# 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?

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Actually it is because you take $\delta>0$, show that almost half of the total mass of $f_\gamma(x)f_\gamma(y)\,dxdy$ is in $[1-\delta,1]^2\cup [-1,\delta-1]^2$ where you have $\log|x-y|<\log\delta$ and conclude (the positive part is, clearly, bounded). Since you can take $\delta\approx \gamma^{-1}$ here and $\log$ is a superlazy guy when it comes to moving anywhere, the crude asymptotics $-(\frac 12+o(1))\log\gamma$ is obvious too. Let me know if you want more precision because that will take some work... – fedja Nov 2 '13 at 9:25
Yes please fedja I would be glad to read the full answer and vote for it ! – user16215 Nov 2 '13 at 9:32

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.

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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.

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\begin{align} {\cal F}_{1}\pars{\gamma}&= \int_{0}^{1}x^{2\gamma + 1}\ln\pars{x}\,{\dd x \over \pars{1 + \gamma}x^{\gamma + 1}} ={1 \over 1 + \gamma}\lim_{\mu \to \gamma}\partiald{}{\mu}\int_{0}^{1}x^{\mu} \,\dd x = {1 \over 1 + \gamma}\lim_{\mu \to \gamma}\partiald{}{\mu}\pars{1 \over \mu + 1} \\[3mm]&= -\,{1 \over \pars{1 + \gamma}^{3}} \end{align}

Whit this result, ${\rm I}\pars{\gamma}$ $\pars{~\mbox{see expression}\ \pars{1} ~}$ is reduced to $${\rm I}\pars{\gamma}=-\,{1 \over 1 + \gamma} + \half\,\pars{1 + \gamma}^{2}\,{\cal F}_{2}\pars{\gamma}\tag{2}$$ where ${\cal F}_{2}\pars{\gamma}$ is defined in $\pars{1}$.

\begin{align} {\cal F}_{2}\pars{\gamma}&= \int_{0}^{\infty}\,\dd y\,\ln\pars{\verts{1 - y^{2}}}y^{\gamma} \int_{0}^{1}\dd x\,x^{2\gamma + 1}\Theta\pars{{1 \over y} - x}\,\dd x \\[3mm]&= \int_{0}^{\infty}\,\dd y\,\ln\pars{\verts{1 - y^{2}}}y^{\gamma}\bracks{% \Theta\pars{1 - y}\int_{0}^{1}\dd x\,x^{2\gamma + 1}\,\dd x + \Theta\pars{y - 1}\int_{0}^{1/y}\dd x\,x^{2\gamma + 1}\,\dd x} \\[3mm]&= \int_{0}^{\infty}\,\dd y\,\ln\pars{\verts{1 - y^{2}}}y^{\gamma}\bracks{% \Theta\pars{1 - y}\,{1 \over 2\pars{1 + \gamma}} + \Theta\pars{y - 1}\,{1 \over 2\pars{1 + \gamma}y^{2\gamma + 2}}} \\[3mm]&={1 \over 2\pars{1 + \gamma}}\bracks{% \int_{0}^{1}\ln\pars{1 - y^{2}}y^{\gamma}\,\dd y + \int_{1}^{\infty}\ln\pars{y^{2} - 1}y^{-\gamma - 2}\,\dd y} \\[3mm]&={1 \over 2\pars{1 + \gamma}}\braces{% \int_{0}^{1}\ln\pars{1 - y^{2}}y^{\gamma}\,\dd y + \int_{0}^{1}\bracks{\ln\pars{1 - y^{2}} - 2\ln\pars{y}}y^{\gamma}\,\dd y} \\[3mm]&={1 \over 1 + \gamma}\bracks{% \int_{0}^{1}\ln\pars{1 - y^{2}}y^{\gamma}\,\dd y - \int_{0}^{1}\ln\pars{y}y^{\gamma}\,\dd y} \end{align} Since $\ds{\int_{0}^{1}\ln\pars{y}y^{\gamma}\,\dd y = \lim_{\mu \to \gamma}\partiald{}{\mu}\int_{0}^{1}y^{\mu}\,\dd y = -\,{1 \over \pars{1 + \gamma}^{2}}}$, ${\cal F}_{2}\pars{\gamma}$ is reduced to: $${\cal F}_{2}\pars{\gamma}= {1 \over \pars{1 + \gamma}^{3}} + {1 \over 1 + \gamma}\int_{0}^{1}\ln\pars{1 - y^{2}}y^{\gamma}\,\dd y$$

$$\mbox{and}\ \pars{~\mbox{see expression}\ \pars{2}~}\quad {\rm I}\pars{\gamma}=-\,{1 \over 2\pars{1 + \gamma}} + \half\,\pars{1 + \gamma}\int_{0}^{1}\ln\pars{1 - y^{2}}y^{\gamma}\,\dd y\tag{3}$$ The last integral $\pars{~\mbox{in}\ \pars{3}~}$ is evaluated as follows $\pars{~{\rm B}\pars{x,y}\ \mbox{and}\ \Gamma\pars{z}\ \mbox{are the}\ Beta\ \mbox{and}\ Gamma\ \mbox{functions, respectively}~}$: \begin{align} \int_{0}^{1}\!\!\!\!\!\ln\pars{1 - y^{2}}y^{\gamma}\,\dd y &=\half\int_{0}^{1}\ln\pars{1 - y}y^{\pars{\gamma - 1}/2}\,\dd y =\half\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}\pars{1 - y}^{\mu} y^{\pars{\gamma - 1}/2}\,\dd y \\[3mm]&= \half\lim_{\mu \to 0}\partiald{{\rm B}\pars{\mu + 1,\bracks{\gamma + 1}/2}}{\mu} =\half\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% {\Gamma\pars{\mu + 1}\Gamma\pars{\gamma/2 + 1/2} \over \Gamma\pars{\mu + \gamma/2 + 3/2}}} \\[3mm]&=\half\,\Gamma\pars{\half + \half\,\gamma}\lim_{\mu \to 0}\bracks{% {\Gamma\pars{\mu + 1}\Psi\pars{\mu + 1} \over \Gamma\pars{\mu + \gamma/2 + 3/2}} - {\Gamma\pars{\mu + 1}\Psi\pars{\mu + \gamma/2 + 3/2} \over \Gamma\pars{\mu + \gamma/2 + 3/2}}} \\[3mm]&=\half\, \overbrace{\Gamma\pars{1/2 + \gamma/2} \over \Gamma\pars{3/2 + \gamma/2}} ^{\ds{=\ {2 \over 1 + \gamma}}} \bracks{\Psi\pars{1} - \Psi\pars{{3 \over 2} + \half\,\gamma}} = -\,{{\bf C} + \Psi\pars{3/2 + \gamma/2} \over 1 + \gamma} \end{align} where $\Psi\pars{z}$ is the $\it digamma$ function and ${\bf C}$ is the $\it\mbox{Euler-Mascheroni constant}$.

With this result and expression $\pars{3}$ we arrive to:

\begin{align} \color{#00f}{\large{\rm I}\pars{\gamma}} &= \color{#00f}{\large-\,\half\bracks{% {1 \over 1 + \gamma} + {\rm C} + \Psi\pars{{3 \over 2} + \half\,\gamma}}} \\[3mm]&=\color{#00f}{\large% -\,\half\bracks{% {3 \over 1 + \gamma} + {\rm C} + \Psi\pars{\half\,\bracks{1 + \gamma}}}} \end{align}

Also $$\Psi\pars{z} \sim \ln\pars{z} - {1 \over 12 z} - {1 \over 12 z^{2}} + {1 \over 120 z^{4}} + {\rm O}\pars{1 \over z^{6}}\,,\quad \verts{z} \gg 1$$

$$\color{#00f}{\large% {\rm I}\pars{\gamma} \sim -\,\half\,{\rm C} - \ln\pars{2} - \half\ln\pars{1 + \gamma}\,,\qquad\qquad\gamma\ \gg\ 1}$$
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