Inspired by Surya Prakash

Since $\int_0^\infty x^{\mu-1} e^{-\beta x} \sin \delta x\,dx \; = \; \frac{\Gamma(\mu)}{(\beta^2 + \delta^2)^{\frac12\mu}} \sin\left(\mu \tan^{-1}\tfrac{\delta}{\beta}\right)$ for $\mu > -1$ and $\beta > |\delta|$ , we see that (putting $x = y^{\frac32}$ ), $\begin{array}{rcl} \displaystyle\int_0^\infty e^{-\sqrt{3}x^{\frac23}} \sin\big(x^{\frac23}\big)\,dx& = & \displaystyle\tfrac32 \int_0^\infty y^{\frac12}\,e^{-\sqrt{3}y} \sin y \,dy \\ & = & \displaystyle\tfrac32 \frac{\Gamma(\frac32)}{(\sqrt{3}^2 + 1^2)^{\frac34}} \sin\left(\tfrac32\tan^{-1}\tfrac{1}{\sqrt{3}}\right) \\ & = & \displaystyle \tfrac32 \frac{\frac12\sqrt{\pi}}{2^{\frac32}} \sin\tfrac14\pi \\ & = & \tfrac{3}{16}\sqrt{\pi} \end{array}$