<s>[INST] Find the value of $A/B$ for this definite integration. $$\int_{0}^{\infty}e^{-\sqrt 3 x^{\frac{2}{3}}}\sin(x^{\frac{2}{3}})\mathrm{d}x$$
Given that the integral above is equal to $\frac{3\pi^A}{B}$ for rational numbers $A$ and $B$, find the value of $A/B$.
I thought of taking $x^{2/3}$ as t but that didn't help. Any other suggestion?
 [/INST] Letting $t=x^{\frac13}$ and noting 
$$ (-\frac{1}{8} e^{-\sqrt{3} t^2} (\sqrt{3} \sin(t^2)+\cos(t^2)))'=te^{-\sqrt 3 t^2}\sin(t^2)$$
then we have
\begin{eqnarray}
&&\int_{0}^{\infty}e^{-\sqrt 3 x^{\frac{2}{3}}}\sin(x^{\frac{2}{3}})\mathrm{d}x\\
&=&3\int_{0}^{\infty}e^{-\sqrt 3 t^2}\sin(t^2)t^2\mathrm{d}t\\
&=&-\frac{3}{8}\int_{0}^{\infty}td(e^{-\sqrt{3} t^2} (\sqrt{3} \sin(t^2)+\cos(t^2)))\\
&=&\frac{3}{8} \int_{0}^{\infty}e^{-\sqrt{3} t^2} (\sqrt{3} \sin(t^2)+\cos(t^2)))dt.
\end{eqnarray}
Noting that
$$ \int_{0}^{\infty}e^{-\sqrt{3} t^2+it^2}=\int_{0}^{\infty}e^{-(\sqrt{3}-i) t^2}=\frac{\sqrt{\pi}}{2\sqrt{\sqrt{3}-i}}=\frac{\sqrt\pi}{2}\left(\frac{1}{4}(1+\sqrt{3})+i\frac{1}{4}(\sqrt{3}-1)\right) $$
we have
$$ \int_{0}^{\infty}e^{-\sqrt{3} t^2}\cos(t^2)dt=\frac{\sqrt\pi}{8}(1+\sqrt{3}), \int_{0}^{\infty}e^{-\sqrt{3} t^2}\sin(t^2)dt=\frac{\sqrt\pi}{8}(\sqrt{3}-1).$$
Thus
\begin{eqnarray}
&&\int_{0}^{\infty}e^{-\sqrt 3 x^{\frac{2}{3}}}\sin(x^{\frac{2}{3}})\mathrm{d}x\\
&=&\frac{3}{8}\left(\sqrt{3} \frac{\sqrt\pi}{8}(\sqrt{3}-1)+\frac{\sqrt\pi}{8}(1+\sqrt{3}))\right)\\
&=&\frac{3\sqrt\pi}{16}.
\end{eqnarray}
</s>