I am trying to figure out whether the following integral is convergent or divergent:
$$\int_0^\infty \frac{\sin^2(x) }{(1 + x)^2} dx$$
At this point, I know that the above integral is equal to:
$$\lim_{t\rightarrow\infty}\int_0^t \frac{\sin^2(x) }{(1 + x)^2} dx$$
But I am not sure how to proceed (not sure how to integrate the function).
Best Answer
Finding an antiderivative of $\sin^2x\over (1+x)^2$ would not be easy, so we will use the comparison test for integrals with unbounded regions of integration. Since $\sin^2 x$ is nonnegative and bounded by $1$, $$0\le { \sin^2x\over(1+x)^2}\le {1\over(1+x)^2}.$$ The given integral converges if the integral $$ \int_0^\infty {1\over(1+x)^2}\,dx $$ converges.
Now $$ \eqalign{ \int {1\over(1+x)^2}\, dx\buildrel{u=1+x}\over{ =}\int {1\over u^2} \, du={-1 \over 1+x}+C. } $$
So $$\eqalign{ \int_0^\infty {1\over(1+x)^2}\, dx& =\lim_{b\rightarrow\infty}\int_0^b {1\over(1+x)^2}\, dx\cr &=\lim_{b\rightarrow\infty} {-1 \over 1+x}\Bigl|_0^b\cr &=0-(-1)\cr &=1. } $$
Thus $\int_0^\infty {1\over(1+x)^2}\, dx$ converges; and so, as mentioned above, $\int_0^\infty {\sin^2 x\over(1+x)^2}\, dx$ converges.