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