Using Laplace transform to find $\int_{0}^{\infty}\frac{\sin^2 t}{t^2}dt$

Question

Find
$$\int_{0}^{\infty}\frac{\sin^2t}{t^2}dt$$ using Laplace transforms.

Answer 1

Let $f(s):=\int_0^\infty\frac{\sin^2 t}{t^2}e^{-st}dt$ so $$f^{\prime\prime}(s)=\int_0^\infty\sin^2 t\:e^{-st}dt=-\frac14\int_0^\infty(e^{-(s+2i)t}+e^{-(s-2i)t}-2e^{-st})dt=\frac12\left(\frac{1}{s}-\frac{s}{s^2+4}\right).$$Integrating viz. $f^{\prime}(\infty)=0$,$$f^{\prime}(s)=\frac14\ln\frac{s^2}{s^2+4}.$$Integrating again viz. $f(\infty)=0$, $$f(s)=\frac{s}{4}\ln\frac{s^2}{s^2+4}+\frac{\pi}{2}-\arctan\frac{s}{2}.$$Imposing $s\to 0^+$, the original integral is $\frac{\pi}{2}$.