The Convergence of the improper integral I would like to prove if the integral $$ \int_0^{\infty} \frac{\sin x}{x}$$ converges absolutely or not.
One way to do so, is to say:
$ \forall ~ x , | \sin x | \leq 1$, we have $$ \frac{|\sin x|}{x} \geq \frac{\sin^2 x}{x} = \frac{1- \cos (2x)}{2x} $$
then using the integration by parts $$ \int_0^{\infty} \frac{1- \cos (2x)}{2x}= \frac{1}{2} \ln x \Big|_0^t - \frac{1}{4} \frac{\sin (2x)}{x} \Big|_0^t - \frac{1}{4} \int_0^t \frac{\sin (2x)}{x^2}$$
First: why $\Big( \frac{1}{2} \ln x \Big|_0^t - \frac{1}{4} \frac{\sin (2x)}{x} \Big|_0^t - \frac{1}{4} \int_0^t \frac{\sin (2x)}{x^2} \Big)$ diverges?
Second: for the same question I found many answers on the site like this. Why they divide the interval $[0, \infty[$ as $$ \int_{\pi}^{(N+1)\pi}\left|\frac{\sin x}x\right|dx =\sum_{k=1}^N\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}x\right|dx$$
 A: I will be answering only the first question here. 
The main problem here is that in proving stuff you need analysis. You can't just use any theorem without knowing when it doesn't apply just because it seems 'natural'. 
Now in this case  you assumed the linearity of the Riemann integral
$$ \int_0^{t} \frac{1- \cos (2x)}{2x} dx=\int_0^{t} \frac{1}{2x}dx - \int_{0}^{t}\frac{\cos(2x)} {2x} dx$$
The only problem here is that the Riemann integral is only linear for functions that are Riemann integrable.In this case both integrands on the right side of the equation are not Riemann integrable over the integration domain hence the linearity property does not apply. So the divergence of right side does not imply the divergence of left side since the equality of left and right side doesn't hold. 
But we can modify this so that it is linear by taking the lower bound to be $a>0$  and hence 
$$ \int_a^{t} \frac{1- \cos (2x)}{2x} dx=\int_a^{t} \frac{1}{2x}dx - \int_{a}^{t}\frac{\cos(2x)} {2x} dx$$
Also note that the each of the terms on the right side individually  are discontinuous at a=0 but the sum is not  This  shows the non linearity of the limit in the general case as
$\lim_{x\to a} (f(x) +g(x)) =\lim_{x\to a}f(x) +\lim_{x\to a}g(x) $
Is only true if both f and g are continuous at a or in other word the  limit exists and is finite for  f and g.
If you want to use this way to prove that the improper integral 
$$ \int_0^{\infty} \frac{1- \cos (2x)}{2x} dx$$
converges then you would have to show that the double limit 
$$\lim_{a\to 0}\lim_{t\to\infty} (\int_a^{t} \frac{1}{2x}dx - \int_{a}^{t}\frac{\cos(2x)} {2x} dx)$$
Exists(more  strict  condition than for single limit as a double limit should be Invariant of the order of which you apply it or more generally the path you take in the a-t plane to get to the said limit points for it to exist) and is finite. 
A: Related to your first question, if you instead integrate from $1$ to $\infty,$ then the resulting expression looks like $$\frac{1}{2}\ln t + \text{ a convergent expression,}$$ so it diverges to $\infty$ as $t\to\infty.$
In your second question, that is done because the numerator $|\sin x|$ is periodic with period $\pi$ and for $x \in [k\pi, (k+1)\pi],$ we have $\frac{1}{x} \geq \frac{1}{(k+1)\pi}.$  Together, this allows us to use the known fact that the harmonic series diverges.  (If I had done it, I would have written it differently, but the main point still applies)
A: We have that
$$\int_0^{\infty} \frac{\sin x}{x}=\int_0^{1} \frac{\sin x}{x}+\int_1^{\infty} \frac{\sin x}{x}$$
and the first one is a proper integral while for the second by parts
$$\lim_{N\to \infty }\int_1^{N} \frac{\sin x}{x}=\lim_{N\to \infty } \left[\frac{-\cos x}x\right]_1^{N}- \lim_{N\to \infty }\int_1^{N}\frac{\cos x}{x^2}=\frac{\cos 1}1- \lim_{N\to \infty }\int_1^{N}\frac{\cos x}{x^2}$$
and $\int_1^{\infty}\frac{\cos x}{x^2}$ converge absolutely.
