Definite integration of trig functions: $\int_{0}^{\infty} \frac{\sin 2020x}{x} \prod_{k=1}^{n} \cos(kx) \, \mathrm d x$

Question

Given $$\int_{0}^{\infty} \frac{\sin x}{x} \operatorname d x=\frac{\pi}{2}$$
Find the natural values of $n$ which satisfy , $$\int_{0}^{\infty} \cos x \cdot \cos 2 x\cdot \cos 3 x\cdots\cos n x \cdot \frac{\sin 2020 x}{x} d x= \frac{\pi}{2}$$ My approach: Consider the given integral as $I(n)$. Now if I find values of $I(n)$ corresponding to $n=1,2,3\dots$ I am getting value $\pi / 2$. So I can see natural values of $n$ less than $2020 $ satisfy the equation. I don't know the answer to the problem. Is there some elegant approach to solve this question?

Answer 1

This integral is closely related to a Borwein integral.

We can use contour integration, and specifically Jordan's lemma, to show that $$\int_{0}^{\infty} \frac{\sin (mx)}{x} \prod_{k=1}^{n} \cos(kx) \, \mathrm dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin(mx)}{x} \prod_{k=1}^{n} \cos(kx) \, \mathrm dx=\frac{\pi}{2} $$ if $m$ is strictly greater than $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}.$

For example, $$\int_{0}^{\infty} \frac{\sin (15x)}{x} \prod_{k=1}^{n} \cos(kx) \, \mathrm dx = \frac{\pi}{2}$$ for $n=1, 2, 3$, and $4$.

But $$\int_{0}^{\infty} \frac{\sin (15x)}{x} \prod_{k=1}^{5} \cos(kx) \, \mathrm dx = \frac{31 \pi}{64}. $$

For a large value of $m$ (like $m=2020$), this might be difficult to confirm numerically because the integral does not converge absolutely.

A slight variation of this integral is mentioned here as an example of a pattern that eventually fails.

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Integrate the function $$f(z) = \frac{e^{imz}}{z} \prod_{k=1}^{n} \cos(kz)= \frac{1}{2^{n}}\frac{e^{imz}}{z} \prod_{k=1}^{n} \left(e^{ikz}+ e^{-ikz} \right), \quad m > \frac{n(n+1)}{2} , $$ around the contour $$[-R, -r] \cup r e^{i[\pi, 0]} \cup [r, R] \cup Re^{i[0, \pi]},$$ where $ r e^{i[\pi, 0]}$ is the upper half of the small circle $|z|=r$ (traversed clockwise), and $Re^{i[0, \pi]}$ is the upper half of the large circle $|z|=R$ (traversed counterclockwise.)

If we expand the product, we get a linear combination of exponential functions with exponents that range from $-in(n+1)z/2$ to $in(n+1)z/2$.

If we then multiply by $e^{imz}$, every exponential function becomes of the form $e^{iaz}$ where $a >0$.

Therefore, by Jordan's lemma, the integral vanishes on the large semicircle as $R \to \infty$.

And since $f(z)$ has a simple pole at the origin, the integral goes to $- i \pi \operatorname{Res}[f(z), 0]$ on the small semicircle as $r \to 0 $.

So we have $$\operatorname{PV} \int_{-\infty}^{\infty}\frac{e^{imx}}{x} \prod_{k=1}^{n} \cos(kx) \, \mathrm dx - i \pi \operatorname{Res}[f(z), 0] =0, \tag{1}$$

where $$\operatorname{Res}[f(z), 0] = \lim_{z \to 0} e^{imz} \prod_{k=1}^{n} \cos(kz) =1. $$

Equating the imaginary parts on both sides of equation $(1)$, we get

$$\int_{-\infty}^{\infty} \frac{\sin (mx)}{x} \prod_{k=1}^{n} \cos(kx) \, \mathrm dx = \pi, $$ and the result follows.

Answer 2

If you have $\int_0^\infty \frac {sin x}{x} = \frac \pi 2$

Then I would try to show that $\int_0^\infty \frac {\sin mx}{x} \frac \pi 2

Next $\cos mx\sin nx = \frac 12 \sin (n+m)x + \frac 12\sin (n-m)x$

$\int_0^\infty \frac {\cos mx\sin nx}{x}\ dx = \frac \pi 2$

Thus:
$\Pi \cos nx \frac {\sin mx}{x} = \sum \frac{\sin kx}{a_k}$ where $\sum \frac{1}{a_k} = 1$