I have to get to the second part of the ecuation with laplace and I don't know how to do it step by step, help please! thanks!!
$$\int\limits_{-\infty }^{+\infty }{\left( \frac{\sin x}{x} \right)^{3}dx}=2\int\limits_{0}^{+\infty }{\frac{x^{2}}{2}\cdot \frac{6}{\left( x^{2}+1 \right)\left( x^{2}+9 \right)}dx}$$
Keep in mind that the Laplace transform takes the following form:
$$F(s)= \int_{0}^{\infty} f(x) e^{-sx}dx. $$
Let us make use of the following set of properties :
$$ \begin{align} \int_0^\infty F(u)g(u) \, du & = \int_0^\infty f(u)G(u) \, du \\[6pt] L[f(t)] & = F(s) \\[6pt] L[g(t)] & = G(s)\end{align} $$
Let
$$ G(x)=\frac{1}{x^3} \implies g(x)=\frac{x^2}{2}, $$
so
$$ f(x)= \sin(x)^3 \implies F(x) = {\frac {6}{ \left( {x}^{2}+1 \right) \left( {x}^{2}+9 \right) }}. $$
Putting those together and considering our function is even, we get:
$$\int_{-\infty}^\infty \frac{\sin^3 x}{x^3}= 2\int_0^\infty \frac{\sin^3 x}{x^3} \, dx = 2\int\limits_{0}^{+\infty }{\frac{x^{2}}{2}\cdot \frac{6}{\left( x^{2}+1 \right)\left( x^{2}+9 \right)}dx} $$.