TITLE: Calculating $\int_0^\infty \frac{\sin(x)}{x} \frac{\sin(x / 3)}{x / 3} \frac{\sin(x / 5)}{x / 5} \cdots \frac{\sin(x / 15)}{x / 15} \ dx$
QUESTION [14 upvotes]: 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.

REPLY [4 votes]: $\newcommand{\S}{\operatorname{sinc}}$
A 'brute force' solution to a beautiful problem. I don't claim this answer is as insightful as the others and the question is somewhat old, but I feel it is relevant and unique enough to merit posting.
As usual, we will let $\S(x) = \sin(x)/x$ with $\S(0)=1$.
The big idea: use angle-addition, Taylor series, and integration by parts to 'revert' the integral to a bunch- $128$, to be precise- of weighted integrals of the form $\int_0^{\infty}\sin(a x)/x\,dx$, each of which is $\pi/2$. Then we just add up the weights to produce the answer.


Angle-addition.

Recall that
\begin{align}
2\sin(\theta)\sin(\phi)=\cos(\theta-\phi)-\cos(\theta+\phi)\\
2\sin(\theta)\cos(\phi)=\sin(\theta-\phi)-\sin(\theta+\phi)\\
2\cos(\theta)\cos(\phi)=\cos(\theta-\phi)+\cos(\theta+\phi)\\
\end{align}
For instance, this tells us $\S(x)\S(x/3) = \frac{3}{2}x^{-2}(\cos(2x/3)-\cos(4x/3))$. You probably see what lies ahead, even if it's somewhat unappetizing. We have
$$
\S(x)\S(x/3)\cdots \S(x/15) =
$$
$$ \frac{1\cdot 3\cdots 15}{2^7 x^8} \sum_{e_i\in\{\pm 1\}} (-1)^{\#(e_i=-1)} \cos\left( x(1+e_1 \cdot 1/3 + e_2\cdot  1/5 +\cdots +e_7 \cdot 1/15)\right) 
$$Thus, if we denote $W=(1\cdot 3\cdot \cdots \cdot 15)/2^7$, we have to evaluate $128$ integrals that look like $\displaystyle{W \int_0^{\infty}\frac{\cos(a x)}{x^8}\,dx}$, for some constant $a$. The trouble is that none of them are improperly integrable near $x=0$, which brings us to the second step in the process.

Taylor series.

The idea here is to add and subtract the same number of lower-order terms of the Taylor series of each $\cos(a x)$ to make the integrals converge. For example, $\displaystyle{\int_0^{\infty}\frac{\cos(x)}{x^8}\,dx}$ does not converge but $\displaystyle{\int_0^{\infty}\frac{\cos(x)-(1-x^2/2+x^4/24-x^6/720)}{x^8}\,dx}$ does. Since the sum of the arguments of the cosines is zero (check! A short explanation is that the $+$ and $-$ signs are equal in number) we will add and subtract up to the degree six terms to make the integrals convergent. Therefore, we have to evaluate
$$
\int_0^{\infty}\frac{\cos(a x)-(1-(ax)^2/2+(ax)^4/24-(ax)^6/720)}{x^8}\,dx
$$

Integration by parts.

We have
$$
\int_0^{\infty}\frac{\cos(a x)-(1-(ax)^2/2+(ax)^4/24-(ax)^6/720)}{x^8}\,dx
$$
\begin{align}
&=\left.-\frac{\cos(a x)-(1-(ax)^2/2+(ax)^4/24-(ax)^6/720)}{8x^7}\right|_0^{\infty}\\ &+\frac{1}{7}\int _0^{\infty} \frac{-a\sin(ax)+a^6 x^5/120 - a^4 x^3/6 + a^2 x }{x^7}\,dx
\end{align}The boundary terms vanish: at $\infty$, the denominator is a greater power and at $0$ the cosine series is $O(x^8)$. Continue this procedure until the denominator is just $x$:
$$
\cdots = \frac{a^7}{7!}\int_0^{\infty}\frac{\sin(ax)}{x}\,dx = \frac{a^7}{7!}\cdot\frac{\pi}{2}
$$

Add 'em up.

All that's left is to add up these values for all $128$ values of $a$ corresponding to the $2^7$ terms of the sum. This becomes
$$
W \cdot \frac{\pi}{2}\cdot \frac{1}{7!} \cdot \sum_{e_i\in\{\pm 1\}} (-1)^{\#(e_i=-1)} (1+e_1 \cdot 1/3 + e_2\cdot  1/5 +\cdots +e_7 \cdot 1/15)^7
$$A somewhat tedious but routine calculation gives the final answer of
$$
\frac{467807924713440738696537864469 \pi}{935615849440640907310521750000}
$$

For clarity, I'll do a smaller example (the three sinc case) that will evaluate to $\pi/2$.
$$
\int _0^{\infty} \S(x)\S(x/3)\S(x/5)\,dx = 15 \int_0^{\infty} \frac{\sin(x)\sin(x/3)\sin(x/5)}{x^3}\,dx
$$
$$
=\frac{15}{4} \int_0^{\infty} \frac{-\sin(7/15 x)+\sin(13/15 x)+\sin(17/15 x)-\sin(23/15 x)}{x^3}\,dx
$$
$$
=\frac{15}{4} \int_0^{\infty} \frac{-\sin(7/15 x)+\sin(13/15 x)+\sin(17/15 x)-\sin(23/15 x) }{x^3}+\frac{7/15 x-13/15 x-17/15 x+23/15 x}{x^3}\,dx
$$I'll just pick one of these, say the first one; the others are exactly the same.
$$
=\int_0^{\infty} \frac{-\sin(7/15 x)+7/15 x}{x^3}\,dx
$$
$$
= \left.\frac{-\sin(7/15 x)+7/15 x}{-2x^2}\right|_0^{\infty} + \int_0^{\infty} \frac{-7/15 \cos(7/15 x)+7/15}{2x^2}\,dx
$$
$$
=0+0 + \frac{(7/15)^2}{2} \int _0^{\infty}\frac{\sin(7/15 x)}{x}\,dx =\frac{(7/15)^2}{2}  \cdot \frac{\pi}{2}
$$So the total is
$$
\frac{15}{4}\cdot \frac{\pi}{2}\left(\frac{(7/15)^2}{2}-\frac{(13/15)^2}{2}-\frac{(17/15)^2}{2}+\frac{(23/15)^2}{2}\right)= \frac{\pi}{2}
$$

Of course, this method is merely computational: it has none of the theoretical trappings of the other answers. Nevertheless, the method could easily be generalized to other products and is quite amenable to reproduction on software. For instance, Mathematica thought long and hard about the original integral, but was able to perform the $128$ sub-integrals much quicker.
As a final remark, I'm reminded of the idea of quadrature from numerical analysis. It seems these integral-sinc-products, and more importantly their corresponding series, are 'exact' for the first seven until an error is introduced in the eighth. Perhaps I will investigate this further.