Evaluate $\int\limits_{x=\alpha}^{2} x e^{-a x^2}\frac{p/x^2+c}{(p/x^2-t)^2}dx$

Question

Evaluate $$I(\eta)=2 a \int\limits_{x=\alpha}^{2} x e^{-a x^2}\frac{p/x^2+c}{(p/x^2-t)^2}dx$$ where $a,p,c,t>0,N\in{\mathbb{N}}, \alpha>0$.

Any idea how to evaluate this integral? This comes from the expected value of $\frac{p/X^2+c}{(1/X^2-t)^2}$, where the pdf of $X$ is $f_X(x)=2 a x e^{-a x^2}$ (Rayleigh distribution).

Answer 1

\begin{align} I &= \frac{2ap}{p^2} \int_\alpha^2 xe^{-ax^2} \frac{\frac{1}{x^2+c}}{\frac{1}{(x^2-t)}}dx = \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \frac{(x^2-t)^2}{x^2+c}dx\\ &= \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \frac{x^4-2tx^2+t^2}{x^2+c}dx \\ &= \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \left(x^2-(2t+c) + \frac{t^2+c(2t+c)}{x^2+c} \right)dx \\ \end{align} We use long division. Now, let $M = t^2 + c(2t+c)$ and $N = (2t+c)$ then \begin{align} I &= \frac{2a}{p} \int_\alpha^2 x e^{-ax^2} \left(x^2+N + \frac{M}{x^2+c} \right)dx \\ &= \frac{2a}{p} \left( \int_\alpha^2 x^2 (x e^{-ax^2})dx + N\int_\alpha^2 (x e^{-ax^2})dx + M \int_\alpha^2 \frac{1}{x^2+c} (x e^{-ax^2})dx\right) \end{align} After this you can use integration by parts and integration by substitution to get the value of the integral.