if we have
$ \begin{align} \mathcal{L}_{I_r}(s)&=\mathbb{E}_{\Phi,\{{g_i}\}}\bigg[\prod\limits_{i\in\Phi}\mathbb{E}_{g_i}(\mathrm{exp}[-sg_iR_i^{-\alpha}]) \bigg]\\ &=\mathbb{E}_{\Phi}\bigg[\prod\limits_{i\in\Phi}\frac{\mu}{\mu+sR_i^{-\alpha}} \bigg]\\ &=\mathrm{exp}\bigg(-2\pi\lambda\displaystyle\int_r^{\infty} \bigg(1-\frac{\mu}{\mu+sv^{-\alpha}}\bigg)v\mathrm{d}v \bigg) \end{align}$
where we can assume that $g_i\sim \mathrm{exp}(\mu)$, and plugging $s=\mu Tr^{\alpha}$ which gives
$\begin{align} \mathcal{L}_{I_r}(\mu Tr^{\alpha})&=\mathrm{exp}\bigg(-2\pi\lambda\displaystyle\int_r^{\infty} \frac{T}{T+(v/r)^{\alpha}}v\mathrm{d}v \bigg) \end{align}$
employing change of variables $u=\bigg(\frac{v}{rT^{\frac{1}{\alpha}}}\bigg)^2$ results in
$\mathcal{L}_{I_r}(\mu Tr^{\alpha})=\mathrm{exp}(-\pi r^2 \lambda\rho(T,\alpha))$, where
$\rho(T,\alpha)=T^{2/\alpha}\displaystyle\int_{T^{-2/\alpha}}^{\infty}\frac{1}{1+u^{\alpha/2}}\mathrm{d}u$
If we rewrite
$\mathcal{L}_{I_r}(\mu Tr^{\alpha})=\exp\bigg(-2\pi\lambda\displaystyle\int_r^{\infty}\frac{1}{1+\bigg( \frac{v}{r}\bigg)^{\alpha}\frac{1}{T}} v\mathrm{d}v\bigg)$ with change of variables $u=\bigg(\frac{v}{rT^{\frac{1}{\alpha}}}\bigg)^2$
we can get $u^{\alpha/2}=\bigg(\frac{v}{r}\bigg)^{\alpha}\frac{1}{T}$, then
$\mathcal{L}_{I_r}(\mu Tr^{\alpha})=\exp\bigg(-2\pi\lambda\displaystyle\int_r^{\infty}\frac{1}{1+u^{\alpha/2}} v\mathrm{d}v\bigg)$
write
$v=rT^{-1/\alpha}u^{1/2}$
$\mathrm{d}v=\frac{1}{2}rT^{-1/\alpha}u^{-1/\alpha}\mathrm{d}u$ then
$\mathcal{L}_{I_r}(\mu Tr^{\alpha})=\exp\bigg(-2\pi\lambda\displaystyle\int_r^{\infty}\frac{1}{1+u^{\alpha/2}} rT^{1/\alpha}u^{1/2}\frac{rT^{1/\alpha}}{2\sqrt{u}}\mathrm{d}u\bigg)$
$\mathcal{L}_{I_r}(\mu Tr^{\alpha})=\exp\bigg(-\pi\lambda r^2 T^{2/\alpha}\displaystyle\int_r^{\infty}\frac{1}{1+u^{\alpha/2}} \mathrm{d}u\bigg)$
How to change the integral limits from $\displaystyle\int_r^{\infty}$ to $\displaystyle\int_{T^{-2/\alpha}}^{\infty}$