It seems to me that this expectation is infinite.
My argument takes a smaller function that is discreet. f(x) = 1 if x>.5, f(x)=2 if x>.25 and so on.
We can use linearity of expectation to calculate E(f(x))=1+1/2*1+1/4*2+1/8*4... (where each term is the probability of passing a threshold and the increase for that threshold).
Is this correct? The reason I suspect if might not be: if I take a 0-1 uniform random matrix and do elimination, the second pivot seems to have distribution x4-(x3/x1)*x2, where x1,x2,x3 and x4 are uniform 0-1 and, therefore, infinite expected value (or at least infinite modulus)
$$\int_0^1 \frac1x \, dx = \lim_{t \to 0^+} \int_t^1 \frac1x \, dx = \lim_{t \to 0^+}-\log(t)= \infty$$