What is an integrable singularity and how is it determined?

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An integrable singularity is defined by two key conditions: first, the integral remains well-defined when an arbitrary neighborhood of the singularity is excluded, and second, the limit of the integral exists as the size of that neighborhood approaches zero. For instance, the integral of \(\int_0^1 \log{x} \, dx\) equals -1, demonstrating that the singularity at \(x=0\) is integrable. In contrast, \(\int_0^1 \frac{1}{x} \, dx\) diverges to infinity, indicating a non-integrable singularity.

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Hello.

What is an integrable singularity? Is it a certain order of the singularity?

\int_0^1\frac{1}{x}=\infty (not integrable)
\int_0^1\log{x}=-1 (integrable)
 
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An integrable singularity is such that
1. When an arbitrary neighbourhood of the singularity is excluded the integral is well.defined.
and
2. When the size of that neighbourhood is shrunk to 0, we get a definite limit.

For example, let's tackle your second one:

NOw, let e be a number greater than 0, and consider:
\int_{\epsilon}^{1}\log(x)=-1-\epsilon\log(\epsilon)+\epsilon

Since the two last terms go to zero as e goes to 0, we have that the improper integral can be given the unique value -1. The singularity was integrable.
 

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