Bourbaki

Exercise 4, § 1

Functions of a Real Variable, VII, 327

Show that the function f(x)=12Γ(x/2)Γ(x+12)f(x) = \frac{1}{\sqrt{2}} \frac{\Gamma(x/2)}{\Gamma\left(\frac{x+1}{2}\right)} is the unique convex solution of the equation u(x+1)=1/x u(x)u(x+1) = 1/x\,u(x) (remark that this equation implies u(x+2)=xx+1 u(x)u(x+2) = \frac{x}{x+1}\,u(x) and apply exerc. 1 a)).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.