Answer
Diverges
Work Step by Step
We need to use formula: $\int_0^{+\infty} \sin x dx=\lim\limits_{M \to +\infty}\int_0^{M} \sin x dx$
Therefore, $\lim\limits_{M \to +\infty}\int_0^{M} \sin x dx=\lim\limits_{M \to +\infty}[-\cos x ]_0^{M}-\lim\limits_{M \to +\infty}[\cos x ]_0^{M}$
Now, we will evaluate the limits .
$-\lim\limits_{M \to +\infty}[\cos M-1 ]=-(\cos \infty-1)$
We conclude that the limit does not exist. So, the improper integral diverges.
