Is there a measure $\nu$ on $[0,\infty)$ such that $ \ln x=\int_{0}^{\infty}d\nu\left(y\right)/\left(x+y-1\right)? $ Thanks for any helpful answers!
Is there a measure
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0For which $x$ such an equation has to be true? – 2012-10-11
1 Answers
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It seems no. Let $\nu:=\nu^+-\nu^-$ the Jordan Hahn decomposition of $\nu$. There should be an $x_0>1$ such that $\int_{\Bbb R}\frac{d\nu^+(y)}{x+y-1}$ or $\int_{\Bbb R}\frac{d\nu^-(y)}{x_0+y-1}$ is finite (it's an implicit assumption as we need the integration to a signed measure make sense).
As $\log x$ is finite, we get that for $x>x_0$, the integrals $\int_{\Bbb R}\frac{d\nu^+(y)}{x+y-1}$ and $\int_{\Bbb R}\frac{d\nu^-(y)}{x+y-1}$ are finite. Taking the limit $x\to +\infty$ yields a contradiction.
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0By dominated convergence the RHS in the displayed equality in the OP converges to $0$, but not the LHS. – 2013-02-13