Let λ\lambda be a positive real constant. For reals x∈(0,1)x\in(0,1), consider the uu-substitution u=λx(λ−1)x+1,u = \frac{\lambda x}{(\lambda-1)x+1}, defined so that the odds u:1−uu:1-u and λx:1−x\lambda x:1-x are equal.

For what function f(x)f(x) is duf(u)\frac{du}{f(u)} equal to dxf(x)\frac{dx}{f(x)}?

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[Source: Communicated by Wayne Zhao; possibly due to Feynman?]

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