Variance of a density that has variable exponents as well as variable base I have been wracking my brain on this one, it was given to me by a friend and I am wondering how to solve it. The joint density is: 
$$f(x,y)=c x^{2-x}y^{x-1}~\text{on}~ 0<x<y<1$$
I am trying to find Var[X], but I am even having trouble finding $c$. I have tried modifying the question to perhaps be changeable into a Gamma function, I have tried integrating it in Mathematica, etc... without much success.
Feel free to provide feedback for part of the solution, or present a problem that is similar but solvable. 
 A: Since all you are concerned about is $\sigma^2_x$, you can start by integrating out $y$.  This will give you the marginal density of $x$:
$$f(x) = cx^{2-x}{y^x\over x}|_0^1 = cx^{1-x}$$
Now for the constant of integration.  It so happens that:
$$\int_0^1x^{a-x}\text{d}x = \sum_{n=1}^{\infty}{1 \over (a+n)^n}$$
for $a > -1$, so, substituting $a = 1$ gives us the inverse of $c$.
On to the variance!  We use the relationship $\sigma^2_x = \mathbb{E}x^2 - (\mathbb{E}x)^2$.
$$\mathbb{E}x = c\int_0^1x^{2-x}\text{d}x = {\sum_{n=1}^{\infty}{1 \over (2+n)^n} \over \sum_{n=1}^{\infty}{1 \over (1+n)^n}}$$
$$\mathbb{E}x^2 = c\int_0^1x^{3-x}\text{d}x = {\sum_{n=1}^{\infty}{1 \over (3+n)^n} \over \sum_{n=1}^{\infty}{1 \over (1+n)^n}}$$
Defining $A_i = \sum_{n=1}^{\infty}1 / (i+n)^n$ for brevity, we engage in a little algebra:
$$\sigma^2_x = {A_3 \over A_1} - {A_2^2 \over A_1^2} = {A_1A_3-A_2^2 \over A_1^2}$$
All the series are obviously convergent, and pretty quickly so.  Resorting to the WolframAlpha online integral calculator, we find that:
$$A_1 \approx 0.6284737$$
$$A_2 \approx 0.4046685$$
$$A_3 \approx 0.295079$$
leading to $\sigma^2_x \approx 0.054921$.
