Ratio of conditional expectations $\frac{E[X^{\epsilon-1}|X\geq rY)}{E[Y^{\epsilon-1}|X\leq rY)}$ if $X$ and $Y$ are Fréchet

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Let $X$, $Y$ be two independent continuous random variables with support on $\mathbb{R}_+$. Specifically, suppose that $X$ and $Y$ are distributed Fréchet with identical location parameter $0$ and shape parameter $\theta>1$, but different scale parameters $s_X$ and $s_Y$, respectively. Let $t$ denote a positive scalar, and $\epsilon$ a positive parameter (where $\theta+1-\epsilon>0$). Compute the ratio $\frac{E[X^{\epsilon-1}|X\geq rY)}{E[Y^{\epsilon-1}|X\leq rY)}$.

My approach was as follows. Consider initially just the numerator. Let the unconditional distribution of $X$ be denoted $F_X$ and let the distribution conditional on $X\geq rY$ be $\tilde{F}_X$. I'd then compute $$\int_0^{\infty} x^{\epsilon-1} d\tilde{F}_X(x).$$ How to derive this conditional distribution? I started with the conditioning probability, i.e., $\pi_X \equiv Pr\{X \geq rY\}$. Given the Fréchet assumption can be shown to be $\pi_X = \frac{1}{1+\phi}$, where $\phi=\left(\frac{s_X}{rs_Y}\right)^{-\theta}$. I then wanted to find $\tilde{F}_X$ as $$Pr\{X\leq\tilde{x}|X\geq rY\} = \frac{1}{\pi_1} \times \int_{0}^{\tilde{x}} \Pr\{Y\geq r^{-1} x \} dF_X(x).$$ The right-hand side can the be written as $$\frac{1}{\pi_1} \times \int_{0}^{\tilde{x}} (1-F(xr^{-1})) dF_X(x).$$ But I got stuck when trying to compute this expression when substituting for the distributional expressions and suspect I'm not approaching this problem in the most efficient way in any case.