)(c_4+c_2)(c_4-c_1)c_2\,dx \end{equation} Taking the average of the two expressions for $I$ yields $$I=\frac{1}{2^4}\int_0^{\pi}(c_4^2+c_3c_1)(c_4+c_2)c_2\,dx.$$ A product to sum identity gives $c_3c_1=\frac{1}{2}(c_4+c_2)$, so substituting this in and making a scaling substitution we obtain \begin{align*} I&=\frac{1}{2^6}\int_0^{2\pi}(2c_2^2+c_2+c_1)(c_2+c_1)c_1\,dx\\ &=\frac{1}{2^6}\left[\int_0^{\pi}(2c_2^2+c_2+c_1)(c_2+c_1)c_1\,dx+\int_\pi^{2\pi}(2c_2^2+c_2+c_1)(c_2+c_1)c_1\,dx\right]\\ &=\frac{1}{2^6}\left[\int_0^{\pi}(2c_2^2+c_2+c_1)(c_2+c_1)c_1\,dx+\int_0^{\pi}(2c_2^2+c_2-c_1)(c_2-c_1)(-c_1)\,dx\right]\\ &=\frac{1}{2^5}\int_0^{\pi}((2c_2^2+c_2)c_1+c_1c_2)c_1\,dx\\ &=\frac{1}{2^4}\int_0^{\pi}(c_2+1)c_2c_1^2\,dx\\ &=\frac{1}{2^5}\int_0^{\pi}(c_2+1)^2c_2\,dx\\ &=\frac{1}{2^5}\int_0^{\pi}(c_2^3+2c_2^2+c_2)\,dx\\ &=\frac{1}{2^5}(0+\pi+0)=\frac{\pi}{32} \end{align*} Observe that the only non-trivial parts of the above calculation were the two uses of product to sum identities and the observations which led to cancellations — someone who has trained to do algebraic manipulations quickly should be able to do most of the above in their head and obtain the final solution in a minute or two.