limit of a singular integral 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?
 A: OK, let's try to get a few terms. We obviously have total mass $1$, so let's get some idea of how it is distributed. Clearly, what we are really interested in is $U(h)=\int_{0}^{1-h}|x^2-h^2|^\gamma$. Then $I(\gamma)=c(\gamma+1)^2\int_0^1 U(h)\log(2h)\,dh$ with some $c>0$. The only really interesting regions are $h\approx 0,x\approx 1$ and $x\approx 0, h\approx 1$ (the rest is $O(e^{-\delta\gamma})$ and the main series is in inverse powers). Now we play the usual Laplace game on both ends:
Let $x\approx 1, h\approx 0$. Put $y=1-h-x$ (I like small numbers starting at $0$ when integrating junk like that). Then we have to find
$$
\int_0 e^{-\gamma\log (1-2h-2y+y^2)}\,dy.
$$ 
Expand, as usual,
$$
\log(1-z)=\sum_k\frac{z^k}k
$$
up to any order you fancy (I'll use 3 dropping $O(h^4+y^4))$. This gives
$$
2h+2y-y^2+\frac 12[4h^2+4y^2+8hy-4hy^2-4y^3]+\frac 83[y^3+3y^2h+3yh^2+h^3]+O(h^4+y^4)\\
=2(h+y)+(2h^2+y^2+4hy)+(\frac 13y^3+6y^2h+8yh^2+\frac 83h^3)+O(h^4+y^4)\,.
$$
Now, keep the linear terms in the exponent and decompose $e$ to the rest into the power series once more (I knew you would love this...). We get
$$
U_1(h)=e^{-2\gamma h}\int_0[(2h^2+y^2+4hy)+(\frac 13y^3+6y^2h+8yh^2+\frac 83h^3)+O(h^4+y^4)]e^{-2\gamma y}\,dy
$$
Now, we have our beautiful cookbook Laplace replacement $\int_0 y^ke^{-2\gamma y}\,dy\to \Gamma(k+1)(2\gamma)^{-k-1}$ and get 
$$
U_1(h)=e^{-2\gamma h}\left[\frac{2h^2}{2\gamma}+\frac 2{(2\gamma)^3}+\frac{4h}{(2\gamma)^2}+\frac 2{(2\gamma)^4}+\frac {12h}{(2\gamma)^3}+\frac {8h^2}{(2\gamma)^2}+\frac 83\frac {h^3}{2\gamma}
+O\left(\frac {h^4}{2\gamma}+\frac{1}{(2\gamma^5)}\right)\right]
$$
Now is time to integrate $U_1(h)\log (2h)$ near $0$ in the same spirit except the cookbook Laplace replacement now is 
$$
\int_0\log(2h)h^ke^{-2\gamma h}\,dh\to -\log\gamma \frac{\Gamma(k+1)}{(2\gamma)^{k+1}}+\frac{\Gamma'(k+1)}{(2\gamma)^{k+1}}\,.
$$
Once you do it (preferably without stupid mistakes), you need to consider the second important region, put everything together, and get the result with the error $O(\gamma^{-4}\log \gamma)$. If that is enough, we are done. If not, make  longer decompositions. I think you should have no principal difficulties now.
A: 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. 
