Answer
see below answer
Work Step by Step
a)
since $\lim\limits_{x \to {\infty}}$$\frac{x^{a}}{\ln{x}}$ = $\infty$, there must be some number $M\gt0$ such that, for all $x\gt{M}$
$\frac{x^{a}}{\ln{x}}\gt{2}$
but this mean that, for all $x\gt{M}$
$x^{a}$$\gt$$2\ln{x}$
b)
for all $x\gt{M}$, we have $x^{a}$$\gt$$2\ln{x}$ Then
$-x^{a}$$\lt$$-2\ln{x}$ = $\ln{x^{-2}}$
so that
$e^{-x^{a}}$$\lt$$e^{\ln{x}^{-2}}$ = $x^{-2}$
c)
by the above calculations, we can use the comparison test on the interval $[M,\infty)$
$\int_{M}^{\infty}$$\frac{dx}{x^{2}}$ converges then $\int_{M}^{\infty}$$e^{-x^{a}}dx$ also converges
since $e^{-x^{a}}$ is continuous on $[1,M]$, we have that
$\int_{M}^{\infty}$$e^{-x^{a}}dx$ converges then $\int_{1}^{\infty}$$e^{-x^{a}}dx$ also converges
