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.


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.

  • $\begingroup$ The integrand is not odd, however, it is a simple exercise to show that the integral is odd (the substitution $x\mapsto-x$ plays a big role here). $\endgroup$ – robjohn Feb 16 '18 at 18:49

Consider a contour that looks like this:enter image description here

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 $\arg(a+1),\arg(a-1) \in [0,2\pi)$.

  • $\begingroup$ Wow, thanks so much! Do you mind explaining to me how you got from your first line to your second line when calculating the integral over the upper and lower segments.? Somehow you changed the sign on the second integral, but I am unsure why we can do this. Edit: Is it related to this paragraph? "So now we must deal with the straight segments on opposite sides of the branch cut. We selected the branch with arg(1−z),arg(1+z)∈[0,2π). The discontinuity over the branch cut is captured by arg(1+z). " This is the only other part that I did not really understand. $\endgroup$ – Ken Feb 4 '18 at 18:59
  • $\begingroup$ I'm good, thanks! Sorry I didn't realize that I had to choose an answer (I'm new to this website). I've selected your answer. Thanks again for all the help. $\endgroup$ – Ken Feb 5 '18 at 2:03
  • $\begingroup$ As I mentioned in a comment to Jack D'Aurizio's answer, this works fine for $a\gt1$, but not for $a\lt-1$, and it's questionable how you want to extend $\sqrt{a^2-1}$ for other $a\in\mathbb{C}\setminus[-1,1]$. $\endgroup$ – robjohn Feb 16 '18 at 16:20
  • $\begingroup$ @robjohn You are absolutely correct if you were to interpret my rearrangement of the residue as using the principal branch with $\arg(a^2-1)\in(-\pi,\pi)$. That was not my intention, however, and I should have been much more clear about that. My intention was to express the value of the integral using branch $\arg(a-1),\arg(a+1)\in [0,2\pi)$. That way, I don’t need to write the value of the integral using cases. For $a<-1$, this should agree with your answer below. $\endgroup$ – bames Feb 16 '18 at 21:22

I believe that to use Complex Analysis is just an overkill here. We want to compute $$ -\int_{-\pi/2}^{\pi/2}\frac{d\theta}{a-\sin\theta}=-\int_{0}^{\pi/2}\frac{2a\,d\theta}{a^2-\sin^2\theta}=-\int_{0}^{\pi/2}\frac{2a\,d\theta}{a^2-\cos^2\theta}$$ for some $a$ such that $a>1$, and by substituting $\theta=\arctan u$ in the last integral we get $$ -\int_{0}^{\pi/2}\frac{2a\,d\theta}{a^2-\cos^2\theta} = -\int_{0}^{+\infty}\frac{2a\,du}{a^2(u^2+1)-1}=\color{red}{-\frac{\pi}{\sqrt{a^2-1}}} $$ without hunting any contour. The $a<-1$ case can be dealt in a similar way.

  • 1
    $\begingroup$ The question states that $a\in\mathbb{C}\setminus[-1,1]$. Does this answer handle $a\lt-1$? $\endgroup$ – robjohn Feb 16 '18 at 16:00
  • $\begingroup$ Jack, also, the question explicitly asks how to evaluate the integral using residues. Whether or not you think it is overkill, that is the question. (I happen to disagree, but that too is irrelevant.) $\endgroup$ – Ron Gordon Feb 16 '18 at 17:15
  • $\begingroup$ @RonGordon: the OP still is free to compute $\int_{-\infty}^{+\infty}\frac{a}{a^2(u^2+1)-1}$ by using a standard semi-circle contour. I agree this approach deviates from the exercise requests, but I also think our hands should not be tied by exercises asking us to follow slightly inefficient approaches. $\endgroup$ – Jack D'Aurizio Feb 16 '18 at 17:38
  • $\begingroup$ @JackD'Aurizio: the problem may have been given to practice contour integration. In that respect, a real method answer is not appropriate. However, a real method answer does show an alternative (and possibly simpler) approach, and would serve to validate an answer gotten via contour integration. In any case, this answer does not work for $a\lt-1$. $\endgroup$ – robjohn Feb 16 '18 at 19:02

$\sqrt{1-z^2}$ 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 $$ \begin{align} I(-a) &=\int_{-1}^1\frac{\mathrm{d}x}{(x+a)\sqrt{1-x^2}}\\ &=-\int_{-1}^1\frac{\mathrm{d}x}{(x-a)\sqrt{1-x^2}}\\[6pt] &=-I(a) \end{align} $$

  • $\begingroup$ Minor nitpick, but the way you have defined your contour, it is your integral over $\gamma_1$ that should go to $0$ and not the one over $\gamma_2$. $\endgroup$ – bames Feb 17 '18 at 1:20

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