I would like to understand whether this improper integral
$$\int_{-\infty}^{\infty} \frac{\sin^2 x}{x}dx$$
converges or diverges.
On one hand, if I fix $\alpha>0$, I note that
$$\int_{-\alpha}^{\alpha} \frac{\sin^2 x}{x}dx=0$$
since the integrand is odd, and the integration interval is symmetric with respect to the origin. On the other hand, however, Mathematica, when given the command
Integrate[x*(Sin[x]/x)^2, {x, -[Infinity], [Infinity]}],
responds that the integral does not converge on $(-\infty, \infty)$. Does anyone have any suggestions to offer?
Best Answer
By definition, for any function $f$ the improper integral $\displaystyle{\int_{-\infty}^{+\infty}} f(x)\ dx$ converges iff both $\displaystyle{\int_{-\infty}^{c} f(x)\ dx}$ and $\displaystyle{\int_{c}^{+\infty}f(x)\ dx}$ converge for some number $c$. That is not the case with $f(x) = \dfrac{\sin^2(x)}{x}$ (see this question).