By Graham Everest

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**Example text**

I: II : All these steps are very similar to the argument for another (better) proof will be given. ∞ {t} 0 ts+1 dt. 2 The Gamma function has the following properties: 1. For all s with (s) > 0, Γ(s + 1) = sΓ(s). 2. For all integers N ≥ 0, Γ(N + 1) = N !. Proof of 1: ∞ Γ(s + 1) = ∞ e−t ts dt = [−e−t ts ]∞ 0 +s 0 0 34 e−t ts−1 dt. The bracket term vanishes at t = 0 because 1. by induction together with Γ(1) = 1. (s) > 0. Item 2. follows from ✷ Now write 1 Γ(s) = Γ(s + 1). s (46) The right-hand side is defined for (s) > −1 apart from s = 0, where it has a simple pole with residue Γ(1) = 1.

Now define for all f ∈ S the Fourier transform of f by ˆf(y) := f(x) exp(−2πixy) dx. R This integral exists for the same reason as above: |ˆf(y)| ≤ |f| < ∞, R in fact ˆf ∈ S again (apply equation (50) with m + n to get the bound for ˆf(n) ). Thus f → ˆf is a linear map from S to S. Periodicity: A function g on R is periodic if g(x) = g(x + m) for all m ∈ Z and for all x ∈ R. Suppose g is periodic and piecewise continuous. Then its k-th Fourier coefficient is defined for k ∈ Z as 1 g(x)e−2πikx dx.

1 + m−1 = lim s m s 1 + 1s . . 1 + m−1 s s 1 1 1 (1 + 1)2( 1 + 2 . . (m − 1) 1 + m−1 = lim m s (1 + s)(2 + s) . . (m − 1 + s) 47 s where we have just expanded the fraction by 2 · 3 · . . · (m − 1). Now collect the integers in the numerator into one product, and the other factors into a product m−1 1+ n=1 1 n s = ms by the trick we did in equation (70). This completes the proof of the corollary. 99 ∞ 1 = Γ(s) = f(s) e−t ts−1 dt. (73) 0 Thus, we have three representations of the Gamma function - the definition, and the ones given in theorems W and E.