Improper integrals: convergent or divergent? 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).
 A: 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.
A: I am sure what you are interested in is whether the given integral has a finite value. Consider a similar expression for the cosine function and find their sum. So you have;
$\int_0^\infty \frac{\sin^2x}{(1+x)^2}dx+\int_0^\infty \frac{\cos^2x}{(1+x)^2}dx=\int_0^\infty \frac{1}{(1+x)^2}dx$. Evaluating the last integral gives;
$\lim_{t\rightarrow \infty} \int_0^t \frac{1}{(1+x)^2}dx=\lim_{t\rightarrow \infty}[1-\frac{1}{1+t}]=1$, which is finite number. Hence, the sum converges and so (since everything is non-negative) each integral must converge.
In other words the sum has a finite value and each integrand is non-negative so each integral must have a finite value.
