How do you test for convergence: #int (((sin^2)x)/(1+x^2)) dx#?

Question

Ezekiel Alexander · Accepted Answer

Since #0<=\frac{\sin^{2}(x)}{1+x^[2}}<=\frac{1}{1+x^{2}}# for all #x#, the improper integral #\int_{-\infty}^{\infty}\frac{\sin^{2}(x)}{1+x^[2}}dx# will converge if the improper integral #\int_{-\infty}^{\infty}\frac{1}{1+x^[2}}dx# converges. The fact that this last integral converges can be seen by direct calculation: #\int_{-\infty}^{\infty}\frac{1}{1+x^[2}}dx=lim_{b->\infty}\int_{-b}^{b}\frac{1}{1+x^{2}}dx# #=lim_{b->\infty}(arctan(x))|_{x=-b}^{x=b}# #= lim_{b->\infty}(arctan(b)-arctan(-b))=\frac{\pi}{2}-(-\frac{\pi}{2})=\pi# Therefore, #\int_{-\infty}^{\infty}\frac{\sin^{2}(x)}{1+x^[2}}dx# converges. Actually, as an extra-technical point, the integral #\int_{-\infty}^{\infty}\frac{\sin^{2}(x)}{1+x^[2}}dx# converges if both of the following integrals converge: #\int_{0}^{\infty}\frac{\sin^{2}(x)}{1+x^[2}}dx# and #\int_{-\infty}^{0}\frac{\sin^{2}(x)}{1+x^[2}}dx#. But these both converge by similar direct calculations.

Christopher Allers · Answer

undefined To test for the convergence of the integral 
$$ \int \frac{\sin^2(x)}{1+x^2} \, dx $$

We can use the Comparison Test or the Limit Comparison Test. Here, we will use the Comparison Test.

1. **Comparison Test**:

Since $ |\sin(x)| \leq 1 $ for all $ x $, we have:
$$ \frac{\sin^2(x)}{1+x^2} \leq \frac{1}{1+x^2} $$

Now, consider the integral:
$$ \int \frac{1}{1+x^2} \, dx $$

This integral is a standard integral which converges. Therefore, by the Comparison Test, if we can show that 
$$ \int \frac{1}{1+x^2} \, dx $$
converges, then 
$$ \int \frac{\sin^2(x)}{1+x^2} \, dx $$
also converges.

The integral $$ \int \frac{1}{1+x^2} \, dx $$ can be evaluated as:
$$ \int \frac{1}{1+x^2} \, dx = \arctan(x) + C $$

This integral converges as $ x $ approaches infinity.

Thus, by the Comparison Test, 
$$ \int \frac{\sin^2(x)}{1+x^2} \, dx $$
also converges.