Is it safe to generalise?

Look here for a detailed analysis of these so-called Borwein integrals.

As for the bonus question, we are interested in knowing when the identity $\int_0^\infty \prod_{k=0}^n \mathrm{sinc}\big(\tfrac{x}{100k+1}\big)\,dx \; =\; \tfrac12\pi$ stops holding. Generalising the work in the given link, this integral is equal to $\tfrac12\pi\int_{-1}^1 \big(f_{101} \star f_{201} \star \cdots \star f_{100n+1}\big)(x)\,dx$ and this will be equal to $\tfrac12\pi$ so long as $D_n < 1$ , where $D_n \; = \; \sum_{k=1}^n \frac{1}{100k+1}$ This is because the support of the convolution $f_{101} \star f_{201} \star \cdots \star f_{100n+1}$ is $[-D_n,D_n]$ . Using the estimates $\int_1^{n+1}\frac{dx}{100x+1} \; \le \; D_n \; \le \; \int_0^n \frac{dx}{100x+1}$ we deduce that $D_n$ approximates $1$ in the range $2.7 \times10^{41} = \tfrac{1}{100}\big(e^{100}-1\big) \; \le \; n \; \le \; \tfrac{101}{100}\big(e^{100}-1\big) = 2.7 \times 10^{43}$ Evaluating $D_n$ exactly in terms of the polygamma function, Mathematica tells me that the witching value of $n$ is in the order of $1.53412 \times 10^{43}$ .

Here is a good article on the Borwein integrals with lots of graphs that make it easier (and more fun) to understand what is going on, at least for me.

@Otto Bretscher Sir, I know that you already know this, but still, it is worth mentioning the person who discovered these amazing patterns!!! That somebody was Borwein.......