The Laplace transform can be evaluated as an integral of a product of two Meijer G-functions. For $\beta >0$, we obtain

$${\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-st}}{1+{t}^{\beta}}dt={\int}_{0}^{\mathrm{\infty}}{G}_{1,1}^{1,1}\left({t}^{\beta}\right|\genfrac{}{}{0ex}{}{0}{0}){G}_{0,1}^{1,0}\left(st\right|\genfrac{}{}{0ex}{}{-}{0})dt=\frac{1}{s}{H}_{2,1}^{1,2}\left({s}^{-\beta}\right|\genfrac{}{}{0ex}{}{(0,1),(0,\beta )}{(0,1)}).$$

For $\beta <0$

$${\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-st}}{1+{t}^{\beta}}dt={\int}_{0}^{\mathrm{\infty}}{G}_{1,1}^{1,1}\left({t}^{-\beta}\right|\genfrac{}{}{0ex}{}{1}{1}){G}_{0,1}^{1,0}\left(st\right|\genfrac{}{}{0ex}{}{-}{0})dt=\frac{1}{s}{H}_{2,1}^{1,2}\left({s}^{\beta}\right|\genfrac{}{}{0ex}{}{(1,1),(0,-\beta )}{(1,1)}),$$

which, incidentally, is the same as formally extending the first result to negative $\beta $.

The resulting Fox H-function can be converted to a G-function when $\beta $ is rational.

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