Why do the Borwein integrals stop being $\frac{\pi}{2}$?

Solution 1:

Sort of an answer.

These integrals are known as the Borwein Integrals, found by David and Johnathan Borwein in Some remarkable properties of sinc and related integrals (2001).

According to wikipedia, if we have a sequence of nonzero reals $a_0,a_1,...,a_n$, we may evaluate $$\int_0^\infty \prod_{k=0}^{n}\frac{\sin a_k x}{a_kx}dx=\frac{\pi}{2a_0}C_n,$$ where $$C_n=\frac{1}{2^nn!\prod_{k=1}^{n}a_k}\sum_{\gamma\in\{\pm1\}^n}\varepsilon_\gamma b_\gamma^n\text{ sgn}(b_\gamma),$$ $$\gamma=(\gamma_1,\gamma_2,...,\gamma_n)\in\{-1,1\}^n,\qquad \varepsilon_\gamma=\gamma_1\gamma_2\cdots\gamma_n,$$ and $$b_\gamma=a_0+a_1\gamma_1+a_2\gamma_2+\dots+a_n\gamma_n.$$ Then, apparently, when $a_0>|a_1|+|a_2|+\dots+|a_n|,$ we have $C_n=1$. I am not sure how this follows from the explicit evaluation, but I'll update when I find out.

Anyway, taking $a_k=\frac1{2k+1}$ and $J_n=\int_0^\infty\prod_{k=0}^{n}\sin(a_kx)/(a_kx)\, dx$, we get $$J_1=\frac\pi2\cdot\frac{1}{2\cdot\tfrac13}\left(\frac43-\frac23\right)=\frac\pi2,$$ $$J_2=\frac\pi2\cdot\frac1{2^2\cdot2\cdot\tfrac13\tfrac15}\left(\left(\frac{23}{15}\right)^2-\left(\frac{17}{15}\right)^2-\left(\frac{13}{15}\right)^2+\left(\frac{7}{15}\right)^2\right)=\frac\pi2,$$ and so on. As you can see, there are $2^n$ terms for $J_n$, and I don't wanna have to write out anymore of them.

Finally, when $n=7$, we have $$\sum_{k=1}^{7}|a_k|=\sum_{k=1}^{7}\frac1{2k+1}=\frac{46207}{45045}\approx 1.0218>a_0=1,$$ and indeed, $$J_7\approx \frac\pi2-2.31\cdot10^{-11}.$$