Question

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

Applications of integrals
How integrals are convergent?
$$\displaystyle{\int_{{{0}}}^{{\infty}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}$$

2020-11-13
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.