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:
$$I = \int_0^\infty \frac{\sin^4 x}{x^3} \, dx $$
Related problem. Recalling the Laplace transform of a function $f(x)$
$$ F(s)=\int_{0}^{\infty} f(x)e^{-sx}dx .$$
Let
$$ G(u)=\frac{1}{u^3} \implies g(u)=\frac{u^2}{2!}, $$
and
$$ f(u)= \sin(u)^4 \implies F(u)=\frac{24}{u(u^2+4)(u^2+16)}. $$
Now,
$$ \int_0^\infty \frac{\sin^4 x}{x^3} \, dx= \frac{24}{2}\int_0^\infty \frac{u^2}{u(u^2+4)(u^2+16)} \, dx = \ln(2)$$
Note:
1) To evaluate the last integral, you can write the integrand as
$$\frac{u}{(u^2+4)(u^2+16)}= \frac{1}{12}{\frac {u}{{u}^{2}+4}}-\frac{1}{12}{\frac {u}{{u}^{2}+16}} $$
2) To find Laplace transform of $\sin^4(x)$, first, write
$$ \sin^4(x) = \frac{1}{(2 i)^4} ( e^{i x}-e^{- i x} )^{4}, $$
then use the binomial theorem to expand the above expression, and finally use the Laplace transform of $e^{ax}$
$$ F(s)=\frac{1}{s-a}. $$
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
$$\begin{align}\int_0^{\infty} du \, F(u) g(u) &= \lim_{R \to \infty} \left (\int_0^R du\, \frac{u}{u^2+4} - \int_0^R du\, \frac{u}{u^2+16} \right ) \\ &= \frac12 \lim_{R \to \infty}\left [ \log{(R^2+4)} - \log{4} - \log{(R^2+16)} + \log{16}\right]\\ &= \frac12 [\log{2^4} - \log{2^2} ] \\ &= \log{2}\end{align}$$
