Since 0<=\frac{\sin^{2}(x)}{1+x^[2}}<=\frac{1}{1+x^{2}}0≤sin2(x)1+x2≤11+x2 for all xx, the improper integral \int_{-\infty}^{\infty}\frac{\sin^{2}(x)}{1+x^[2}}dx∫∞−∞sin2(x)1+x2dx will converge if the improper integral \int_{-\infty}^{\infty}\frac{1}{1+x^[2}}dx∫∞−∞11+x2dx 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.