integrale definito [0, 1] di ln (Γ(x)) in dx
₀∫¹ log (Γ(x+t))/(✓2π) dx = t log t-t with t≥0 prendo t =0 u ottengo
₀∫ᵗ log(Γ(x)) dx = ₀∫¹ log✓(2π) = log ✓(2π)
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integrale definito [0, 1] di ln (Γ(x)) in dx