Explain why $\int_0^\infty\frac{\sin{4x}}{x}\prod\limits_{k=1}^n \cos\left(\frac{x}{k}\right) dx\approx\frac{\pi}{2}$ Why do we have, for every $n\in\mathbb N$, $$\int_0^\infty\left(\prod_{k=1}^n \cos\left(\frac{x}{k}\right)\right)\frac{\sin{4x}}{x}dx\approx\frac{\pi}{2}\ ?$$
 A: Hint. One may use the transformation
$$
\begin{align} \prod_{k=1}^n \cos \frac{x}{k} & = \frac{1}{2^n}\sum_{e\in S} \cos\left[\left(\frac{e_1}1+\cdots+\frac{e_n}n\right)x\right] \quad \text{where }S=\{1,-1\}^n 
\end{align}
$$ and the addition formula
$$
2\sin a \cos b = \sin(a + b) + \sin(a - b).
$$
Thus, for every positive natural   integer $n$, the integral

$$
I_n=\int_0^\infty\left(\prod_{k=1}^n \cos\frac{x}{k}\right)\frac{\sin{4x}}{x}\:dx
$$ 

is such that $$2^nI_n=\sum_{e\in S} \int_0^\infty\frac{\sin\left[\left(4+\frac{e_1}1+\cdots+\frac{e_n}n\right)x\right]}{2x}dx+\sum_{e\in S} \int_0^\infty\frac{\sin\left[\left(4-\frac{e_1}1-\cdots-\frac{e_n}n\right)x\right]}{2x}dx.
$$
By symmetry of the set $S$ the two sums coincide hence
$$
2^nI_n=\sum_{e\in S} \int_0^\infty\frac{\sin\left[\left(4+\frac{e_1}1+\cdots+\frac{e_n}n\right)x\right]}{x}dx=\sum_{e\in S} \frac\pi2\cdot
\mathrm{sgn}\left(4+\frac{e_1}1+\cdots+\frac{e_n}n\right)$$
where we have used that, for every $\alpha \in \mathbb{R}$, $\alpha \ne0$,
$$
\int_0^\infty \frac{\sin (\alpha x)}x\:dx=\frac \pi2 \cdot \text{sgn}(\alpha).
$$
If $n\leq30$, then $$\left|\frac{e_1}1+\cdots+\frac{e_n}n\right|<4$$
for every $(e_1,\ldots,e_n)$ in $S$ hence all the signs are $+1$ and

$$
I_n=\frac \pi2.
$$ 

If $n\ge31$, then one only gets the semi-explicit formula

$$
I_n=\frac\pi2\cdot\frac{1}{2^n}\sum_{e\in S}\mathrm{sgn}\left(4+\frac{e_1}1+\cdots+\frac{e_n}n\right)
$$

which is nevertheless enough to show that $$0<I_n<\frac\pi2.$$
