3.1. Subcase 2.2: k = 2 j + 1 and sj < 0 At the end of section III in Ref. the claim is made that . . . , all one center contributions of matrix elements of operator r -2 j-1 vanish if js -1 ≥ 0. This statement is wrong as will be shown below and casts doubt on the validity of Eqn. (40') in combination with u s 1 ′ ( j, a, b). To illustrate this, we calculate the integral of the operator r -3 ( j = 1) between two s-orbitals with exponent β = 1 positioned at the origin (s = 0 so that js -1 = 0). According to the statement above it should vanish, but it clearly does not. For simplicity, the cutoff function (1exp -r 2 ) 2 is used (α = 1, q = 2). The resulting integral is one-dimensional: