This equation is in https://ieeexplore.ieee.org/document/7413975
In this equation (10) and (11), $h_x$ is Rayleigh fading power. But I don't know why the (10) is transformed to (11). $\mathbb{E}$ is the expected value. I think other symbols do not seem to have any effect to the transformation.
Anyone to answer me?
Thank you.
\begin{align*}& \mathcal{L}_{I_{X}}(s)\notag\\ & =\mathbb{E}[\exp(-sI_{X})] \tag{8}\\ & =\mathbb{E}\left[\prod_{x\in\Phi_{X}}\exp(-sPh_{x}(A\vert x_{\text{rx}}-x\vert)^{-\alpha})\right] \tag{9}\\ & \overset{(a)}{=}\mathbb{E}_{\Phi_{X}}\left[\prod_{x\in\Phi_{X}}\mathbb{E}_{h_{x}}\{\exp(-sPh_{x}(A\vert x_{\text{rx}}-x\vert)^{-\alpha})\}\right] \tag{10}\\ & =\mathbb{E}_{\Phi_{X}}\left[\prod_{x\in\Phi_{X}}\frac{1}{1+sP(A\vert x_{\text{rx}}-x\vert)^{-\alpha}}\right] \tag{11}\\ & \overset{(b)}{=}\exp\left(-p\lambda_{X}\int_{-\infty}^{+\infty}\frac{1}{1+(A\vert x_{\text{rx}}-x\vert)^{\alpha}/sP}\text{d}x\right) \tag{12}\\ & \overset{(c)}{=}\exp\left(-p\lambda_{X}(sP)^{1/\alpha}\frac{2}{A}\int_{0}^{+\infty}\frac{1}{1+u^{\alpha}}\text{d}u\right) \tag{13}\\ & =\exp\left(-p\lambda_{X}(sP)^{1/\alpha}\frac{2}{A}\pi/\alpha\csc(\pi/\alpha)\right) \tag{14}\end{align*}