Show that there exists $T>0$ such that $\frac{e^{t}}{t^{1-\alpha}}-\int_{0}^{t}\frac{e^{\xi}}{\xi^{1-\alpha}}d\xi<0$ for all $t\geq T$

Question

Let $\alpha\in\,]0,1[$. I want to show that there exists $T>0$ such that the positive function

$F(t):=e^{-t}\int_{0}^{t}\frac{e^{\xi}}{\xi^{1-\alpha}}d\xi$ decays monotonically to zero.

I calculated $$\lim_{t\rightarrow \infty} F(t)=0.$$ Also, $$F^{\prime}(t)=\frac{f(t)}{e^t}.$$

It remains then to show that there exists $T>0$ such that $f(t):=\frac{e^{t}}{t^{1-\alpha}}-\int_{0}^{t}\frac{e^{\xi}}{\xi^{1-\alpha}}d\xi<0$ for all $t\geq T$.

Notice that

$$f^{\prime}(t)=-(1-\alpha)\frac{e^t}{t^{2-\alpha}}<0.$$ So, $f$ is decreasing. Therefore, it suffices to show that there exists $T>0$ such that $$f(T)<0.$$

Testing the function $f$ numerically, it seems that there does exist $T=T(\alpha)$ such that $f(T(\alpha))<0$ for all $\alpha\in\,]0,1[.$

Answer 1

For $t>0$ we have $f’’(t)<0$, so $f’(t)$ is decreasing. Then for each $t\ge 1$ the value $$f(t)=f(1)+\int_{1}^t f’(s)ds\le f(1)+f’(1)(t-1)$$ is negative for all sufficiently large $t$.