How integrals are convergent? int_(0)^(oo)sin17x dx

Step 1
Since there are multiple questions, so we will be answering only the first one.
To deal with such type of Improper Integrals, we will replace the infinity with a variable (usually t), do the integral and then take the limit of the result as t goes to infinity.
We will call these integrals convergent if the associated limit exists and is a finite number (i.e. it’s not plus or minus infinity) and divergent if the associated limit either doesn’t exist or is (plus or minus) infinity.
${\displaystyle {\int }_0^{\mathrm{\infty }}\sin 17xdx=\underset{t\to \mathrm{\infty }}{lim}{\int }_0^t\sin 17xdx}$
${\displaystyle \underset{t\to \mathrm{\infty }}{lim}{\int }_0^t\sin 17xdx=\underset{t\to \mathrm{\infty }}{lim}\left[\frac{1}{17}(-\cos \left(17t\right)+1)\right]}$
=limit does not exist
Thus, ${\displaystyle {\int }_0^{\mathrm{\infty }}\sin 17xdx}$ is not convergent.