I'm asked to find the value of


where $a$ is complex and $a\not\in[-1, 1]$.

I think I should use Cauchy's integration formula but don't know how to apply it.


Consider the following contour integral:

$$\oint_C dz \frac{1}{(z-a) \sqrt{z^2-1}} $$

where $C$ is the following contour for $a \gt 1$:

enter image description here

The outer circle has radius $R$ and the small circular arcs about the branch points at $z=\pm 1$ have radius $\epsilon$. The contour integral is then

$$\int_{-R}^{-1-\epsilon} \frac{dx}{(x-a)\sqrt{x^2-1}} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{1}{(-1+\epsilon e^{i \phi}-a) \sqrt{(-1+\epsilon e^{i \phi})^2-1}}\\ + \int_{-1+\epsilon}^{1-\epsilon} \frac{dx}{(x-a) (i) \sqrt{1-x^2}}+ i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{1}{(1+\epsilon e^{i \phi}-a) \sqrt{(1+\epsilon e^{i \phi})^2-1}}\\ + \int_{1-\epsilon}^{-1+\epsilon} \frac{dx}{(x-a) (-i) \sqrt{1-x^2}} + + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{1}{(-1+\epsilon e^{i \phi}-a) \sqrt{(-1+\epsilon e^{i \phi})^2-1}}\\+ \int_{-1-\epsilon}^{-R} \frac{dx}{(x-a)\sqrt{x^2-1}} + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac1{(R e^{i \theta}-a) \sqrt{R^2 e^{i 2 \theta}-1}}$$

Note that, as $R \to \infty$ and $\epsilon \to 0$, all integrals vanish except the third and fifth, which combine to be, as the contour integral,

$$-i 2 \int_{-1}^1 \frac{dx}{(x-a) \sqrt{1-x^2}} $$

The contour integral is, by the residue theorem, equal to $i 2 \pi$ times the residue at the pole $z=a$, so that

$$\int_{-1}^1 \frac{dx}{(x-a) \sqrt{1-x^2}} = -\frac{\pi}{\sqrt{a^2-1}} \quad a \gt 1$$

Of course, this holds for $a$ complex and not on the negative real axis. You can show that, when $a$ is real and $a \lt 1$, by either reversing the above diagram or introducing semicircular detours in the paths to the left off the branch point at $z=-1$, we have

$$\int_{-1}^1 \frac{dx}{(x-a) \sqrt{1-x^2}} = -\frac{\pi}{\sqrt{a^2-1}} \quad a \in \mathbb{C} \cap a \notin [-1,1]$$

  • $\begingroup$ Actually, your first integral should 'not be there' - correct? The top and bottom paths should cancel. $\endgroup$ – peter a g Dec 8 '15 at 15:07
  • $\begingroup$ @peterag: thanks for the catch! Yes, the two integrals along the out paths cancel because the effects of the two branch cuts indeed cancel away from $[-1,1]$. $\endgroup$ – Ron Gordon Dec 8 '15 at 15:31
  • $\begingroup$ My pleasure of course - so it might be clearer to include the cancellation explanation in your 'Note that, as..." sentence. $\endgroup$ – peter a g Dec 8 '15 at 15:35

I would do it like this:

First, multiply and divide by $x+a$, and use the fact that the $x$-part in the numerator leads to an odd integrand and thus to a zero contribution. Using the fact that the remaining integrand is even, your integral equals $$ \int_{0}^{1}\frac{2a}{(x^2-a^2)\sqrt{1-x^2}}\,dx $$ In this integral, we let $$ t=\frac{x}{\sqrt{1-x^2}}. $$ This transforms the integral into $$ -\int_0^{+\infty} \frac{2a}{a^2+(a^2-1)t^2}\,dt. $$ The primitive will now be an arctan, and I leave the final calculation to you.

I get, as a final answer, $$ -\frac{\pi}{a\sqrt{1-1/a^2}}.$$


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