I was looking for a simple way to evaluate the integral $\int_0^\infty \frac{\sin x}{x}dx$ ( a belated look at this question). There are symmetries to be exploited, for one thing. So I had an idea, but the idea depends, optimistically, on two expressions for $\pi$ that I cannot prove, and my question is whether anyone sees reasonable proofs of either expression (1, 2 below).
Put
$$\int_0^\infty \frac{\sin x}{x}dx = \int_0^\pi \sum_{k=0}^{k=\infty} \left\{\frac{\sin x}{x + 2 k \pi }-\frac{\sin x }{x+2k\pi+\pi}\right\} \, dx$$
The intuition behind this step is that as we make our way around the unit circle from 0 to $\pi$, for each angle $\alpha$ we have an angle $\alpha + \pi$ for which the value of $\frac{\sin x} {x}$will be smaller and negative. No need to integrate beyond $\pi$, and for angles greater than $2\pi$ we are simply dividing up the original integral's range by adding multiples of $\pi$ to the value of $x$.
WLOG we can form Riemann sums by dividing the range of integration into an even number of equi-angle subranges $dx_n$. For example, if $n = 6$, we have
$$\int_0^\pi \sum_{k=0}^{k=\infty}\sin x \left\{\frac{1}{x + 2 k \pi }-\frac{1 }{x+2k\pi+\pi}\right\} \, dx$$
$$\approx \left\{ \sum_{k=0}^\infty \sum_{n=1}^6 \sin \left( \frac{n\pi}{6}\right)\left(\frac{1}{\pi}\right) \left(\frac{1}{\frac{n}{6} + 2k}-\frac{1}{\frac{n}{6}+2k+1}\right) \right\} \left\{ \frac{\pi}{6} \right\}.$$
Taking this as true, we note that pairs of summands of the Riemann sums symmetric about $\frac{\pi}{2}$ sum to $1$, and the value of the summand at $\frac{\pi}{2}$ is $1/2$. Again, taking this as true, we have that
$$\sum f(x_n) = \frac{1}{2} + 1\cdot \frac{n-2}{2} + \sin\pi\cdot f(\pi) = \frac{n-1}{2},$$ and $$\sum f(x_n) \, dx_n = \left(\frac{n-1}{2}\right) \cdot \frac{\pi}{n}.$$
So for the approximation above for $n = 6$, we expect a value of about $\frac{5 \pi}{12}$ for high $k$.
In words, we are only counting n - 1 summands, since the last summand at $\pi$ is $0$. Since we divided the interval into $n$ subintervals, we are stuck with the sum above and the conditional limit (conditioned at least on the truth of the two limits below):
$$\int_0^\infty \frac{\sin x}{x} \, dx = \lim_{n = 1}^\infty \frac{n-1}{2}\cdot \frac{\pi}{n} = \frac{\pi}{2}.$$
The problem is this: we need to prove, to begin with, that
$$\sum_{k=0}^\infty \left(\frac{ 1}{\pi/2 + 2\pi k} - \frac{1}{\pi/2 + 2k\pi + \pi}\right) = \frac{1}{2}$$
or
$$ \frac{1}{\pi}\sum \frac{1}{1/2 + 2k}- \frac{1}{1/2 + 2k\pi +1}= \frac{1}{2},$$
that is,
(1) $\sum_{k=0}^\infty \frac{4}{3+16k+16k^2}= \frac{\pi}{2}$.
We must also prove for $0< j < n$ that
(2) $\sin (\frac{j\pi}{n})( \sum_{k=0}^{\infty} (\frac{ 1}{\frac{j\pi}{n} + 2\pi k} - \frac{1}{\frac{j\pi}{n} + 2k\pi + \pi}) + \sum_{k=0}^{\infty} (\frac{ 1}{\frac{(n-j)\pi}{n} + 2\pi k} - \frac{1}{\frac{(n-j)\pi}{n} + 2k\pi + \pi}))= 1$.
or , factoring $\frac{1}{ \pi}$ out of the sum and mutiplying:
$$\sin \left( \frac{j\pi}{n}\right)\left( \sum_{k=0}^\infty \left(\frac{1}{\frac{j}{n} + 2 k} - \frac{1}{\frac{j}{n} + 2k + 1}\right) + \sum_{k=0}^\infty \left( \frac{ 1}{\frac{(n-j)}{n} + 2 k} - \frac{1}{\frac{(n-j)}{n} + 2k + 1}\right)\right)= \pi.$$
(1) might be a special case of something at Wolfram's site, but I didn't spot it. (2) looks messy, but maybe induction?
If this is otherwise correct, I think one could work back from the relations for $\pi$ and establish the value of the integral. It is important not to forget that k starts at 0.
Edit: the following is equivalent to (2) and maybe easier to scan. For $0<r<1$,
$$(2) \sum_{k=0}^{\infty}\frac {2(\sin \pi r)(1+4k+4k^2-r+r^2) }{(1+2k+r)(1+2k-r)(2+2k-r)(2k+r)} = \pi$$
This was derived assuming r rational, but I don't think it matters.
