# 009B Sample Final 3, Problem 7

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

Solution:

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