In the book An Intermediate Course in Probability by Allan Gut there is this exercise
For a positive, (absolutely) continous random variable $X$ we define the Laplace transform as,. $$L_{X}(s)=E\,e^{-sX}=\int_{0}^{\infty}e^{-sx}f_{X}(x)\,dx, \quad s>0.$$ Suppose that $X$ is positive and stable with index $\alpha \in (0,1)$, which means that $$L_{X}(s)=e^{-s^{\alpha}},\quad s>0.$$ Further, let $Y\in \text{Exp}(1)$ be independent of $X$. Show that $$(\frac{Y}{X})^{\alpha}\in \text{Exp}(1)\qquad (\text{which means that}\quad (\frac{Y}{X})^{\alpha})\stackrel d= Y.$$
I am thinking that I should solve this by using
$$ X^{-\alpha}=\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}\lambda^{\alpha-1}e^{-\lambda X}\, \mathrm{d}\lambda,\quad x>0,\quad \text{and}\\ Y^{\alpha}=\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}\lambda^{\alpha -1}e^{-\frac{\lambda}{Y}}\, \mathrm{d}\lambda,\quad y>0.\\ (\frac{Y}{X})^{\alpha}=(\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}\lambda^{\alpha-1}e^{-\lambda X}\, \mathrm{d}\lambda)\times(\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}\lambda^{\alpha -1}e^{-\frac{\lambda}{Y}}\, \mathrm{d}\lambda),\quad X,Y\text{ independent.}\\ f_{(\frac{Y}{X})^{\alpha}}(x,y)= (\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}\lambda^{\alpha-1}e^{-\lambda f_{X}(x)}\, \mathrm{d}\lambda)\times (\frac{1}{\Gamma(\alpha)}\int_{0}^{\infty}\lambda^{\alpha -1}e^{-\frac{\lambda}{e^{-y}}}\, \mathrm{d}\lambda)$$
But here I am stuck. I do not know how to go from here. How to evaluate this Intergal? Anyone who can leave a hint? I would be very thankful.
Write out the cdf for $(Y/X)^\alpha$, then find the pdf by applying the Leibniz rule and use the formula for the Laplace transform of $t f(t)$: $$F_{(Y/X)^\alpha}(x) = \operatorname P(Y < x^{1/\alpha} X) = \int_0^\infty d\tau_1 f_X(\tau_1) \int_0^{x^{1/\alpha} \hspace{2px} \tau_1} d\tau_2 f_Y(\tau_2), \\ f_{(Y/X)^\alpha}(x) = \frac d {dx} F_{(Y/X)^\alpha}(x) = \int_0^\infty d\tau_1 f_X(\tau_1) \frac d {dx} (x^{1/\alpha} \tau_1) f_Y(x^{1/\alpha} \tau_1) = \\ \frac {x^{1/\alpha - 1}} \alpha \int_0^\infty \tau_1 f_X(\tau_1) e^{-x^{1/\alpha} \hspace{2px} \tau_1} d\tau_1 = -\frac {x^{1/\alpha - 1}} \alpha \frac d {ds} \int_0^\infty f_X(\tau_1) e^{-s \tau_1} d\tau_1 \bigg\rvert_{s = x^{1/\alpha}} = \\ -\frac {x^{1/\alpha - 1}} \alpha \frac d {ds} e^{-s^\alpha} \bigg\rvert_{s = x^{1/\alpha}} = e^{-x}.$$