I have to compute the MEF for Pareto-type distributions, that is distributions $F_{X}$ such that
$$1-F_{X}(x)=x^{-1 / \gamma} \ell_{F(x)} \quad \gamma >0,$$ where $\ell_{F(x)}$ is a slowly varying function $$\lim _{x \rightarrow \infty} \frac{\ell_{F(t x)}}{\ell_{F(x)}}=1 \quad \forall t>0.$$
Before computing the MEF I was told to note that
$$\lim _{x \rightarrow \infty} \frac{1-F_{X}(t x)}{1-F_{X}(x)}=\lim _{x \rightarrow \infty} \frac{(t x)^{-1 / \gamma} \ell_{F}(t x)}{x^{-1 / \gamma \ell_{F}(x)}}=\lim _{x \rightarrow \infty} t^{-1 / \gamma} \frac{\ell_{F}(t x)}{\ell_{F}(x)}=t^{-1 / \gamma}.$$
And told to use the formula $$e(t)=\frac{\int_{t}^{\infty}\left[1-F_{X}(x)\right] d x}{1-F_{X}(t)}.$$
That is the information I was given beforehand. So here is my attempt at computing the MEF: $$e(t)=\frac{\int_{t}^{\infty} 1-F_{X}(x) d x}{1-F_{X}(t)}=\int_{t}^{\infty} \frac{1-F_{X}(x)}{1-F_{X}(t)} d x.$$ Letting $u=\frac{x}{t} \Longrightarrow d u=\frac{d x}{t}.$
$$\int_{t}^{\infty} \frac{1-F_{X}(x)}{1-F_{X}(t)} d x=t \int_{1}^{\infty} \frac{1-F_{X}(t u)}{1-F_{X}(t)} d u=t\left[\int_{1}^{\infty} u^{-1 / \gamma} d u+\int_{1}^{\infty}\left(\frac{1-F_{X}(t u)}{1-F_{X}(t)}-u^{-1 / \gamma}\right) d u\right]. $$ And that is how far i got. What should my next step be? and does the last step even make sense?
I was given a hint that $$ \lim_{t \to \infty}\int_{1}^{\infty}\left(\frac{1-F_{X}(t u)}{1-F_{X}(t)}-u^{-1 / \gamma}\right) d u =0, $$ but have no idea how to apply it. Can you help me onto the next step?
Thanks in advance :).