Is my calculator giving rounding errors?

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 $\begin{aligned} B_n & = \tfrac12\pi - \pi\int_1^\infty F_n(x)\,dx \; = \; \tfrac12\pi - \pi\int_1^{c_n} \frac{(2n+1)!!}{2^n (n-1)!}(c_n-x)^{n-1}\,dx \\ & = \tfrac12\pi - \pi \frac{(2n+1)!!}{2^n n!}(c_n-1)^n \end{aligned}$ In our case the key value is $n=7$ , since $c_6 < 1 < c_7$ , and hence $\tfrac12\pi - B_7 \; = \; \pi \frac{15!!}{2^7 7!}(c_7-1)^7 \; = \; \frac{6879714958723010531}{935615849440640907310521750000}\pi \; = \; 2.31 \times 10^{-11}$ while each of $B_1,B_2,B_3,B_4,B_5,B_6$ is equal to $\tfrac12\pi$ .