First of all, according to the integral test for numeric series, we have theorem $T$:
$T$: Let $f(x)$ be continuous, positive and decreasing in $[1, \infty)$ and $a_n$ a sequence such that $f(n)=a_n$. Then
$i$) If $\int_1^\infty f(x) dx$ converges, then $\sum_{n=1}^\infty a_n$ converges.
$ii$) If $\int_1^\infty f(x) dx$ diverges, then $\sum_{n=1}^\infty a_n$ diverges.
$Problem:$
I need to prove the convergence or divergence of
$a)$ $$\int_1^\infty \frac{e^x}{x^x}dx$$
$b)$ $$\int_2^\infty \frac{1}{x\ln^x(x)}dx$$
$Proof$:
$a)$ $i$) Let $f(x)= \frac{e^x}{x^x}$ and $a_n=\frac{e^n}{n^n}=(\frac{e}{n})^n$ so that $f(n)=a_n$.
$ii$) Because
$$\lim_{n\to\infty} \sqrt[n]{a_n}=\lim_{n\to\infty} \frac{e}{n}=0<1$$
then, according to the root test for series, $\sum_{n=1}^\infty a_n$ converges.
$iii$) Because of $T$, if $\int_1^\infty \frac{e^x}{x^x}dx$ diverges, then $\sum_{n=1}^\infty a_n$ should diverge. But $\sum_{n=1}^\infty a_n$ converges. Then $\int_1^\infty \frac{e^x}{x^x}dx$ doesn't diverge. Then it must converge.
$Conclusion$: $\int_1^\infty \frac{e^x}{x^x}dx$ is convergent.
$b)i)$ Let $f(x)=\frac{1}{x\ln^x(x)}$ and $a_n=\frac{1}{n\ln^n(n)}$ so that $f(n)=a_n$.
$ii)$ Because
$$\lim_{n\to\infty} \sqrt[n]{a_n}=\lim_{n\to\infty} (\frac{1}{\sqrt[n]{n}}*\frac{1}{\ln(n)})=\lim_{n\to\infty} \frac{1}{\sqrt[n]{n}} * \lim_{n\to\infty} \frac{1}{\ln(n)}$$
where both limits exist and equal $0$, then
$$\lim_{n\to\infty} \frac{1}{\sqrt[n]{n}} * \lim_{n\to\infty} \frac{1}{\ln(n)}=1*0=0<1$$
and, according to the root test for series, $\sum_{n=1}^\infty a_n$ converges.
$iii$) Because of $T$, if $\int_2^\infty \frac{1}{x\ln^x(x)}dx$ diverges, then $\sum_{n=1}^\infty a_n$ should diverge. But $\sum_{n=1}^\infty a_n$ converges. Then $\int_2^\infty \frac{1}{x\ln^x(x)}dx$ doesn't diverge. Then it must converge.
$Conclusion$: $\int_2^\infty \frac{1}{x\ln^x(x)}dx$ is convergent.