E(0, infinity)f^{n}(a)(2-a)^n/(n!)

Each term in the sum is multiplied by the value of the "n'th" derivative of the function defined at a. So, the first term is the "0th" derivative, or the function defined at a: ln(abs(a-1)); the second term is the 1st derivative defined at a: 1/(a-1); the third is the 2nd derivative defiend at a: -1/(a-1)^2 ... etc.