Denote $f\_{\gamma}(x) =\frac { (1+\gamma)}{2} |x|^{\gamma}$. We consider:
$$I(\gamma) = \int\_{-1}^1\int\_{-1}^1 \ln (|x-y|) f\_{\gamma}(x) f\_{\gamma}(y) dx dy$$
I would like to know the limit of $I(\gamma)$ when $\gamma \to \infty$, and if this limit is infinite I would like to have an equivalent.


My first idea was to say that $f\_{\gamma}(x)$ converges to $\frac 1 2 (\delta\_{-1}+\delta\_1)$, and replacing this into the integral we obtain something which seems infinite. But I know the convergence is not uniform, so this argument is not valid.


Any idea?



limit of a singular integral
The exact result of the integral is:


$I(\gamma)=\frac{-2 \gamma -2 \pi (\gamma +1) \gamma \cot \left(\frac{\pi \gamma
 }{2}\right)+(\gamma +1) \gamma \left(2 H\_{-\gamma -2}-H\_{-\frac{\gamma
 }{2}-\frac{1}{2}}+H\_{-\frac{\gamma }{2}}-4 H\_{\gamma +1}+\log (4)\right)+2}{4
 \gamma (\gamma +1)}$


where $H\_r$ are the harmonic numbers.


This gives easily the asymptotics:


$I(\gamma)\approx-\frac{\log (\gamma )}{2}-\frac{\gamma \_{\text{E}}}{2}+\frac{\log (2)}{2}-\frac{3}{2 \gamma }$


where $\gamma \_{\text{E}}$ is the Euler-Mascheroni constant. 


