<s>[INST] How to calculate $\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}}$ when $x>1$? Numerically, it looks that the limit is 
$$\lim_{x\to 1}\int_{0}^{1}\frac{dy}{\sqrt{1-y^{2}}}\frac{y^{3/2}}{\sqrt{x - y}} = \frac{1}{\sqrt{2}}\log(1 - x) + cte $$,
but I have not been able to demonstrate it analytically. Does anyone have a idea on how to deal with this limit?
 [/INST] Here is how you advance, compute the Taylor series of $\frac{1}{\sqrt{x-y}}$ at the point $x=1$
$$ \frac{1}{\sqrt{x-y}} = {\frac {1}{\sqrt {1-y}}}-\frac{1}{2}\, \left( 1-y \right) ^{-3/2} \left( x-1
 \right) +O \left(  \left( x-1 \right) ^{2} \right) $$
$$ \implies \frac{1}{\sqrt{x-y}} \sim {\frac {1}{\sqrt {1-y}}}. $$
So, when $x$ close to $1$, our integral behaves as
$$\int_{0}^{1} \frac{ y^{3/2} }{\sqrt{1-y^2}\sqrt {x-y}}dy \sim \int_{0}^{1} \frac{ y^{3/2} }{\sqrt{1-y^2}\sqrt {1-y}}dy$$
$$=\int _{0}^{1}\!{\frac{ \left( -u+1 \right)^{3/2}}{\sqrt {-u+2}u}}{du}\sim \int _{0}^{1}\!{\frac {1}{\sqrt {2}u}}{du}.$$
The last integral does not converge. Note that, we used the change of variables $1-y=u$ and
$$ \frac{ ( -u+1 )^{3/2}}{\sqrt{-u+2}u} \sim \frac{1}{\sqrt{2}\,u} \quad \mathrm{when}\quad u\sim 0.$$
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