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En utilisant la fonction θ de Jacobi, dont on rappelle que:
θ(x)=∑k=−∞+∞e−πk2x2
et pour x>0
xθ(x)=θ(1/x)
Montrez que :
πx/2Γ(x/2)ζ(x)=x−11−x1+∫1+∞(tx−1+t−x)(θ(t)−1)dt
en déduire que :
πx/2Γ(x/2)ζ(x)=π(1−x)/2Γ((1−x)/2)ζ(1−x)
Solutions
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