Calculating $\int_0^\infty \frac{\sin(x)}{x} \frac{\sin(x / 3)}{x / 3} \frac{\sin(x / 5)}{x / 5} \cdots \frac{\sin(x / 15)}{x / 15} \ dx$ I found the following result on this webpage:
$$\int_0^{\infty } \left(\prod _{k=0}^7 \frac{\sin \left(\frac{x}{2 k+1}\right)}{\frac{x}{2 k+1}}\right) \, \mathbb{d}x= \frac{\pi}{2} - \frac{6879714958723010531}{935615849440640907310521750000} \pi $$
However, I can't determine how to prove it.
 A: The great reference is:
Borwein, David; Borwein, Jonathan M.
"Some remarkable properties of sinc and related integrals." 
Ramanujan J. 5 (2001) 73–89. 
DOI 10.1023/A:1011497229317
A: I have a visual explanation, why its below $\frac{\pi}{2}$
In 2001, the Borwein Brothers surprised their readers with the trick you referred to:

The pattern suggests that the definite integral of the sinc function ($=\pi/2$), does not alter when this sinc is multiplied with other sincs having lower frequencies. But unexpectedly, they show that the eighth integral is about one billionth of a percent smaller. How is that possible?
French mathematician Fourier allows us to look behind the scenes. From his perspective, we see a rectangular pedestal that gradually erodes: although in the beginning only its corners are rounded, inevitably the middle of the plateau will be affected as well, thereby lowering its actual height.
To understand this, let's first recollect some Fourier theory:
Five familiar Fourier facts

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*The Fourier transform of a sinc, is a rectangular function (SINC-RECT pair). For the Borwein Integrals, it is instructive to see each sinc-function as the frequency spectrum of a rectangular signal.


*The 'Uncertainty Principle' tells us: 'the wider the sinc, the narrower the block'. Thus the decreasing frequency of each new sinc factor, translates in a decreasing width of its corresponding rectangular pulse.


*The Convolution Theorem tells us: a multiplication of spectra is equivalent to the convolution of two signals. Therefore, we can understand the multiplication of these sincs, as a convolution of their corresponding block pulses.


*The inverse Fourier Transform tells us, that the integral of a spectrum is the central value in the time domain (similar as the integral of a signal, is the central component in the Frequency domain). Therefore these Borwein Integrals are proportional to the central value of the convolution outcome of the corresponding block pulses.


*The convolution of two equivalent block pulses (equal width), results in a triangular function (an isosceles): the left and right corner of the rectangle are completely eroded (see animation below), but the center of the plateau is preserved
(which btw explains why $\int_0^{\infty} \operatorname{sinc}(x)^2 dx = \frac{\pi}{2}$ (see: integralOfSquaredSincs))

Source: Convolution
Convolving a wide rectangle with a narrow one
With this theory in mind, let's look at the second Borwein Integral. The product of the first two sincs, is equivalent to convolving a rectangle having unit width (image below, left lower graph) with a (1/3)-kernel (upper graph in the middle). The result is a trapezoid, in which only two third of the initial unit plateau width is preserved, and one third is eroded:

​Why? Because in the interval [-1/3, 1/3], the (1/3)-kernel was completely embedded within the unit block. Just outside this domain, however, an increasing part of the kernel is not 'matched', resulting in two downward slopes moving out of the center. However, since the center of the initial plateau is not (yet) affected, the second Borwein integral equals still $\frac{\pi}{2}$.
The result after some more convolutions:
In the third Borwein integral, the resulting trapezoid is convolved with an (1/5)-kernel, which takes away another 20% of the original plateau width. Therefore the new residual plateau is reduced to 1 - 1/3 - 1/5 = 7/15 part of the initial plateau width:

​
With each new convolution, the residual plateau gets smaller and smaller:

​*(which is not that well visible, because each convolution also 'smoothens' the outcome, exactly how the probability distribution of a coin flipping experiment gets smoother when more coins are flipped)*
Not enough plateau left to completely match the (1/15)-kernel
Once the outcome of the (1/11)-convolution in turn is convolved with the (1/13)-kernel, the residual plateau equals:
$1 - \frac{1}{3}-\frac{1}{5}-...-\frac{1}{13}= \frac{2021}{45045}<\frac{1}{15}$
which is for the first time smaller than the next convolution kernel.
So for the eighth Borwein integral, the remaining plateau is too small to completely match the (1/15)-kernel in the center, resulting in a slightly lower integration outcome for the main frequency component.
And since this main frequency component is proportional to this eighth integral:
$\int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3}\cdots\frac{\sin(x/15)}{x/15}dx$
we now see why it's for the first time slightly below $\frac{\pi}{2}$.

*

*Borwein, David; Borwein, Jonathan M. (2001), "Some remarkable properties of sinc and related integrals" Link to article
