Is $\int_{-\infty}^{\infty} \sin x \, \mathrm{dx}$ divergent or convergent?

Your first claim was correct: the limit does not exist. $t$ and $a$ are unrelated, so there's no good reason you should be able to set $t=a=b$ and take a limit. For $\int_{-\infty}^\infty \sin x dx$ to be defined, both $\int_{-\infty}^0 \sin x dx$ and $\int_{0}^{\infty} \sin x dx$ must exist: but as you saw, neither do.

What you calculated is instead called the Cauchy Principal Value; indeed, the Cauchy principal value of $$\int_{-\infty}^\infty \sin(x) dx$$ is $0$ (as it is for every odd function).

The problem with $\int_0^t\sin(x)dx $ is that this (as a function of $t$) oscillates around $0$. With each period of the integrand you first add, then remove the same amount indefinitely. Therefore it does not converge.

$\int \frac{1}{x}$ is different. Both $\int_0^1 \frac{1}{x}$ and $\int_1^\infty \frac{1}{x}$ diverge, without oscillating. The finite integral $\int_1^t \frac{1}{x}$ is, e.g, positive for all $t$.

Maybe it is helpful.

The inverse Fourier transformation of the Dirac-delta function is $$ \delta (t) = \frac{1}{2\pi} \int_{-\infty}^\infty \mathrm{e}^{\mathrm{i} \omega t} \mathrm{d} \omega $$

By letting $\omega = x$, it is shown that $$ \int_{-\infty }^\infty \sin x\mathrm{d} x= \Im(2\pi \delta (t=1)) = 0 $$