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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!

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A signed measure? –  copper.hat Oct 10 '12 at 17:25
    
Dear copper.hat, it should be a non-negative measure. Thanks. –  k-6 Oct 10 '12 at 17:32
    
If $\nu \leq 0$, then $\ln 1 = 0 = \int_0^\infty \frac{d \nu (y)}{y}$ would give $\nu = 0$ on $[0, \infty)$? –  copper.hat Oct 10 '12 at 17:34
    
Oops, it should be a non-negative measure. Thanks. –  k-6 Oct 10 '12 at 17:35
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Hmm, if $\nu$ is non-negative, the same reasoning would give $v = 0$ as in the above case. (I don't know if a suitable signed measure exists, but a non-negative or non-negative one does not exist.) –  copper.hat Oct 10 '12 at 17:40

1 Answer 1

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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excuse me so what happens once again when $x\rightarrow\infty$? –  Seyhmus Güngören Feb 13 '13 at 11:13
    
By dominated convergence the RHS in the displayed equality in the OP converges to $0$, but not the LHS. –  Davide Giraudo Feb 13 '13 at 11:24

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