I wish to find the distribution of $Y=\sum_{n=1}^{N}\frac{1}{X_{n}^{\alpha}}$, where $\alpha$ is a positive number, and $X_n$ has following distribution: \begin{equation} f_{X_n}(x)=\frac{2x}{R^2}, \ \ \ H\leq x\leq \sqrt{H^2+R^2}, \end{equation} $H$ and $R$ are positive real numbers.
My initial idea is to first find the distribution of $\frac{1}{X_{n}^{\alpha}}$, which is \begin{equation} f_{\frac{1}{X_{n}^{\alpha}}}(z)=\frac{2}{\alpha R^2}z^{-\frac{2}{\alpha}-1}, \ \ \ (H^2+R^2)^{-\alpha/2}\leq z\leq H^{\alpha}. \end{equation} Then calculate the Laplace transformation of $\frac{1}{X_{n}^{\alpha}}$, which is \begin{equation} \begin{aligned} \mathcal{L}_{\frac{1}{X_{n}^{\alpha}}}(s)&=\frac{2}{\alpha R^2}\int_{(H^2+R^2)^{-\alpha/2}}^{H^{\alpha}}e^{-sz}z^{-\frac{2}{\alpha}-1}\, dz \\ &=\frac{2}{\alpha R^2}(H^2+R^2)\text{E}_{\frac{2}{\alpha}+1}(s(H^2+R^2)^{-\alpha/2})-\frac{2H^2}{\alpha R^2}\text{E}_{\frac{2}{\alpha}+1}(sH^2), \end{aligned} \end{equation} where $\text{E}_{(\cdot)}(\cdot)$ is exponential integral. Normally we just calculate $\mathcal{L}_{Y}(s)=(\mathcal{L}_{\frac{1}{X_{n}^{\alpha}}}(s))^N$, then perform inverse Laplace transformation on $\mathcal{L}_{Y}(s)$, and we can get the distribution of $Y$. But to the best of my knowledge, there exists no inverse Laplace transformation to exponential integral.
Is there any other way to get the distribution of $Y=\sum_{n=1}^{N}\frac{1}{X_{n}^{\alpha}}$? Thanks.