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What to do when integration boundary is on pole. I want to integrate $(dx/x)\log(x-1)$ from $0$ to, lets say, "$a$", where "$a$" is arbitrary $a>1$, $a\le 2$. $x$ is real.

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This integral will diverge, because $\log(x-1)/x \approx \log(-1)/x$ as $x \to 0$ and $\int_0^a dx/x$ diverges.

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    What about if the integrand is just log(x-1)?2012-03-02
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    @Potato: $\int_0^1\log(x)\;\mathrm{d}x=-1$; that is, the integral converges near $0$.2012-03-02
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    Right. I think the asker may be mistaken and have meant that. I'm not sure.2012-03-02
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    $\frac{\log(1-x)}{x}$ has a removable singularity at $0$, but that is not a pole.2012-03-02