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?
 A: Sub $x=u^3$; then the integral becomes
$$3 \int_0^{\infty} du \, u^2 \, e^{-\sqrt{3} u^2} \sin{u^2} $$
This integral is quite doable, and we can apply a number of shortcuts to get to the result more quickly.  First, consider the integral
$$I(a) = \int_0^{\infty} du \,  e^{-a u^2} e^{i u^2} $$
where $a \gt 0$, which we may show through Cauchy's theorem via extension to the complex plane is
$$I(a) = \sqrt{\frac{\pi}{a-i}}  = \frac12 \sqrt{\pi} (a^2+1)^{-1/4} e^{i (1/2) \arctan{(1/a)}}$$
The result we seek is  $-3 \operatorname{Im}{I'(\sqrt{3})} $.  So let's start with
$$I'(\sqrt{3}) = -\frac14 \sqrt{\pi} (\sqrt{3}-i)^{-3/2} = -\frac14 \sqrt{\pi} 2^{-3/2} e^{i (3/2) \arctan{(1/\sqrt{3})}}$$
Thus the integral is three times the negative imaginary part of this quantity, or

$$\int_0^{\infty} dx \, e^{-\sqrt{3} x^{2/3}} \sin{x^{2/3}} = \frac{3\sqrt{\pi}}{16} $$

so your $A=1/2$ and $B=16$, so $A/B=1/32$.
A: 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}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\expo{-\root{3}
x^{2/3}}\sin\pars{x^{2/3}}\,\dd x}
\,\,\,\stackrel{x^{2/3}\,\,\, \mapsto\ x}{=}\,\,\,
{3 \over 2}\int_{0}^{\infty}x^{1/2}\,\expo{-\root{3}
x}\sin\pars{x}\,\dd x
\\[5mm] = &\
{3 \over 2}\,\Im\int_{0}^{\infty}x^{1/2}\,
\expo{-\pars{\root{3} - \ic}x}\,\,\dd x
\\[5mm] = &\
{3 \over 2}\,\Im\int_{0}^{\infty}x^{\color{red}{3/2} - 1}\,
\bracks{\sum_{k = 0}^{\infty}\color{red}{\pars{\root{3} - \ic}^{k}}\,
{\pars{-x}^{k} \over k!}}\,\,\dd x
\end{align}
With Ramanujan's Master Theorem:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}\expo{-\root{3}
x^{2/3}}\sin\pars{x^{2/3}}\,\dd x} =
{3 \over 2}\,\Im\bracks{\Gamma\pars{3 \over 2}\pars{\root{3} - \ic}^{-3/2}}
\\[5mm] = &\
{3\root{\pi} \over 4}\,\Im\bracks{2\expo{-\ic\pi/6}}^{-3/2} =
{3\root{\pi} \over 4}\,2^{-3/2}\,\sin\pars{\pi \over 4}
\\[5mm] = &\
\bbx{3\root{\pi} \over 16} \approx 0.3323 \\ &
\end{align}
