Show that the integral of $1/(1-x^x)$ from $0$ to $1/e$ is divergent How can you show that $\int_0^{1/e}\frac{dx}{1-x^x}$ diverges? Do you have to substitute $x = \frac1u$?

Using the inequality: $e^{-y} > 1 - y, 0 < y < 1$. Put $u = x\ln x$. Observe that $x \to 0^{+} \implies u \to 0^{-}$. Thus you can write $u = -y, 0 < y < 1$. The inequality $e^{-y} > 1 - y$ can be proved easily on $y \in (0,1)$. Thus $\dfrac{1}{1- x^x} = \dfrac{1}{1-e^{x\ln x}} = \dfrac{1}{1-e^{-y}} > \dfrac{1}{y} = -\dfrac{1}{x\ln x}\implies \displaystyle \int_{0}^{1/e}\dfrac{dx}{1-x^x}> \displaystyle \int_{0}^{1/e} \dfrac{dx}{-x\ln x}= +\infty$ by a simple substitution $m = \ln x$, proving divergence.
