I have come across another improper integral I wish to evaluate via residues.
The integral is:
$$\int_{-\infty}^\infty{\frac{\sin(x)^2}{x^2}}dx$$
$\sin(z)$ behaves in an uneasy way so I tried using the function $\frac{{e^{iz}}^2}{z^2}$ with a half circle on the upper complex plane with radius R and a half-circle of radius 1/R which arcs below $0$.
The problem is the small semi-circles integral does not go to $0$ and in fact doesn't exist.
What other types of contours or function substitutions should be used here?
Best Answer
Note that $ \cos(2x)=1-2\sin(x)^2 $, this suggest to consider the integral
$$ \int_{C} \frac{ {\rm e}^{2 i z} - 1 }{ z^2} dz \,.$$