Intuitively, why does $\int_{-\infty}^{\infty}\sin(x)dx$ diverge? [duplicate]

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.

It has to do with how improper integrals of this type are defined. Normally, one defines (for continuous functions $f(x)$, for instance) $$\int_0^\infty f(x) \, dx = \lim_{b\to\infty} \int_0^b \! f(x)\, dx,$$ whenever that limit exists. A natural extension is to then define $$\int_{-\infty}^0 f(x) \, dx = \lim_{a\to\infty} \int_a^0 \! f(x)\, dx,$$ (or if you prefer, $\int_{-\infty}^0 \! f(x) \, dx = \int_0^\infty f(-x)\,dx$) whenever that limit exists.

Finally we define $\int_{-\infty}^{\infty}\! f(x)\, dx = \int_{-\infty}^0 \!f(x)\, dx + \int_0^{\infty} \! f(x)\, dx$, because we want all the typical rules of integration to hold for infinite integrals whenever they make sense. As a result, we see that these two limits are taken independently of one another.

What your intuition is currently pointing to is this:

$$\lim_{n\to\infty} \int_{-n}^n \sin(x) \, dx = 0$$ which is true, however if we adopted this definition for double-sided infinite integrals we would no longer be able to split them in the middle, for instance, and so we give up a different intuitive rule which, arguably, is more widely applicable.