I would like an explanation (justification) about the use of a laplace transform to solve the following integral. Excuse my poor English.
$I = \displaystyle \int_{-\infty}^{\infty} \frac{\sin^4(\frac{t}{2})}{t^2} \, dt. \tag*{}$ Using the substitution $x = \frac{t}{2}$, we obtain $\begin{align*} I &= \displaystyle \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^4{x}}{x^2} \, dx\\ &= \int_0^{\infty} \frac{\sin^4{x}}{x^2} \, dx, \text{ even integrand}. \end{align*} \tag*{}$
One way to proceed from here is to note that (via trig. identities) $\begin{align*} \displaystyle \mathcal{L} \{\sin^4{t}\} &= \frac{1}{8} \cdot \mathcal{L}\{3 - 4\cos(2x) + \cos(4x)\}\\ &= \frac{1}{8} \Big(\frac{3}{s} - \frac{4s}{s^2 + 2^2} + \frac{s}{s^2 + 4^2}\Big), \tag*{} \end{align*}$ while $\displaystyle \mathcal{L}^{-1} \Big\{\frac{1}{s^2}\Big\} = t. \tag*{}$ Then, we obtain $\begin{align*} I &= \displaystyle \int_0^{\infty} \sin^4{x} \cdot \frac{1}{x^2} \, dx\\ &= \int_0^{\infty} \mathcal{L} \{\sin^4{x}\} \cdot \mathcal{L}^{-1}\Big\{\frac{1}{x^2} \Big\} \, dx\\ &= \int_0^{\infty} \frac{1}{8} \Big(\frac{3}{x} - \frac{4x}{x^2 + 2^2} + \frac{x}{x^2 + 4^2}\Big) \cdot x \, dx\\ &= \int_0^{\infty} \Big(\frac{2}{x^2 + 4} - \frac{2}{x^2 + 16}\Big) \, dx\\ &= \Big[\arctan\Big(\frac{x}{2}\Big) - \frac{1}{2} \arctan\Big(\frac{x}{4}\Big)\Big] \Bigg|_0^{\infty}\\ &= \frac{\pi}{4}. \end{align*} \tag*{}$
Hence, we conclude that $\displaystyle \int_{-\infty}^{\infty} \frac{\sin^4(\frac{t}{2})}{t^2} \, dt = \frac{\pi}{4}. \tag*{}$