Intuitively, why does $\int_{-\infty}^{\infty}\sin(x)dx$ diverge? According to Wolfram Alpha, $\int_{-\infty}^{\infty}\sin(x)dx$ does not converge.

This makes no sense to me, intuitively, which I'll justify with a plot:



As we see, the positive and negative areas 'cancel out', so, for any $\alpha \in \mathbb{R}$, $\int_{-\alpha}^{\alpha}\sin(x)dx=0$ (I'm just thinking geometrically- in no way is this supposed to be a rigourous justification).

So, why is is that $\int_{-\infty}^{\infty}\sin(x)dx$ diverges?

A natural and intuitive reason, more than a rigourous one, would be best.

Thanks
The integral is
$$
\lim_{a \to \infty}\lim_{b \to \infty}\int_{-a}^b \sin x \, d x
$$
If you keep one fixed and vary the other limit of integration, there is no limit, the integral fluctuates between $-1$ and $1$. The value doesn't have to blow up for there not be a limit.
Since we know that $\sin x$ is a odd function. And we also know from the property of definite integral that if a function is odd then 
limit from (-a) tends to (+a) integral f(x) dx = 0
and we know that sin x is a odd function 
therefore 
limit from (-infinite ) tends to (+infinite ) 
Integral sin x dx = 0