Question about an integrable singularity 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.
 A: 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 $\int_{-1}^1 \frac{dx}{\sqrt{|x|}} = 4$.
A: 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.
A: 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 
      
$$
A: Why you take $1-x=y^{2}$ in a substitution ?? is something wrong in further calculations If i take $1-x=y$
