Area under distributions defined by $p(x)=x^{\frac{1}{x}-a}$ I've been watching 3Blue1Brown's livestream, specifically the one about power towers. I was playing with the inverse function of this system, $x^\frac{1}{x}$, and noticed that it's able to create a very interesting family of distributions by modifiying the function to $x^{\frac{1}{x}-a}$. Additionally, it seems that these functions all have finite area on $[0, \infty]$. I had a few questions about these distributions:


*

*Do they have a name?

*Is it possible to find the area in terms of $a$ so it can be normalized? I.e., can we find $g(a)=\int_{0}^{\infty}{x^{\frac{1}{x}-a}} \ d x$?

*What value of $a$ gives the minimum area, and what is that area? I.e., can we find $b$ s.t. $g'(b)=0$ and compute $g(b)$? Playing around in Desmos suggests we might have $b=e$, but it's not close enough to be sure.

*If we normalize our original function using the answer to 2. and treat it like a probability density function, what is the expected value of $x$?


Here's my Desmos playground: https://www.desmos.com/calculator/7ayroajui4
 A: NOT A FULL ANSWER
$$\int^\infty_0 x^{\frac{1}{x}-a}dx$$
substituting $u=\frac{1}{x}$
$du=-\frac{1}{x}dx$, $dx=-x^2du$
$$\int^{0^+}_\infty -u^{-u+2-a}du$$
$$\int^\infty_0u^{-u}u^{2-a}du$$
we do know that
$$\int^1_0x^{-x}dx=\frac{1}{1^1}+\frac{1}{2^2}+\frac{1}{3^3}...$$
which could help us solve it for a=-2:
$$\sum^\infty_{n=1}\frac{1}{n^n} + \int^\infty_1x^{-x}dx$$
if we want to find for which a the integral diverges a good check is:
$$\lim_{x\to \infty} x^{\frac{1}{x}-a}$$
$$\lim_{x\to \infty} e^{ln(x)(\frac{1}{x}-a)}$$
$e^x$ is a continuous function
$$e^{\lim_{x \to \infty}ln(x)(\frac{1}{x}-a)}$$
little bit of product ruling later,
$$e^{-\infty a}$$
which tells us it converges for $a>0$ and unintegratable to infinity for a=0
when looking at the function's graph, it's obvious it has a maximum for some a.
$$\frac{d}{dx}x^{\frac{1}{x}-a} = -x^{\frac{1}{x}+a-2}(ln(x)-ax-1)$$
from this we get two 0's (0,a) and ($-\frac{W(-ae)}{a}$,a)
W - lambert W function.
The second one is the maximum of the function.(a<0)
Hope it helps.
