Computing an integral depending on two parameters A few days ago I stumbled upon this integral
$$
f(y,c) :=\int_{-c-\sqrt{2}}^{-c+\sqrt{2}}\sqrt{2-(x+c)^2}\; \ln\left|\frac{x+y}{x-y}\right| \;\mathrm{d}x
$$
where $c>0$ and $y\in(-c-\sqrt{2},-c+\sqrt{2})$. After thinking for a while, it doesn't seem to be easy to solve (at least via standard tricks).
Thus, I wonder whether $f(y,c)$ admits a closed-form expression. Even an expression computed via symbolic math softwares would be fine (I don't currently have any of these softwares installed in my laptop...) Thanks for your help.
 A: First generalise your integral to depend on $a$, and change variables so that $x \to a(x - c)$. Then $$f(y,c,a) := a^2\int_{-1}^{1}\sqrt{1-x^2}\; \ln\left|\frac{x-\frac{c-y}{a}}{x-\frac{c+y}{a}}\right| \;\mathrm{d}x,$$ and choosing $a=\sqrt{2}$ reduces to your integral.
Now define $$g(A):= \int_{-1}^{1}\sqrt{1-x^2}\; \ln\left|x-A\right| \;\mathrm{d}x.$$ Your original $f$ can then be expressed as $$f(y,c,a) = a^2\left(g\left(\frac{c-y}{a}\right) - g\left(\frac{c+y}{a}\right)\right),$$ so we've reduced to a one-dimensional problem. Given the absolute value in the log, there are different cases depending on whether $|A|>1$ or not.
When $|A|\ge1$ the $\ln(0)$ singularity at $x=A$ lies outside of the range of integration, and we have that $$g(A)= \int_{-1}^{1}\sqrt{1-x^2}\; \ln\left(x+|A|\right) \;\mathrm{d}x.$$ In this region Mathematica gives a solution in terms of a hypergeometric function, $$g(A) = -\frac{\pi}{16}\left(\frac{1}{A^2}\;{}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix};\frac{1}{A^2}\right)  + 4\ln\left(\frac{1}{A^2}\right)\right).$$
We can write this expression in terms of elementary functions using the following identity for $\;{}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix};z\right)$, that $$z\;{}_3F_2\left(\begin{smallmatrix}1,1,\frac{3}{2}\\2,3 \end{smallmatrix};z\right) = -8\left(\frac{1-\sqrt{1-z}}{z} + \mathrm{arctanh}\left(\sqrt{1-z}\right) -\ln(2\sqrt{e})\right) -4\ln z.$$Inserting this into the expression for $g(A)$ we find that$$g(A) = \frac{\pi}{2}\left(A^2 - \sqrt{A^2(A^2-1)} + \mathrm{arctanh}\left(\sqrt{1-\frac{1}{A^2}}\right) - \ln(2\sqrt{e})\right),$$ for $|A|\ge1$.
For $|A|<1$ there is a singularity in the middle of the range of integration. To treat this carefully I believe you'd have to write$$g(A)= \int_{-1}^{A}\sqrt{1-x^2}\; \ln\left(A-x\right) \;\mathrm{d}x +  \int_{A}^{1}\sqrt{1-x^2}\; \ln\left(x-A \right) \;\mathrm{d}x$$ and take limits up to $A$ from the left in the first integral and from the right in the second. I don't know much about this, but an obvious question to ask as this point is; can we extend the expression for $g(A)$ when $|A|\ge1$ to $|A|<1$?
It's clear that this is not going to be true in the most naive way because $g(A)$ is real for $|A|<1$, but the expression for  $|A|\ge1$ is complex when $|A|<1$, eg. from the $\sqrt{A^2-1}$. However, all is not lost; numerically in Mathematica I find that
the integral matches up with the real part of the functional form I gave. So I have compelling numerical evidence that $$\int_{-1}^{1}\sqrt{1-x^2}\; \ln\left|x-A\right| \;\mathrm{d}x = \frac{\pi}{2}\Re\left(A^2 - \sqrt{A^2(A^2-1)} + \mathrm{arctanh}\left(\sqrt{1-\frac{1}{A^2}}\right) - \ln(2\sqrt{e})\right)$$ for all $A\in\mathbb{R},$ solving your problem in terms of elementary functions for all regions of parameter space.
