I saw this in one of the most famous answers on Mathoverflow [1]. The CAS expert (Jacques Carette) who wasted three days trying to locate the "problem" in the software chimes in in the comments.

[1] https://mathoverflow.net/questions/11517/computer-algebra-er...

The poor Maple developer who tried to fix the "bug" is active on MathOverflow and commented briefly on his experience of this in a comment below this answer: https://mathoverflow.net/a/11607

Unfortunately I can't think through it now, but I would bet there is a fun connection to be made with 3*5=15. Something about roots more frequently getting repeated, or zeros in the s-plane going to a higher order maybe? I'd be delighted if someone can chime in with the connection, if there is one.

You can compute this integral as follows. First, you remember that sinc(x) = sin(x)/x is the Fourier transform of 1/2 * indicator function of [-1, 1]. In general you have:

  sinc(x/n) = n/2 \int_{-1/n}^{1/n} d t e^{i t x}.

Next, notice that

  \int_0^\infty sinc(x) \dots sinc(x/n) = 1/2 \int_{-\infty}^\infty sinc(x) \dots sinc(x/n).

Next, replace the sinc functions by their Fourier representation, change the order of integration and use the formula

  \int_{-\infty}^\infty d x e^{i a x} = 2 pi \delta(a),

where \delta(a) is the Dirac delta. You are left with an integral of a delta function over a product of intervals. What remains to be done is just some tedious computation.

\int_{-1}^1 d t_0 \dots \int_{-1/(2 k + 1)}^{1/(2 k + 1)} d t_k 1/2 \dots (2 k + 1)/2 2 \pi \delta(t_1 + \dots + t_k).

Next, the question is whether t_1 + \dots + t_k = 0 has a solution for t's in the integration domain. This holds until k = 7. To see this, compute

1/3 + 1/5 = 8/15 < 1

1/3 + 1/5 + 1/7 = 71/105 < 1

1/3 + 1/5 + 1/7 + 1/9 = 248/315 < 1

1/3 + 1/5 + 1/7 + 1/9 + 1/11 = 3043/3465 <1

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 = 43024/45045 < 1

but

1/3 + 1/5 + 1/7 + 1/9 + 1/11 + 1/13 + 1/15 = 46027/45045 > 1.