$\int_{0}^{\infty }{{{x}^{n}}\sin \left( {{x}^{1/4}} \right)\exp \left( -{{x}^{1/4}} \right)dx}=0$ I'd like to show that for all positive integers $n$ we have
$$I\left( n \right)=\int_{0}^{\infty }{{{x}^{n}}\sin \left( {{x}^{1/4}} \right)\exp \left( -{{x}^{1/4}} \right)dx}=0.$$
This is true after some computer experiments, besides after setting $x={{u}^{4}}$ we get
$$I\left( n \right)=4\int_{0}^{\infty }{{{u}^{4n+3}}\sin \left( u \right)\exp \left( -u \right)du}.$$
But how to proceed? Integration by parts maybe?

Let 
$$I_n = \int_0^\infty x^n \sin (\sqrt[4]{x}) \exp (\sqrt[4]{x}) \, dx, \qquad n \in \mathbb{N}.$$
After enforcing a substitution of $x \mapsto \sqrt[4]{x}$ one has
$$I_n = 4 \int_0^\infty x^{4n + 3} e^{-x} \sin x \, dx.$$
The following useful property for the Laplace transform will now be employed to evalaute the integral:
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$
Noting that
$$\mathcal{L} \{\ x^{4n + 3} \sin x \}(t) = \frac{(4n + 3)!}{(1 + t^2)^{2n + 2}} \sin \left [4(n + 1) \tan^{-1} \left (\frac{1}{t} \right ) \right ],$$
and
$$\mathcal{L}^{-1} \{e^{-x} \} (t)= \delta (t - 1),$$
where $\delta (x)$ is the Dirac delta function, 
then
\begin{align}
I_n &= 4\int_0^\infty x^{4n + 3} \sin x \cdot e^{-x} \, dx\\
&= 4\int_0^\infty \mathcal{L} \{x^{4n + 3} \sin x\} (t) \cdot \mathcal{L}^{-1} \{e^{-x} \} (t) \, dt\\
&= 4(4n + 3)! \int_0^\infty \frac{1}{(1 + t^2)^{2n + 2}} \sin \left [4(n + 1) \tan^{-1} \left (\frac{1}{t} \right ) \right ] \cdot \delta (t - 1) \, dt\\
&= \frac{4(4n + 3)!}{2^{2n + 2}} \sin [4(n + 1) \tan^{-1} (1)]\\
&= \frac{(4n + 3)!}{4^n} \sin ((n + 1)\pi)\\
&= 0,
\end{align}
as required to show.

A direct approach
Here is an approach where the results of the above Laplace transform and its inverse are not quoted in advance. 
From 
$$I_n = 4 \int_0^\infty e^{-x} x^{4n + 3} \sin x  \, dx,$$
we re-write this as
$$I_n = -4 \operatorname{Im} \int_0^\infty x^{4n + 3} e^{-(1 + i)x} \, dx.$$
Integrating by parts $(4n + 3)$ times gives
\begin{align}
I_n &=-4 \, \operatorname{Im} \left [\frac{(-1)^{4n + 3} (4n + 3)!}{(1 + i)^{4n + 3}} \right ] \int_0^\infty e^{-(1 + i)x} \, dx\\
&= 4 \, \operatorname{Im} \left [\frac{(-1)^{4n + 4} (4n + 3)!}{(1 + i)^{4n + 4}} \right ]\\
&= 4(4n + 3)! \operatorname{Im} \left [\frac{1}{(1 + i)^{4n + 4}} \right ]\\
&= 0,
\end{align}
where the last line is due to the fact that
$$\frac{1}{(1 + i)^{4n + 4}} = \frac{1}{2^{2n + 2}} \left [\cos ((n + 1)\pi) + i \sin ((n + 1)\pi) \right ].$$
