TITLE: Branch cuts for the contour integral $\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx$
QUESTION [6 upvotes]: How can the branch cut be handled in the contour integral, for $|b| \leq 1, \, a > 1$,
$$\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx \quad ?$$
If $a=1$ can the value of the integral be shown to be zero?

Edit: The contour involves a branch cut from $-a + \sqrt{a^{2}-1}$ to $\infty$ along the $x$-axis, a branch cut from $- a - \sqrt{a^{2}-1}$ to $- \infty$ along the $x$-axis. There are two poles $- z_{b} = e^{i(\theta_{b} - \pi)}$ and $- \frac{1}{z_{b}} = e^{-i(\theta_{b}+\pi)}$ where $\theta_{b} = \cos^{-1}(b)$. The resulting proposed value is:
\begin{align}
\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx = \frac{2 \pi}{\sqrt{1-b^{2}}} \, \tan^{-1}\left[ \frac{\sqrt{1-b^{2}} \, (a - \sqrt{a^{2}-1})}{1 - b \, (a - \sqrt{a^{2}-1})} \right].
\end{align}

REPLY [6 votes]: I offer this solution as an alternative, but equivalent, perspective as the one given by @Sangchul Lee.  My idea, although a bit messy, is to consider a contour integral over a single contour that encompasses the three branch points and the pole.
So, let's consider
$$\oint_C dz \frac{\log{(z+a)}}{(z+b)\sqrt{z^2-1}} $$
where $a \gt 1$, $|b| \lt 1$, and $C$ is the following contour:

Here I have constructed the contour assuming $a=2$ and $b=0.2$, but the contour hopefully makes clear what is going on for any of the allowed values of $a$ and $b$.  The outer circular arcs have a large radius $R$ and the circular detours have small radius $\epsilon$.
The next step is to write out the contour integral in terms of integrals over the various pieces of the contour $C$. So, without further ado, here are the 16 (!) pieces of the contour integral.  Note that I am beginning at the small circle about $z=-1$ and am traversing the contour $C$ counterclockwise.
$$i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log{\left (-1+\epsilon e^{i \phi}+a \right )}}{\left (-1+\epsilon e^{i \phi}+b \right )\sqrt{\left (-1+\epsilon e^{i \phi} \right )^2-1}} +\int_{-1+\epsilon}^{-b-\epsilon} dx \frac{\log{(x+a)}}{(x+b) \sqrt[+]{x^2-1}}\\ + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log{\left (-b+\epsilon e^{i \phi}+a \right )}}{\left (\epsilon e^{i \phi} \right )\sqrt[+]{\left (-b+\epsilon e^{i \phi} \right )^2-1}}+\int_{-b+\epsilon}^{1-\epsilon} dx \frac{\log{(x+a)}}{(x+b) \sqrt[+]{x^2-1}}\\ + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log{\left (1+\epsilon e^{i \phi}+a \right )}}{\left (1+\epsilon e^{i \phi}+b \right )\sqrt{\left (1+\epsilon e^{i \phi} \right )^2-1}}+\int_{1+\epsilon}^{R} dx \frac{\log{(x+a)}}{(x+b) \sqrt{x^2-1}}\\ +i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{\log{\left ( R e^{i \theta}+a\right )}}{\left ( R e^{i \theta}+b\right ) \sqrt{R^2 e^{i 2 \theta}-1}} + e^{i \pi} \int_R^{a+\epsilon} dx \frac{\log^{(+)}{(a-x)}}{(b-x) \sqrt{x^2-1}}\\ + i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{\log{\left (\epsilon e^{i \phi} \right )}}{\left (-a+\epsilon e^{i \phi}+b \right )\sqrt{\left (-a+\epsilon e^{i \phi} \right )^2-1}}+ e^{-i \pi} \int_{a+\epsilon}^R dx \frac{\log^{(-)}{(a-x)}}{(b-x) \sqrt{x^2-1}}\\ +i R \int_{-\pi}^{0} d\theta \, e^{i \theta} \frac{\log{\left ( R e^{i \theta}+a\right )}}{\left ( R e^{i \theta}+b\right ) \sqrt{R^2 e^{i 2 \theta}-1}}+\int_R^{1+\epsilon} dx \frac{\log{(x+a)}}{(x+b) \sqrt{x^2-1}}\\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\log{\left (1+\epsilon e^{i \phi}+a \right )}}{\left (1+\epsilon e^{i \phi}+b \right )\sqrt{\left (1+\epsilon e^{i \phi} \right )^2-1}}+\int_{1-\epsilon}^{-b+\epsilon} dx \frac{\log{(x+a)}}{(x+b) \sqrt[-]{x^2-1}} \\+ i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{\log{\left (-b+\epsilon e^{i \phi}+a \right )}}{\left (\epsilon e^{i \phi} \right )\sqrt[-]{\left (-b+\epsilon e^{i \phi} \right )^2-1}}+\int_{-b-\epsilon}^{-1+\epsilon} dx \frac{\log{(x+a)}}{(x+b) \sqrt[-]{x^2-1}}$$ 
I used the notations $\log^{(+)}$ and $\log^{(-)}$ to denote the respective values of the log along the upper and lower branches above and below the negative real axis when $x \lt -a$.  I also used the notations $\sqrt[+]{}$ and $\sqrt[-]{}$ to denote the respective values of the square root along the upper and lower branches above and below the positive real axis when $x \in (-1,1)$.  
What are the values of these functions just above and below the branch cuts?  We may define
$$\log^{(+)}{(a-x)} = \log{(x-a)}+i \pi$$
$$\log^{(-)}{(a-x)} = \log{(x-a)}-i \pi$$ 
$$\sqrt[+]{x^2-1} = -i \sqrt{1-x^2} $$
$$\sqrt[-]{x^2-1} = +i \sqrt{1-x^2} $$
Note that, for the square roots, we define the signs this way because of the way we defined the branch cut for the log.  That is, we define the upper branch of the log as having an argument of $\pi$ and the lower branch of the log as having an argument of $-\pi$.  For continuity, we then define the lower branch of the square root as having an argument of $0$, while the upper branch of the square root as having an argument of $2 \pi$.  Taking the square roots sorts out the signs.
Now, we may simplify things greatly by observing that all integrals about the circular arcs vanish as $R \to \infty$ and $\epsilon \to 0$.  This goes for the two arcs about the pole $z=-b$ because, while the two arcs are traversed in the same direction, they involve opposite signs of the square root.  I leave it to the reader to verify that this is indeed true.
That leaves the following for the contour integral:
$$\int_a^{\infty} dx \frac{\log{(x-a)}+i \pi - \log{(x-a)} + i \pi}{(b-x) \sqrt{x^2-1}} + PV \int_{-1}^1 dx \frac{\log{(x+a)}}{(x+b) (-i) \sqrt{1-x^2}} \\+ PV \int_1^{-1} dx \frac{\log{(x+a)}}{(x+b) (i) \sqrt{1-x^2}}$$
By Cauchy's theorem, the contour integral is zero.  (We have already taken care of the pole at $z=-b$ by excluding it from the interior of $C$ using those semicircular detours.)  Thus,
$$PV \int_{-1}^1 dx \frac{\log{(x+a)}}{(x+b) \sqrt{1-x^2}} = \pi \int_a^{\infty} \frac{dx}{(x-b) \sqrt{x^2-1}} $$
Now we may evaluate the integral using a few substitutions.  Begin by subbing $x=\cosh{y}$; the integral is then
$$\begin{align} \int_a^{\infty} \frac{dx}{(x-b) \sqrt{x^2-1}} &= \int_{\log{(a+\sqrt{a^2-1})}}^{\infty} \frac{dy}{\cosh{y}-b} \\ &= 2 \int_{a+\sqrt{a^2-1}}^{\infty} \frac{du}{u^2-2 b u+1}\\ &= 2 \int_{a+\sqrt{a^2-1}}^{\infty} \frac{du}{(u-b)^2+1-b^2}\end{align}$$
Therefore

$$PV \int_{-1}^1 dx \frac{\log{(x+a)}}{(x+b) \sqrt{1-x^2}} = \frac{2 \pi}{\sqrt{1-b^2}} \arctan{\left (\frac{\sqrt{1-b^2}}{a-b+\sqrt{a^2-1}} \right )}$$