How much larger is the denominator? (Try 2 -- copy error in entering the answer)

it has been a few years since I posted this proof as a solution to a previous question, so it seems worth digging it out again.

These are what are called the Borwein integrals. For any $k > 0$ define the function $f_k(x) \; = \; \tfrac{k}{2} \chi_{[-k^{-1},k^{-1}]}(x)$ noting that $\int_\mathbb{R} f_k(x)\,dx \; = \; 1 \hspace{2cm} (\mathcal{F}f_k)(y) \; =\; \tfrac{1}{\sqrt{2\pi}} \mathrm{sinc}\big(\tfrac{y}{k}\big)$ for any $k > 0$ , where $\mathcal{F}$ denotes the Fourier transform, and $\mathrm{sinc}(x) \; = \; \frac{\sin x}{x}$ Thus $\prod_{j=1}^n \mathrm{sinc}\big(\tfrac{y}{2j+1}\big) \; = \; (2\pi)^{\frac12n} \prod_{j=1}^n (\mathcal{F}f_{2j+1})(y) \; = \; \sqrt{2\pi} \big[\mathcal{F} (f_3 \star f_5 \star \cdots \star f_{2n+1})\big](y)$ where $\star$ denotes the convolution operation $(f \star g)(x) \; =\; \int_{\mathbb{R}} f(y)g(x-y)\,dy$ for suitable functions $f,g$ . Note that all functions we shall be dealing with are "suitable". Thus $\begin{aligned} B_n \; = \; \int_0^\infty \prod_{j=0}^n \mathrm{sinc}\big(\tfrac{x}{2j+1}\big)\,dx & = \; \tfrac12\times2\pi\big\langle \mathcal{F}f_1 \,,\, \mathcal{F}(f_3 \star f_5 \star \cdots f_{2n+1})\big\rangle \\ & = \; \pi\big\langle f_1\,,\, f_3 \star f_5 \star \cdots \star f_{2n+1} \big\rangle \\ & = \; \tfrac12\pi \int_{-1}^1 \big(f_3 \star f_5 \star \cdots \star f_{2n+1}\big)(x)\,dx \end{aligned}$ Now define $c_n \; = \; \sum_{j=1}^n \frac{1}{2j+1}$ and let $F_n(x) \; = \; (f_3 \star f_5 \star \cdots \star f_{2n+1})(x)$ It is a simple matter of induction to note that:

• $F_n$ is an even function of $x$ ,
• $F_n(x)$ vanishes for $|x| > c_n$ ,
• we have $F_n(x) \; = \; \frac{(2n+1)!!}{2^n (n-1)!} (c_n - x)^{n-1} \hspace{1cm} c_{n-1} < x < c_n \;.$

Thus, if $n$ is small enough that $c_n < 1$ , then $B_n \; = \; \tfrac12\pi \int_{-1}^1 F_n(x)\,dx \; = \; \tfrac12\pi \int_{\mathbb{R}}F_n(x)\,dx \; = \; \tfrac12\pi \prod_{j=1}^n \int_{\mathbb{R}}f_{2j+1}(x)\,dx \; = \; \tfrac12\pi$ On the other hand, if $n$ is such that $c_{n-1} < 1 < c_n$ , then $\begin{aligned} B_n & = \; \tfrac12\pi - \pi\int_1^\infty F_n(x)\,dx \; = \; \tfrac12\pi - \pi\int_1^{c_n} \frac{(2n+1)!!}{2^n (n-1)!}(c_n-x)^{n-1}\,dx \\ & = \; \tfrac12\pi - \pi \frac{(2n+1)!!}{2^n n!}(c_n-1)^n \end{aligned}$ Since $c_6 < 1 < c_7$ , we have $B_n = \tfrac12\pi$ for $1 \le n \le 6$ , while $B_7 \; = \; \tfrac12\pi - \pi\frac{15!!}{2^7 \,7!}(c_7-1)^7$ and by symmetry $I(8) = 2B_7$ , so that $\frac{I(8)}{\pi} \; = \; 1 - \frac{15!!}{2^6\,7!}(c_7-1)^7 \; = \; \frac{467807924713440738696537864469}{467807924720320453655260875000}$ and the difference between the numerator and the denominator is $\boxed{6879714958723010531}$ .

Of course, if you give Wolfram Alpha the above formula for $B_7$ , it can handle it!

$I : n \to \int_{-\infty }^{\infty } \left(\prod _{m=1}^n \text{sinc}\left(\frac{x}{2 m-1}\right)\right) \, dx$

See Illusive patterns in math explained by ideas in physics .

Image credit: Majumdar and Trizac. ©2019 American Physical Society

$I(8) \Rightarrow \frac{467807924713440738696537864469 \pi }{467807924720320453655260875000}$

$f = I(8)$ and $\text{Denominator}[f]-\text{Numerator}[f] \Rightarrow 6879714958723010531$

The error happened as a result of choosing an incorrect copying method for the answer text . The first method seems to round to 16 decimal places and second copies as a text string without rounding. I took advantage of this situation to correct a significant, but not problem affecting, error in the last line of the problem statement.