A 'Natural Proof' of the Gauss Multiplication formula II: Theorem and Proof

We are now ready to prove the result:

Theorem (Gauss's Multiplication formula for Γ)

Γ(Nt) = (2π)^((1-N)/2) . N^(Nt - 1/2) . prod_[k=0 to N-1] Γ(t + k/N) ------- (2)

Proof:

From (1) we have that,

√(2π) / Γ(Nt) = prodz_[m=0 to +infinity] (m + Nt) = prod_[k=0 to N-1] prodz_[m=0 to +infinity] ((mN + k) + Nt)

it is easy to see that there is a bijection between the product indices on either side of the last equality since m=0 on the right gives the terms of the product on the left from m=0 up to m=N-1 and so on.

Then,

√(2π) / Γ(Nt) = prod_[k=0 to N-1] prodz_[m=0 to +infinity] N.(m + (t + k/N))

From the lemma we established in the last post we have,

√(2π) / Γ(Nt) = prod_[k=0 to N-1] N^[Z(0, t + k/N)] prodz_[m=0 to +infinity] (m + (t + k/N))

We then recognise the regularised product from (1) again and along with the value of Z(0, t + k/N), have,

√(2π) / Γ(Nt) = prod_[k=0 to N-1] N^[1/2 - (t + k/N)] . √(2π) / Γ(t + k/N) = N^(1/2 - Nt) . (2π)^(N/2) . prod_[k=0 to N-1] 1/Γ(t + k/N)

Which gives the required formula on rearrangement.

QED.

The proof flows naturally from the regularized determinant expression for the Gamma function unlike standard proofs which match up zeros and poles of either side of (2), which is why the term 'natural' was used to describe it. We also note that the simple functional equation Γ(z+1) = z.Γ(z) is obvious from (1).

I am much indebted to Professors Kurokawa and Wakayama for their wonderful paper.

For more on regularization see Basic Analysis Of Regularized Series And Products by Jay Jorgenson and Serge Lang.