009B Sample Final 3, Problem 7 Detailed Solution

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

$\int _{1}^{\infty }{\frac {\sin ^{2}(x)}{x^{3}}}~dx$

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Background Information:
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.

Final Answer:
converges (by the Direct Comparison Test for Improper Integrals)

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