$\int_{0}^{\infty }{{{x}^{n}}\sin \left( {{x}^{1/4}} \right)\exp \left( -{{x}^{1/4}} \right)dx}=0$

A Laplace transform is easier than integrating by parts. Consider the function $$ F_n(s) = \int_{0}^\infty t^{4n+3}e^{-st}dt = \frac{(4n+3)!}{s^{4n+4}}. $$ Then $I(n) = \mathrm{Im}[F_n(1-i)]$. Since $$ F_n(1-i) = \frac{(4n+3)!}{(1-i)^{4n+4}} = \frac{(4n+3)!}{(-4)^{n+1}} \in \mathbb R, $$ $I(n) = 0$.

First define for any integer $n$

$$ {{A}_{n}}=\int_{0}^{\infty }{{{x}^{n}}{{e}^{-x}}\sin \left( x \right)dx}\quad and\quad {{B}_{n}}=\int_{0}^{\infty }{{{x}^{n}}{{e}^{-x}}\cos \left( x \right)dx} $$ Now the substitution $u={{x}^{1/4}}$ reduces the integral in question to $4{{A}_{4n+3}}$.

Using integration by parts any one can verify the recurrence relation: $$ \left\{ \begin{align} & {{A}_{n}}=\frac{n}{2}\left( {{A}_{n-1}}+{{B}_{n-1}} \right) \\ & {{B}_{n}}=\frac{n}{2}\left( {{B}_{n-1}}-{{A}_{n-1}} \right) \\ \end{align} \right. $$ Solving this with initial conditions: ${{A}_{0}}=1/2\quad and\quad {{B}_{0}}=1/2$ you get ${{A}_{n}}=0$ for any integer $n$ such that $n\equiv 3\left( \bmod 4 \right)$ which means that $4{{A}_{4n+3}}=0$.

Using contour integration, with the contour being a 'pie slice', composed of lines $$z_1(t) = t,\qquad t\in[0,R]$$ $$z_2(t) = t(1-i),\qquad t\in[0,\frac{R}{\sqrt{2}}]$$ $$z_3(t) = Re^{it},\qquad t\in[-\frac\pi 4,0] $$ and consideringing $$ \oint_C z^{4n+3} e^{-z} dz = 0$$ you can prove that $$ (1-i)^{4n+4}\int_0^\infty u^{4n+3} e^{-u(1-i)} du = \int_0^\infty u^{4n+3} e^{-u} du $$ so $$ \int_0^\infty u^{4n+3} e^{-u(1-i)} du = \frac{1}{(-4)^{n+1}} \int_0^\infty u^{4n+3} e^{-u} du = \frac{(4n+3)!}{(-4)^{n+1}} \in \mathbb R$$ Therefore $$I(n) = {\rm Im} \Big(\int_0^\infty 4u^{4n+3} e^{iu} e^{-u} du \Big) = {\rm Im} \Big(4\int_0^\infty u^{4n+3} e^{-u(1-i)} du \Big) = 0$$