How integrals are convergent? ∫0∞sin⁡17xdx

Question

How integrals are convergent?  ∫0∞sin⁡17xdx

Sadie Eaton · Accepted Answer

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}^{\infty} \sin 17 x d x = lim_{t \to \infty} \int_{0}^{t} \sin 17 x d x

lim_{t \to \infty} \int_{0}^{t} \sin 17 x d x = lim_{t \to \infty} [\frac{1}{17} (- \cos (17 t) + 1)]

=limit does not exist
Thus,

\int_{0}^{\infty} \sin 17 x d x

is not convergent.

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

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