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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 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 (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 $b-a\not=4$.

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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: 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 $g(\cos\theta)$.

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