# 009B Sample Final 3, Problem 7

Does the following integral converge or diverge? Prove your answer!

${\displaystyle \int _{1}^{\infty }{\frac {\sin ^{2}(x)}{x^{3}}}~dx}$
Foundations:
Direct Comparison Test for Improper Integrals
Let  ${\displaystyle f}$  and  ${\displaystyle g}$  be continuous on  ${\displaystyle [a,\infty )}$
where  ${\displaystyle 0\leq f(x)\leq g(x)}$  for all  ${\displaystyle x}$  in  ${\displaystyle [a,\infty ).}$
1.  If  ${\displaystyle \int _{a}^{\infty }g(x)~dx}$  converges, then  ${\displaystyle \int _{a}^{\infty }f(x)~dx}$  converges.
2.  If  ${\displaystyle \int _{a}^{\infty }f(x)~dx}$  diverges, then  ${\displaystyle \int _{a}^{\infty }g(x)~dx}$  diverges.

Solution:

Step 1:
We use the Direct Comparison Test for Improper Integrals.
For all  ${\displaystyle x}$  in  ${\displaystyle [1,\infty ),}$
${\displaystyle 0\leq {\frac {\sin ^{2}(x)}{x^{3}}}\leq {\frac {1}{x^{3}}}.}$
Also,
${\displaystyle {\frac {\sin ^{2}(x)}{x^{3}}}}$  and  ${\displaystyle {\frac {1}{x^{3}}}}$
are continuous on  ${\displaystyle [1,\infty ).}$
Step 2:
Now, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {\int _{1}^{\infty }{\frac {1}{x^{3}}}~dx}&=&\displaystyle {\lim _{a\rightarrow \infty }\int _{1}^{a}{\frac {1}{x^{3}}}~dx}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {1}{-2x^{2}}}{\bigg |}_{1}^{a}}\\&&\\&=&\displaystyle {\lim _{a\rightarrow \infty }{\frac {1}{-2a^{2}}}+{\frac {1}{2}}}\\&&\\&=&\displaystyle {{\frac {1}{2}}.}\end{array}}}$
Since  ${\displaystyle \int _{1}^{\infty }{\frac {1}{x^{3}}}~dx}$  converges,
${\displaystyle \int _{1}^{\infty }{\frac {\sin ^{2}(x)}{x^{3}}}~dx}$
converges by the Direct Comparison Test for Improper Integrals.