Why do we have, for every $n\in\mathbb N$, $$\int_0^\infty\left(\prod_{k=1}^n \cos\left(\frac{x}{k}\right)\right)\frac{\sin{4x}}{x}dx\approx\frac{\pi}{2}\ ?$$

  • $\begingroup$ Maybe related to Borwein integrals, en.wikipedia.org/wiki/Borwein_integral $\endgroup$ – Gerry Myerson Jan 22 '17 at 12:11
  • $\begingroup$ This is indeed very similar to a Borwein integral in the sense that it evaluates to the same value for a large number of $n$'s, but this eventually breaks down (and the difference in some cases can be so small that it would be extremely hard to see this numerically). The generalization here is $$\int_0^\infty\left(\prod_{k=1}^n \cos\left(a_k x\right)\right)\frac{\sin(Nx)}{x}dx = \frac{\pi}{2}$$ for all $n$ such that $\sum_{k=1}^n |a_k| < N$. The proof is exactly the same as in the answer below. $\endgroup$ – Winther Jan 22 '17 at 14:59

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 $$0<I_n<\frac\pi2.$$

  • $\begingroup$ What is meant by $e$ and $S$ here. I'm not familiar with the notation $S=\{1,-1\}^n$ $\endgroup$ – ankit Jan 22 '17 at 12:50
  • $\begingroup$ The notation: $e=(e_1,\cdots,e_n)$ where each $e_k$ is equal to $-1$ or $1$. $S$ is the set of all $n$-tuples of that form. $\endgroup$ – Olivier Oloa Jan 22 '17 at 12:55
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    $\begingroup$ The general formula being $$\frac\pi2\cdot P\left(\left|\sum_{k=1}^n\frac{\epsilon_k}k\right|<4\right)$$ where $(\epsilon_k)$ is i.i.d. symmetric Bernoulli. $\endgroup$ – Did Jan 22 '17 at 13:06
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    $\begingroup$ Well, I do not want to dampen your enthusiasm but if you look at it, this is merely a rewriting of your sums over $S$ divided by $2^n$... :-) But yes, this rewriting allows to go further and to show the convergence of the integrals to $$\int_0^\infty\frac{\sin{4x}}{x}\prod_{k=1}^\infty \cos\left(\frac{x}{k}\right)dx=\frac{\pi}{2}p$$ where $$p=P\left(\left|\sum_{k=1}^\infty\frac{\epsilon_k}k\right|>4\right)<1$$ $\endgroup$ – Did Jan 22 '17 at 13:22
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    $\begingroup$ By Bienaymé-Chebychev inequality, $$p>1-\frac{\pi^2}{96}>0.897$$ hence, for every $n\geqslant31$, $$0.897\cdot\frac\pi2<I_n<\frac\pi2$$ $\endgroup$ – Did Jan 22 '17 at 13:53

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