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

How integrals are convergent?
${\int }_{0}^{\mathrm{\infty }}\mathrm{sin}17xdx$
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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.
${\int }_{0}^{\mathrm{\infty }}\mathrm{sin}17xdx=\underset{t\to \mathrm{\infty }}{lim}{\int }_{0}^{t}\mathrm{sin}17xdx$
$\underset{t\to \mathrm{\infty }}{lim}{\int }_{0}^{t}\mathrm{sin}17xdx=\underset{t\to \mathrm{\infty }}{lim}\left[\frac{1}{17}\left(-\mathrm{cos}\left(17t\right)+1\right)\right]$
=limit does not exist
Thus, ${\int }_{0}^{\mathrm{\infty }}\mathrm{sin}17xdx$ is not convergent.