Show this integral operator is compact for various values of $\alpha$ I am having some problems evaluating a multivariable integral. 
This question is features in Stakgold's book Green's functions and boundary value problems. page 359.
Consider the kernel for $a\leq x,y\leq b$
$$ k(x,y)= \frac{1}{|x-y|^\alpha}$$
Show that this is Hilbert-Schmidt (and thus compact) for $\alpha< 1/2$ (part 1), and show that it is compact for $\alpha <1$ (part 2).
Consider $\alpha< 1/2 $
$$ ||k(x,y)||_{L^2xL^2} = \int^b_a\int^b_a |k(x,y)|dxdy=  \int^b_a\int^b_a \frac{1}{|x-y|^{2\alpha}} dxdy $$
So splitting this up, 
$$ \int^b_a\int^y_a \frac{1}{(y-x)^{2\alpha}} dx+ \int^b_a \frac{1}{(x-y)^{2\alpha}} dx dy $$
Since $\alpha< 1/2 $, $1-2\alpha >0 $, so integrating,
$$ =(2\alpha+1)^{-1}\int^b_a -[(y-x)^{1-2\alpha}]^y_a  +  [(x-y)^{1-2\alpha}]^b_y dy $$
and
$$ =(2\alpha+1)^{-1}\int^b_a (y-a)^{1-2\alpha} +  (b-y)^{1-2\alpha} dy $$
Integrating again, 
$$ =(1-2\alpha)^{-1}(2-2\alpha)^{-1}( [(y-a)^{2-2\alpha}]^b_a + - [(b-y)^{2-2\alpha}]^b_a ) $$
or
$$ =(1-2\alpha)^{-1}(2-2\alpha)^{-1}( (b-a)^{2-2\alpha}+  (b-a)^{2-2\alpha}) $$
So 
$$ ||k(x,y)||_{L^2xL^2} = 2(1-2\alpha)^{-1}(2-2\alpha)^{-1}(b-a)^{2-2\alpha} < \infty$$
Thus $K$ is H.S.
Part 2.
Consider $\alpha <1$. 
$$ ||k(x,y)||_{L^2xL^2} = \int^b_a\int^b_a |k(x,y)|dxdy=  \int^b_a\int^b_a \frac{1}{|x-y|^{2\alpha}} dxdy $$
The hint is to split the kernel $k$ into 
$$ k_\epsilon(x,y) = h(x,y)+l(x,y) $$ 
Where $l$ vanishes for $|x-y|> \epsilon$.  
My idea is to split the kernel into $h(x,y)=\frac{1}{|x-y|^\alpha} $, $l(x,y)=-\frac{1}{|x-y|^\alpha} , |x-y|<\epsilon, l(x,y)=0, |x-y|>\epsilon $
Thus 
$$ ||k_\epsilon(x,y)||_{L^2xL^2} = \int^b_a\int^b_a |k_\epsilon(x,y)|^2dxdy=  \int^b_a\int^b_a |h(x,y)+l(x,y)|^2dxdy $$
$$ = \int^b_a\int^y_a |h(x,y)+l(x,y)|^2dx+\int^b_y |h(x,y)+l(x,y)|^2dxdy $$
$$ = \int^b_a\int^{y-\epsilon}_a \frac{1}{(y-x)^{2\alpha}} dx+\int^b_{y+\epsilon}\frac{1}{(x-y)^{2\alpha}} dxdy $$
For $\alpha=1/2$, we have
$$ = \int^b_a -[\log(y-x)]^{y-\epsilon}_a + [\log(x-y)]^b_{y+\epsilon}dy $$
or 
$$ = \int^b_a -2\log(\epsilon)dy+  \int^b_a\log(y-a)dy+ \int^b_a\log(b-y)dy $$
changing variables,
$$ = \int^b_a -2\log(\epsilon)dy+ 2\int^{b-a}_0\log(z)dy $$
Integrating,
$$ =(b-a)\log( \epsilon^{-2})+ \lim_{\eta\to} 2[z\log(z)-z]^{b-a}_\eta $$
$$ =(b-a)\log( \epsilon^{-2})+ 2[(b-a)\log(b-a)-(b-a)-\lim_{\eta\to 0} (\eta\log(\eta)-\eta)  $$
Taking the limit using L'Hopital's rule, 
$$= (b-a)\log( \epsilon^{-2})+ 2[(b-a)\log(b-a)-(b-a)$$
 A: Consider $\alpha< 1/2 $
$$ ||k(x,y)||_{L^2xL^2} = \int^b_a\int^b_a |k(x,y)|dxdy=  \int^b_a\int^b_a \frac{1}{|x-y|^{2\alpha}} dxdy $$
So splitting this up, 
$$ \int^b_a\int^y_a \frac{1}{(y-x)^{2\alpha}} dx+ \int^b_a \frac{1}{(x-y)^{2\alpha}} dx dy $$
Since $\alpha< 1/2 $, $1-2\alpha >0 $, so integrating,
$$ =(2\alpha+1)^{-1}\int^b_a -[(y-x)^{1-2\alpha}]^y_a  +  [(x-y)^{1-2\alpha}]^b_y dy $$
and
$$ =(2\alpha+1)^{-1}\int^b_a (y-a)^{1-2\alpha} +  (b-y)^{1-2\alpha} dy $$
Integrating again, 
$$ =(1-2\alpha)^{-1}(2-2\alpha)^{-1}( [(y-a)^{2-2\alpha}]^b_a + - [(b-y)^{2-2\alpha}]^b_a ) $$
or
$$ =(1-2\alpha)^{-1}(2-2\alpha)^{-1}( (b-a)^{2-2\alpha}+  (b-a)^{2-2\alpha}) $$
So 
$$ ||k(x,y)||_{L^2xL^2} = 2(1-2\alpha)^{-1}(2-2\alpha)^{-1}(b-a)^{2-2\alpha} < \infty$$
Thus $K$ is H.S.
Part 2.
Consider $\alpha <1$. 
$$ ||k(x,y)||_{L^2xL^2} = \int^b_a\int^b_a |k(x,y)|dxdy=  \int^b_a\int^b_a \frac{1}{|x-y|^{2\alpha}} dxdy $$
The hint is to split the kernel $k$ into 
$$ k_\epsilon(x,y) = h(x,y)+l(x,y) $$ 
Where $l$ vanishes for $|x-y|> \epsilon$.  
My idea is to split the kernel into $h(x,y)=\frac{1}{|x-y|^\alpha} $, $l(x,y)=-\frac{1}{|x-y|^\alpha} , |x-y|<\epsilon, l(x,y)=0, |x-y|>\epsilon $
Thus 
$$ ||k_\epsilon(x,y)||_{L^2xL^2} = \int^b_a\int^b_a |k_\epsilon(x,y)|^2dxdy=  \int^b_a\int^b_a |h(x,y)+l(x,y)|^2dxdy $$
$$ = \int^b_a\int^y_a |h(x,y)+l(x,y)|^2dx+\int^b_y |h(x,y)+l(x,y)|^2dxdy $$
$$ = \int^b_a\int^{y-\epsilon}_a \frac{1}{(y-x)^{2\alpha}} dx+\int^b_{y+\epsilon}\frac{1}{(x-y)^{2\alpha}} dxdy $$
For $\alpha=1/2$, we have
$$ = \int^b_a -[\log(y-x)]^{y-\epsilon}_a + [\log(x-y)]^b_{y+\epsilon}dy $$
or 
$$ = \int^b_a -2\log(\epsilon)dy+  \int^b_a\log(y-a)dy+ \int^b_a\log(b-y)dy $$
changing variables,
$$ = \int^b_a -2\log(\epsilon)dy+ 2\int^{b-a}_0\log(z)dy $$
Integrating,
$$ =(b-a)\log( \epsilon^{-2})+ \lim_{\eta\to} 2[z\log(z)-z]^{b-a}_\eta $$
$$ =(b-a)\log( \epsilon^{-2})+ 2(b-a)\log(b-a)-(b-a)-\lim_{\eta\to 0} (\eta\log(\eta)-\eta)  $$
Taking the limit using L'Hopital's rule, 
$$= (b-a)\log( \epsilon^{-2})+ 2(b-a)\log(b-a)-2(b-a)$$
Now consider $1/2 < \alpha< 1$, 
$$ =(1-2\alpha)^{-1} \int^b_a [-(y-x)^{1-2\alpha}]^{y-\epsilon}_a +
[(x-y)^{1-2\alpha}]^b_{y+\epsilon}
dy $$
or
$$ =(1-2\alpha)^{-1} \int^b_a 
-(\epsilon)^{1-2\alpha}
+(y-a)^{1-2\alpha}
+(b-y)^{1-2\alpha}
-(\epsilon)^{1-2\alpha}
dy $$
Integrating again,
$$ =(1-2\alpha)^{-1} (2-2\alpha)^{-1} (
-2(b-a)(\epsilon)^{1-2\alpha}
+[(y-a)^{2-2\alpha}]^b_a
+[-(b-y)^{2-2\alpha}]^b_a) $$
or since $1/2 < \alpha <1$, $1 > 2-2\alpha > 0$
$$ =(1-2\alpha)^{-1} (2-2\alpha)^{-1} (
-2(b-a)(\epsilon)^{1-2\alpha}
+2(b-a)^{2-2\alpha} $$
So for $\epsilon>0$ and  $1/2 \leq \alpha <1 $, this operator has a Hilbert Schmidt kernel and thus is compact.  We also see that 
$$ ||K-K_\epsilon||=|| k-k_\epsilon||_{L^2\times L^2}=\int^b_a \int_{|x-y|<\epsilon} |k(x,y)|^2 dx dy$$
As $\epsilon\to 0$, the FTC tells us that $K-K_\epsilon \to 0$, and so $K$ is compact since it is uniformly approximated by a compact operator. 
