Is $\infty^0= 1$? for given integral $\int_e^\infty \frac{1}{x(\log x)^p}dx$,
I had derived above integral = $\int_1^\infty {u^{-p}}du = [ {1\over -p+1}u^{-p+1}]_1^\infty = {1\over -p+1}[\infty^{-p+1} -1]$
I need to characterize the range of $p \in \Bbb R$ which makes the given integral converges, however, when p =1, there's $\infty^0$ occurs, but I'd never learned to deal with this notation/character.
Is it converging to $1$? If yes, which logical reasoning could be provided?

In general improper integrals like $\int_a^{\infty}{f(x)dx}$ can be defined as $\lim_{b \to \infty}F(x) - F(a)$ where $F(x)$ is the anti derivative of $f(x)$.
 In this case how ever the anti derivate for the case $p=1$ is not the one deduced from formula for other values of $p$. So you have to take it as special case and say $\int_e^{\infty}\frac{1}{xlogx}=\int_1^{\infty}\frac{1}{u}du=log(u)|_1^{\infty}=\lim_{b \to \infty}log(b) - log(1) =\infty$
