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

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

Question
Applications of integrals
asked 2020-11-12
How integrals are convergent?
\(\displaystyle{\int_{{{0}}}^{{\infty}}}{\sin{{17}}}{x}{\left.{d}{x}\right.}\)

Answers (1)

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.
0

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