The purpose of this problem is to prove the following formula
∀z∈]0,1[,Γ(z)Γ(1−z)=sin(πz)πwhere Γ(z):=∫0+∞tz−1e−tdt is the Euler function. This formula, generalizable to the whole band 0<ℜ(z)<1, is very useful in order to understand how the Riemann ζ-function behaves for ℜ(z)<1.
We start by some preliminaries.
1) a) Show that Γ(x) is defined for all x>0. Show also that the function B such that
B(x,y):=∫01tx−1(1−t)y−1dt,is defined for all x,y>0.
1) b) For x,y>0, show that
Γ(x)Γ(y)=Γ(x+y)B(x,y).
1) c) Let fα~:x∈[−π,π[⟼cos(αx) where α∈R\Z and let fα:R→R be the unique 2π-periodic function such that fα(x)=fα~(x),x∈[−π,π[ and fα is even.
Show that fα can be decomposed as a real Fourier series, determine the coefficients of this decomposition and deduce the relation
sin(απ)π=k=−∞∑k+α(−1)k
Now we can move to the heart of the proof. To do so, define the integral for all z∈]0,1[
Iz:=∫0+∞1+ttz−1dt
2) Show that Iz is well-defined for all 0<z<1.
3) Show that
Iz=∫011+ttz−1+t−zdt=k=−∞∑+∞k+z(−1)k=sin(πz)π4) Show also that
∀x,y>0,B(x,y)=∫0+∞(1+u)x+yux−1duand conclude.