Evaluate $\int_0^\infty \frac{e^{-x}\sin(x)}{\sqrt[3]{x} } dx$ I am having trouble with the following integral

Evaluate
$$\int_0^\infty \frac{e^{-x}\sin(x)}{\sqrt[3]{x} } dx$$

$$\int_0^\infty \frac{e^{-x}\sin(x)}{\sqrt[3]{x} } dx=\int_0^\infty 3ue^{-u^3}\sin(u^3)du$$
let $u=\sqrt[3]{x}\rightarrow dx=3x^{2/3}du$
How does one proceed from here? Thank you for your time.
 A: Let’s use the definition of the gamma function and the complex nature of the sine function:
$$\int_0^\infty \frac{\sin(x)}{e^x\sqrt[3] x}dx=\frac{i}{2}\int_0^\infty x^{-\frac13 } e^{-x}\left(e^{-ix}-e^{ix}\right)dx=\frac i2\int_0^\infty x^{-\frac13} e^{-x(1+i)}dx-\frac i2\int_0^\infty x^{-\frac13}e^{x(i-1)}dx=\frac{i}{2}\left(\frac{Γ\left(\frac23\right)}{(1-i)^\frac23}-\frac{Γ\left(\frac23\right)}{(1+i)^\frac23}\right)=\frac{3\Gamma\left(\frac53\right)}
{4\sqrt[3]2}= \frac{3\Gamma\left(\frac53\right)}
{2^\frac73}=\frac{\pi}{\sqrt 3\sqrt[3] 2 \Gamma\left(\frac13\right)}= 0.53738206038361965917757281552415611212622673003414… $$
There are many other alternate forms, but this one is simplest. Please correct me and give me feedback!
A: You may use Laplace transformation which also uses separation of the real and imaginary part:
Let's consider this one:
$$F(s,q)=\int_0^\infty e^{-st}\sin t\, t^{q-1}dt=\Im\int_0^\infty e^{-t(s-i)}\sin t\, t^{q-1}dt$$
where $s\geqslant0$ and $q>0$. Where $q ∈ \mathbb Q$?
Which we know:
$$F(s,q)=\Im\int_0^\infty e^{-x}x^{q-1}\frac{dx}{(s-i)^q}=\Im\frac{(s+i)^q}{(s^2+1)^q}\int_0^\infty e^{-x}x^{q-1}dx$$
$$F(s,q)=\frac{\Gamma(q)}{(1+s^2)^{\frac{q}{2}}}\sin\Big(q\tan^{-1}\frac{1}{s}\Big)$$
in particular at $s = 1$ and $q = \frac{2}{3}$

