Goal: Evaluate $\int_0^\infty x^{-x} \, \text{d}x$ (by testing via calculator, the integral clearly converges).
My attempt:
$$\int_0^\infty x^{-x} \, \text{d}x = \int_0^\infty e^{-x\ln{(x)}} \, \text{d}x =\int_0^\infty \sum_{n=0}^\infty \frac{(-x\ln(x))^n}{n!} \, \text{d}x \\ = \sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^\infty x^n(\ln(x))^n \,\text{d}x $$
However, the integral within the sum diverges. Is there any way to simplify the integral from here, or is there another method that I’m overlooking?
