[Math] Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx $

Question

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:

$$I = \int_0^\infty \frac{\sin^4 x}{x^3} \, dx $$

Answer 1

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 $e^{ax}$

$$ F(s)=\frac{1}{s-a}. $$