Let A and B each be random real numbers chosen from the uniform interval (0,1).
Call Z the tenths place digit of A^{B}.
Find the probability distribution of Z.
Start with the case of the digit 9. As all the Z values are below 1, what is sought here is the probability that Z > .9. Others can be found by subtracting this (and successibe probabilities) from the probability that Z > .8, etc.
P(Z>.9) = Integral{0 to 1} ln(.9)/ln(A) dA, treating A as the independent variable; use x if you prefer. Note that ln(.9) is negative, avoiding a singularity in the logarithmic integral function at A=1.
This turns out to involve the Logarithmic Integral function li(x). Note that ln(.9) is negative, avoiding a singularity in the logarithmic integral function at A=1.
An alternative, integrating over B, would involve the Exponential integral Ei(x):
Integral{0 to 1} e^(ln(.9)/B) dB.

Posted by Charlie
on 20141029 14:35:18 