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.
