<s>[INST] Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx $ I have a trouble with a integral:
Using this Laplace trasform equation:
$$\begin{align}
 \int_0^\infty F(u)g(u) \, du  & = \int_0^\infty f(u)G(u) \, du \\[6pt]
 L[f(t)] & = F(s) \\[6pt]
 L[g(t)] & = G(s)
 \end{align}$$
Applying to compute this integral: 
$$I = \int_0^\infty \frac{\sin^4 x}{x^3} \, dx $$
 [/INST] Here is how we compute the LT of $\sin^4{t}$ directly.  As mentioned in @Mhenni's solution, write as $\sin^4{t} = (e^{i t}-e^{-i t})^4/(2 i)^4$
$$F(u) = \frac{1}{16} \int_0^{\infty} dt \, (e^{i 4 t} - 4 e^{i 2 t} + 6 - 4 e^{-i 2 t} + e^{-i 4 t}) e^{-u t}$$
Evaluate separately and combine judiciously:
$$F(u) = \frac18 \left [ \frac{u}{u^2+16} - 4 \frac{u}{u^2+4} + \frac{3}{u}\right]$$
As also mentioned above, $g(u) = u^2/2$ (This is a direct look-up in the table, or it may be evaluated using residue theory.)  Now, when we multiply this through, it looks like the integral will be divergent.  However, we may cancel out the $u$ piece by using the fact that $u^2/(u^2+a^2) = (1/a^2) (1-1/(u^2+a^2))$.  We then get
$$\begin{align}F(u) g(u) &= \frac{1}{16} \frac{u^3}{u^2+16} - \frac14 \frac{u^3}{u^2+4} + \frac{3}{16} u \\ &= \frac{u}{16} - \frac{u}{u^2+16} - \frac{u}{4} + \frac{u}{u^2+4}+\frac{3}{16}u\\ &= \frac{u}{u^2+4} - \frac{u}{u^2+16} \end{align}$$
To do the integral, note that we need to be careful because, taken individually, the integrals diverge - but the divergences cancel.  Thus write
$$\begin{align}\int_0^{\infty} du \, F(u) g(u) &= \lim_{R \to \infty} \left (\int_0^R du\, \frac{u}{u^2+4} - \int_0^R du\, \frac{u}{u^2+16}  \right ) \\ &= \frac12 \lim_{R \to \infty}\left [ \log{(R^2+4)} - \log{4} - \log{(R^2+16)} + \log{16}\right]\\ &= \frac12 [\log{2^4} - \log{2^2} ] \\ &= \log{2}\end{align}$$
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