Question about $\frac{\sin(x)}{x}$ and $\frac{\cos(x)}{x}$ So here is my question. 
As known the famous integral 
$$
\int_0^{\infty} \frac{\sin(x)}{x}dx$$
converges an its value is $\frac{\pi}{2}$.
As I was trying to solve a different integral today, after rewriting and integrating by parts I ended up having the following integral on the paper,
$$\int_0^{\infty}\frac{\cos(x)}{x}dx
$$
After several times failling to solve it I "asked" Wolfram-Alpha and i got the answer that
$$\int_0^{\infty}\frac{\cos(x)}{x}dx=\infty
$$
which in my opinion was very surprising because as
$\int_0^{\infty} \frac{\sin(x)}{x}dx$ converges I was expecting so was $\int_0^{\infty}\frac{\cos(x)}{x}dx$. I have to admit that I even didnt manage to prove that it is divergent. Is there an intutitive explanation for $\int_0^{\infty}\frac{\cos(x)}{x}dx=\infty$? Or maybe if someone could provide the prove that it is divergent that will be already "intutitve" enough... 
I appreciate any answers.
Thanks in advance!
 A: You can gain some intuition by looking at the graph of $\sin(x)/x$ and $\cos(x)/x$ respectively and to see how it behaves. 

As you can see $\cos(x)/x$ has as an asymptote the $y$-axis.

A: We have
$$\frac{\cos x}{x}\sim_0\frac1x$$
hence the given integral is divergent.
A: $\cos0 = 1$, so $\cos x>1-\varepsilon$ for $-\delta<x<\delta$.  Therefore
$$
\int_0^\infty \frac{\cos x}{x}\,dx \ge\int_0^\delta \frac{1-\varepsilon}{x} \, dx = \infty.
$$
A: I tried to solve in this way but I couldn't come to conclusion
$$\int_0^\infty\frac{\cos x}{x} \, dx=\int_0^\infty\int_{-\infty}^x\frac{\cos x}{-y^2}\,dy\,dx\\ =\int_{-\infty}^0\int_0^\infty\frac{\cos x}{-y^2}\,dy\,dx+\int_0^\infty\int_y^\infty\frac{\cos x}{-y^2}\,dy\,dx\\ =\int_{-\infty}^0\left(\left.\frac{\sin x}{-y^2}\right|_0^\infty\right)\,dy+\int_0^\infty\left(\left.\frac{\sin x}{-y^2}\right|_y^\infty\right)\,dy\\ =\int_{-\infty}^0(\lim_{x\to-\infty}\frac{\sin x}{y^2})\,dy+\int_0^\infty(\lim_{x\to\infty}\frac{\sin x}{-y^2}-\frac{\sin y}{-y^2})\,dy\\ =\int_0^\infty(\lim_{x\to-\infty}\frac{\sin x}{y^2})\,dy+\int_0^\infty(\lim_{x\to\infty}\frac{\sin x}{-y^2}-\frac{\sin y}{-y^2})\,dy$$
