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.

\(\displaystyle{\int_{{{0}}}^{{\infty}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}=\lim_{{{t}\rightarrow\infty}}{\int_{{0}}^{{t}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}\)

\(\displaystyle\lim_{{{t}\rightarrow\infty}}{\int_{{0}}^{{t}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}=\lim_{{{t}\rightarrow\infty}}{\left[\frac{{1}}{{17}}{\left(-{\cos{{\left({17}{t}\right)}}}+{1}\right)}\right]}\)

=limit does not exist

Thus, \(\displaystyle{\int_{{0}}^{{\infty}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}\) is not convergent.

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.

\(\displaystyle{\int_{{{0}}}^{{\infty}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}=\lim_{{{t}\rightarrow\infty}}{\int_{{0}}^{{t}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}\)

\(\displaystyle\lim_{{{t}\rightarrow\infty}}{\int_{{0}}^{{t}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}=\lim_{{{t}\rightarrow\infty}}{\left[\frac{{1}}{{17}}{\left(-{\cos{{\left({17}{t}\right)}}}+{1}\right)}\right]}\)

=limit does not exist

Thus, \(\displaystyle{\int_{{0}}^{{\infty}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}\) is not convergent.