Integration of $\int_{-1}^{1}\frac{dx}{(x-a)\sqrt{1-x^2}}$ using Residue Calculus I would like to use residue calculus to prove that
$\int_{-1}^{1}\frac{dx}{(x-a)\sqrt{1-x^2}} = \frac{\pi}{\sqrt{a^2-1}}$ , 
where $a$ is a complex constant outside of the real interval $[-1, 1]$.
My idea for a strategy is to use a contour shaped like a dog-bone. It consists of two small circles (of radius $\varepsilon$) around $-1$ and $1$, that are connected by two segments near the real axis (one slightly above, and one slightly below). We know that the poles at $1$ and $-1$ are contained in the contour, and the pole at $z=a$ is not (for small enough $\varepsilon$). I would like to show integral goes to zero on the circle segments but I am having trouble doing this. Since the integrand is odd, we know that the segments along the axis will give us twice the required integral. By the residue theorem, we know that the integral over the contour equals $2\pi i$ times the sum of the residues of the poles contained within the contour. However, I am also having trouble calculating these two residues.
I don't think any other contour would work because we do not 
know where $a$ is on the complex plane. Any help would be greatly appreciated.
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Progress Update: I have figured out how to show that the integral goes to zero on the circle segments. However, I have also realized that the integrand is actually not odd (it was silly of me to make such a mistake). So now I am stuck trying to find a relationship between the integral along the top segment and the integral along the bottom segment. One of these is the desired integral, depending on the direction of the contour. But I do not know how to handle the other segment.
 A: Consider a contour that looks like this:
in the limit as $R$, the radius of $C_R$, goes to $\infty$. The contour should be constructed so that the point $z=a$ is inside the contour. The residue theorem then tells you
$$\oint_C\frac{dz}{(z-a)\sqrt{1-z^2}}=-\frac{2\pi i}{\sqrt{1-a^2}}=\frac{2\pi}{\sqrt{a^2-1}}$$
where $\sqrt{a^2-1}$ is to be evaluated on branch $\arg(a-1),\arg(a+1)\in [0,2\pi)$. You can show that the integral over $C_R$ goes to zero as $R\rightarrow \infty$. The integral over the segments of the contour parallel to the imaginary axis obviously cancel each other. And you stated that you have already shown that the integral over $C_\rho$ goes to zero as $\rho\rightarrow 0$. 
So now we must deal with the straight segments on opposite sides of the branch cut. We selected the branch with $\arg(1+z)\in [0,2\pi)$, $\arg(1-z)\in [-\pi,\pi)$. The discontinuity over the branch cut is captured by $\arg(1+z)$. So parametrize the upper segment using $z = \xi - 1$ and the lower segment by $z = \xi e^{2\pi i}-1$. Then we have:
\begin{align}
\oint_C\frac{dz}{(z-a)\sqrt{1-z^2}}&=\int_2^0\frac{d\xi}{(\xi -a-1)\sqrt{2-\xi}\sqrt{\xi}}+\int_0^2\frac{e^{2\pi i}d\xi}{(\xi e^{2\pi i} -a-1)\sqrt{2-\xi e^{2\pi i}}\sqrt{\xi e^{2\pi i}}}\\
&=\int_2^0\frac{d\xi}{(\xi -a-1)\sqrt{2-\xi}\sqrt{\xi}}-\int_0^2\frac{d\xi}{(\xi -a-1)\sqrt{2-\xi}\sqrt{\xi}} \\
&=\int_1^{-1}\frac{dx}{(x-a)\sqrt{1-x^2}}-\int_{-1}^1\frac{dx}{(x-a)\sqrt{1-x^2}} \qquad \text{(subst. }x=\xi-1\text{)} \\
&= -2\int_{-1}^1\frac{dx}{(x-a)\sqrt{1-x^2}} \\
&= \frac{2\pi}{\sqrt{a^2-1}}
\end{align}
From this you can conclude that
$$\int_{-1}^1\frac{dx}{(x-a)\sqrt{1-x^2}} = -\frac{\pi}{\sqrt{a^2-1}}$$
remembering that $\sqrt{a^2-1}$ is to be evaluated on branch $\arg(a+1),\arg(a-1) \in [0,2\pi)$.
A: $\sqrt{1-z^2}$ can be well-defined on $\mathbb{C}\setminus[-1,1]$ by
$$
\sqrt{1-z^2}=\sqrt2\exp\left(\int_i^z\frac{w\,\mathrm{d}w}{w^2-1}\right)
$$
where the path of integration does not pass through $[-1,1]$. Note that with this definition, $\sqrt{1-z^2}\sim-iz$ as $|z|\to\infty$.
The value along the top of $[-1,1]$ is the classical $\sqrt{1-x^2}$ while along the bottom of $[-1,1]$, the value is $-\sqrt{1-x^2}$.
Define the contours, as $R\to\infty$ and $\epsilon\to0$,
$$
\gamma_1=Re^{2\pi i[0,1]}
$$
and
$$
\gamma_2=[-1+i\epsilon,1+i\epsilon]\cup1+\epsilon e^{[1,-1]\pi i/2}\cup[1-i\epsilon,-1-i\epsilon]\cup-1+\epsilon e^{[3,1]\pi i/2}
$$
The dual contour $\gamma_1\cup\gamma_2$ surrounds $\mathbb{C}\setminus[-1,1]$ counter clockwise, and the only singularity inside is at $z=a$.
Furthermore,
$$
\int_{\gamma_1}\frac{\mathrm{d}z}{(z-a)\sqrt{1-z^2}}
=2\int_{-1}^1\frac{\mathrm{d}x}{(x-a)\sqrt{1-x^2}}
$$
and
$$
\int_{\gamma_2}\frac{\mathrm{d}z}{(z-a)\sqrt{1-z^2}}
=0
$$
Thus,
$$
\begin{align}
\int_{-1}^1\frac{\mathrm{d}x}{(x-a)\sqrt{1-x^2}}
&=\pi i\operatorname*{Res}_{z=a}\left(\frac1{(z-a)\sqrt{1-z^2}}\right)\\
&=\frac{\pi i}{\sqrt{1-a^2}}\bbox[5px,border:2px solid #C0A000]{\sim-\frac\pi{a}\text{ as }|a|\to\infty}\\
&=\bbox[5px,border:2px solid #C0A000]{\left\{\begin{array}{}
-\frac\pi{\sqrt{a^2-1}}&\text{if }a\gt1\\
\frac\pi{\sqrt{a^2-1}}&\text{if }a\lt-1
\end{array}\right.}
\end{align}
$$

The Integral is Odd
By substituting $x\mapsto-x$, we get
$$
\begin{align}
I(-a)
&=\int_{-1}^1\frac{\mathrm{d}x}{(x+a)\sqrt{1-x^2}}\\
&=-\int_{-1}^1\frac{\mathrm{d}x}{(x-a)\sqrt{1-x^2}}\\[6pt]
&=-I(a)
\end{align}
$$
