The so-called Symm's integral equation on an interval $[a,b]$ is defined by 
$$\int\_a^bu(y)\log|x-y|dy=f(x),\,\,x\in[a,b],$$ and $f$ is a given function.


In the introduction of a [paper](https://www.researchgate.net/publication/231939080_Collocation_with_Chebyshev_polynomials_for_Symm%27s_integral_equation_on_an_interval) by I. H. Sloan and E. P. Stephan on Symm's integral equations it is stated that 


"For $b-a\not= 4$ and $f$ suitably smooth, the solution to the Symm's equation is unique with end-point singularities of the form $\frac{1}{\sqrt{(x-a)(b-x)}}.$"


They reference to $\textit{ Lineare IntegralOperatoren, (Teubner-Verlag, Stuttgart, 1970)}$ by K. Jorgens, which I do not have access to. Here are my questions:


$\textbf{Q1.}$ I am very suspicious to their claim. One can take $u$ to be polynomial and after (tedious) computations you can see that $\int\_a^bu(x)\log|x-y|dy=Q(x,\log x)$, where $Q$ is a polynomial of two variable and clearly $\lim\_{x\to0^+}Q(x,\log x)=0$. No singularities what so ever. Is that a misunderstanding or what is the catch in here? 


$\textbf{Q2.}$ Assuming $f$ is sufficiently smooth, what $b-a$ being equal to 4 has to do with end-point singularity or is that "magical" 4 tied to uniqueness? 


$\textbf {Remark:}$ The computations provided by Christian Remling answers this question completely. I need to add that T. Carleman in [his paper](http://gdz.sub.uni-goettingen.de/pdfcache/PPN266833020_0015/PPN266833020_0015___LOG_0012.pdf) (1922) gives the solution to this integral equation as follows:


$u(x)=-\frac{1}{\sqrt{(x-a)(b-x)}}\int\_a^b\frac{\sqrt{(y-a)(b-y)}f'(y)dy}{y-x}-\frac{1}{\sqrt{(x-a)(b-x)}}\frac{1}{\log(b-a)-2\log2}\int\_a^b\frac{f(y)dy}{\sqrt{(y-a)(b-y)}}$


as long as $b-a\not=4$.



Regularity of solutions to certain integral equation
Let's first make the interval equal to $(-1,1)$ by the change of variable
$$
x = \frac{a+b}{2} + \frac{b-a}{2}\, t ;
$$
then the equation becomes
$$
\int\_{-1}^1 u(s)[c+\log |s-t|]\, ds = g(t) , \quad\quad\quad\quad (1)
$$
with $c=\log (b-a)/2$ and $g=2f/(b-a)$ (and I use the sloppy but convenient notation where $u(s)$ really means $u(x(s))$, and similarly for $g$).


Let $v(t)=\sqrt{1-t^2}u(t)$. This is useful because the eigenfunctions of the (Hilbert-Schmidt) operator
$$
(Kv)(t) = \int\_{-1}^1 [c+\log |s-t|]v(s)\, \frac{ds}{\sqrt{1-s^2}}
$$
in $L^2(-1,1)$
are [explicitly known:](https://books.google.com/books?id=nNPOjwXwVgIC&pg=PA17&lpg=PA17&dq=chebyshev%20polynomials%20eigenfunctions%20integral%20operator&source=bl&ots=DF3Xf1HKTY&sig=KlhnkqOdgaVRtkOycaAmmHW-PWE&hl=en&sa=X&ved=0ahUKEwifs73Z4_TOAhVE5SYKHS2zAlUQ6AEIKjAC#v=onepage&q=chebyshev%20polynomials%20eigenfunctions%20integral%20operator&f=false) they are given by the Chebyshev polynomials $T\_n(t)=\cos(n\arccos t)$, and the eigenvalues are $\lambda\_n=-\pi/n$ for $n\ge 1$ and $$
\lambda\_0=\pi(c-\log 2) .
$$


Now we see why $b-a=4$ is special: this leads to $\lambda\_0=0$, and thus solvability of (1) is not guaranteed in this case; we need $g$ to be orthogonal to constants, and then we don't have uniqueness.


In the other case, the unique solution is given by
$$
u(t) = \frac{1}{\sqrt{1-t^2}} \sum\_{n\ge 0} \frac{g\_n}{\lambda\_n} T\_n(t)
$$
This has the asserted behavior near $t=\pm 1$, at least if the expansion coefficients $g\_n$ decay rapidly. This will be the case for smooth (on the circle) $g$ because the $a\_n$'s are also the Fourier coefficients of $g(\cos\theta)$.


