By Paula Ribenboim

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1). According to the Fundamental Theorem of Arithmetic we can represent m and n by the products m = pα1 1 · · · · · pαr r and n = q1β1 · · · · · qrβs of primes p1 < · · · < pr and q1 < · · · < qr where the powers α1 , . . , αr and β1 , . . , βs are positive integers. Here we assume pi = qj since the greatest common denominator between m and n is unity. If at least one of the exponents αj is larger than unity we find from the second case of the definition Eq. 2) of µ the results µ(m) = 0 and µ(m · n) = 0 confirming µ(m · n) = µ(m) · µ(n).

19), interchange summation and integration. With the help of the substitution u ≡ πn2 x we find the formula π s/2 Γ 2s ∞ ∞ n=1 ∞ 2 dx e−πn x xs/2−1 = d(n) d(n)n−s = D(s), n=1 0 which is indeed Eq. 18). 2. Finite Sums and Inverse Mellin Transform Now we go a step further and represent D not just as the product of two functions with one being the Mellin transform but as the inverse Mellin transform of a single function. 21) c−i∞ where the path is to the right of all poles of the integrand. The representation Eq.

Here we have made use of the definition Eq. 2) of the M¨obius function. We now turn to the general case of an integer n > 1. According to the Fundamental Theorem of Arithmetic we can represent n as the product α r−1 · pα ≡ m · pα n = pα1 1 · pα2 2 . . pr−1 of r powers α1 , α2 , . . , αr−1, α of distinct primes p1 , p2 , . . , pr−1 , p. In the last step we αr−1 into one integer have combined the product of the first r − 1 integers pα1 1 , pα2 2 , . . , pr−1 m and p is not a divisor of m. The divisors of n consist of all divisors d of m together with the products of all divisors d of m times p, p2 , .