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I've been trying to understand a concept of an integrable singularity. So far what I have discovered was that the case of the singularity occurs when there is a point where the integration becomes infinite. Am I right? Also, is there anything else that would add to the definition of the integrable singularity? Thanks.

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    If the singularity is integrable then its integration is *not* divergent.2011-09-04

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You can think of integrable singularity as a singularity of integrand that will disappear under suitable change of variables. Consider for example:

$ \int_{0}^1 \frac{1}{\sqrt{1-x} } \mathrm{d} x $ Then for $1-x = y^2$ and $-\mathrm{d}x = 2 y \, \mathrm{d} y$:

$ \int_{0}^1 \frac{1}{\sqrt{1-x} } \mathrm{d} x = \int_{0}^1 \frac{1}{\sqrt{y^2} } 2 y \, \mathrm{d} y = 2 $

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    @eurocoder the zero of the denominator cancels the zero in the numerator. The integrand has the finite right limit after the change of variable.2017-01-02
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I think what you mean is that the integrand approaches infinity, but an improper integral around that point is finite. For example, $1/\sqrt{|x|}$ has an integrable singularity at $0$, since $\int_{-1}^1 \frac{dx}{\sqrt{|x|}} = 4$.

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It means that the indefinite integral of $f(x)$ can be extended across the singularity: there is a continuous function $F(x)$ on an interval containing the singular point in the interior, with F' = f. So as far as integration is concerned the singular point of the integrand is an ordinary one for the integral.