Gamma function integral inequality

Question

In appendix 1 of the following paper, I found \begin{align} \int_s^\infty \exp(-\frac{1}{2}z^\beta)dz&<\frac{1}{\beta}2^{1/\beta}\Gamma\left(\frac{1}{\beta}\right) \end{align} however I can't figure out how they got that inequality. I tried writing out the gamma function as an integral and taking logs of both sides, but that got me nowhere.

Answer 1

With $u=\frac {z^\beta}2$, we have $dz= \frac 1 \beta 2^{\frac 1 \beta}u^{\frac 1 \beta -1}du$ $$\int_s^\infty \exp(-\frac{1}{2}z^\beta)dz=\frac{2^{\frac 1 \beta}}{\beta}\int_{2^{\frac 1 \beta}s^{\frac 1\beta}}^{+\infty}e^{-u}u^{\frac 1 \beta -1}du\leq \frac{2^{\frac 1 \beta}}{\beta}\int_{0}^{+\infty}e^{-u}u^{\frac 1 \beta -1}du=\frac{2^{\frac 1 \beta}}{\beta}\Gamma\left(\frac 1 \beta\right) $$

Answer 2

K. defaoite's comment is correct. In more detail, let $x=z^\beta/2$. Then \begin{align} z&=(2x)^{1/\beta}\\ dx&=\frac{\beta z^{\beta-1}}{2}dz\\ &=\frac{1}{\beta}2^{1/\beta}x^{1/\beta-1}dz\\ dz&=\frac{2}{\beta z^{\beta-1}}dx \end{align} Doing integration by substitution one has that \begin{align} \int_0^{\infty}\exp(-z^\beta/2)dz&=\int_0^\infty \exp(-x)\frac{1}{\beta}2^{1/\beta}x^{1/\beta-1}dx\\ &=\frac{1}{\beta}2^{1/\beta}\Gamma\left(\frac{1}{\beta}\right) \end{align} which is the right hand side of the inequality. It then follows immediately.