# Integral with branch cut ( Problem while calculating residue)

While calculating this integral $\int_{-1}^{1}\frac{dx}{\sqrt{1-x^2}(1+x^2)}$ , I am really struggling to calculate the residue at (-i), I am getting the value of residue as $\frac{-1}{2\sqrt{2}i}$, but for the value of residue is $\frac{1}{2\sqrt{2}i}$ with no minus sign in it. Now, what I might have been doing wrong since I used $\frac{1}{\sqrt{2}(1+(e^{i.3\pi/2})^2}$. I think I might have been doing some wrong in branch cut. Help me to figure it out.

• Please post the complex integral you are using. Impossible to answer otherwise. – David Nov 18 '14 at 2:42
• $\oint \frac{dz}{\sqrt{1-z^2}(1+z^2)}$ – Roshan Shrestha Nov 18 '14 at 2:46
• What exactly does $\sqrt{\phantom1}$ mean here? As you have pointed out yourself, it is necessary to know what branch you are using. – David Nov 18 '14 at 2:50
• If I use the branch at $2\pi$, what should I consider – Roshan Shrestha Nov 18 '14 at 2:52
• In that case, if $z=-i$ then $1-z^2$ is on the branch cut and there are going to be problems. – David Nov 18 '14 at 2:55

What you need here is a dog bone contour.

Let $\Gamma$ be the circular contour and $\gamma$ be the dog bone. On the bottom part of the dog bone, we get the negative of the square root and a negative for the direction.

\begin{align} 2\int_{-1}^1\frac{dz}{(z^2+1)\sqrt{1-z^2}} &= 2\pi i\sum\text{Res}\\ \int_{-1}^1\frac{dz}{(z^2+1)\sqrt{1-z^2}} &= \pi i\sum\text{Res} \end{align} where the residues are at $z=\pm i$. By the estimation lemma (see your other post on branch cuts), as $\epsilon\to 0$, the integration of the small circles go to zero and the integral of $\Gamma$ is zero as $R\to\infty$. Now evaluate the residues and you are done.

Edit:

To add some detail, (u line is the upper line of the dog bone and l line is the lower line) \begin{align} \int_{-1}^1\frac{dz}{(z^2+1)\sqrt{1-z^2}} &= \int_{\Gamma}fdz + \int_{\gamma_1}fdz + \int_{\gamma_2}fdz + \int_{\text{u line}}fdz - \int_{\text{l line}}fdz\tag{1}\\ &= \int_{\text{u line}}f(z)dz + \int_{\text{l line}}f(z)dz\tag{2}\\ 2\int_{-1}^1\frac{dz}{(z^2+1)\sqrt{1-z^2}} &= 2\pi i\sum\text{Res}\\ \int_{-1}^1\frac{dz}{(z^2+1)\sqrt{1-z^2}} &= \pi i\sum\text{Res}\\ &= \pi i\biggl[\lim_{z\to i}\frac{(z-i)}{(z^2+1)\sqrt{1-z^2}} - \lim_{z\to -i}\frac{(z+i)}{(z^2+1)\sqrt{1-z^2}}\biggr] \end{align} The last integral in (1) is subtracted since we pick up the negative from the square root in the lower half. In (2), the integral is now positive due to the negative from the direction of motion. As $R\to\infty$, $\int_{\Gamma}\to 0$, and as $\epsilon\to 0$, $\int_{\gamma_1,\gamma_2}\to 0$.

• I am still having trouble at finding residue at -i. My work is like this $a_{-1}(-i)=\frac{1}{\sqrt{2}.2e^{i3\pi/2}}$, but there is mistake in here and I haven't been able to figure it out. – Roshan Shrestha Nov 18 '14 at 16:06
• In my diagram of contour, there is no dog bone like structure, but the branch cut extends from -1 to 1 – Roshan Shrestha Nov 18 '14 at 16:07
• @RoshanShrestha that is a branch cut. The dog bone goes around the branch cut. You see in the diagram on my post the $-a$ and $a$? Consider that $-1$ to $1$ branch cut would connect them the dog bone goes around it. – dustin Nov 18 '14 at 16:21
• @dustin : I am really weak at this topic, so could you explain why you have used the dogbone contour to be clockwise rather than counterclockwise in this example? Thank you so much! – Heber Feb 12 '18 at 1:26
• @HeberSarmiento I would love to help you but unfortunately, I have been working in software development and data science so I cannot remember it has been awhile. Looking up dog bone contours should give you a good starting point on information. – dustin Feb 12 '18 at 5:44

The easier way is to rely on real analysis. Let $x=\sin(t)$. We then have the integral to be \begin{align} I & = \int_{-\pi/2}^{\pi/2} \dfrac{\cos(t)dt}{\cos(t) (1+\sin^2(t))}\\ & = \int_{-\pi/2}^{\pi/2} \dfrac{dt}{1+\sin^2(t)}\\ & = \sum_{k=0}^{\infty}(-1)^k \int_{-\pi/2}^{\pi/2} \sin^{2k}(t)dt\\ & = 2\sum_{k=0}^{\infty}(-1)^k \dfrac{\pi}{2^{2k+1}} \dbinom{2k}k\\ & = \pi \sum_{k=0}^{\infty} \left(-\dfrac14\right)^k \dbinom{2k}k \end{align} We have $$\sum_{k=0}^{\infty} x^k \dbinom{2k}k = \dfrac1{\sqrt{1-4x}} \text{ for }-1/4\leq x < 1/4$$ Hence, $$I = \dfrac{\pi}{\sqrt2}$$

• Thanks, but am not a student of pure maths, so it would be great to do it in other way around. – Roshan Shrestha Nov 18 '14 at 2:40
• @RoshanShrestha If you are not a pure math student, then I would believe that my technique is much easier than the complex analysis technique. – Adhvaitha Nov 18 '14 at 2:42
• Yeap, but I still have to solve it through complex analysis – Roshan Shrestha Nov 18 '14 at 2:43
• I think the complex analysis technique is easier. I only have to evaluate two residues and I am done without substitutions and a series. – dustin Nov 18 '14 at 3:22
• @dustin "easier" is a relative term. Complex analysis involves more machinery than simple real analysis. – Adhvaitha Nov 18 '14 at 3:24

We will do the residue calculation since it is necessary to conclude on this problem.

Suppose we seek to compute $$Q = \int_{-1}^{+1} \frac{dx}{\sqrt{1-x^2}(1+x^2)}.$$

Re-write this as $$\int_{-1}^{+1} \exp(-1/2\mathrm{LogA}(1+z)) \exp(-1/2\mathrm{LogB}(1-z)) \frac{1}{1+z^2} dz$$ and call the function $f(z).$

We must choose two branches of the logarithm $\mathrm{LogA}$ and $\mathrm{LogB}$ so that the cut is on the real axis from $-1$ to $+1.$ This is accomplished when $\mathrm{LogA}$ has the cut on the negative real axis and $\mathrm{LogB}$ on the positive real axis. The poles are simple so to compute the residues we merely need to evaluate the logarithms at these two points.

For the first one at $\rho_0=i$ we put $$1+i = \sqrt{2} e^{1/4\pi i}$$ and $$1-i = \sqrt{2} e^{7/4\pi i}$$ to get the residue $$\frac{1}{2i} \exp(-1/2\times \log\sqrt{2} -1/2\times 1/4\pi i) \exp(-1/2\times \log\sqrt{2} -1/2\times 7/4\pi i) \\ = \frac{1}{2i} \exp(-\log\sqrt{2} - \pi i) = - \frac{1}{2i\sqrt{2}}.$$

For the second one at $\rho_1=-i$ we put $$1+(-i) = \sqrt{2} e^{-1/4\pi i}$$ and $$1-(-i) = \sqrt{2} e^{1/4\pi i}$$ to get the residue

$$-\frac{1}{2i} \exp(-1/2\times \log\sqrt{2} +1/2\times 1/4\pi i) \exp(-1/2\times \log\sqrt{2} -1/2\times 1/4\pi i) \\ = - \frac{1}{2i} \exp(-\log\sqrt{2}) = - \frac{1}{2i\sqrt{2}}.$$

Now using the dogbone contour shown in the accepted answer and traversed counterclockwise we pick up $2Q.$ This is because for $x\in(-1,1)$ we get above the cut the value $$\exp(-1/2\log (1+x))\exp(-1/2\log (1-x) -1/2\times 2\pi i) = - \frac{1}{\sqrt{1-x^2}}$$ and below the cut $$\exp(-1/2\log (1+x))\exp(-1/2\log (1-x)) = \frac{1}{\sqrt{1-x^2}}.$$

Therefore $$2Q= -2\pi i \left(\mathrm{Res}_{z=i} f(z) + \mathrm{Res}_{z=-i} f(z) + \mathrm{Res}_{z=\infty} f(z)\right).$$

The residue at infinity is zero because $f(z)$ is $O(1/R^3)$ on the circle.

This gives $$2Q = - 2\pi i \left(- \frac{1}{2i\sqrt{2}} - \frac{1}{2i\sqrt{2}}\right) = \pi \frac{2}{\sqrt{2}} \quad\text{or}\quad Q = \frac{\pi}{\sqrt{2}}.$$

In order to be rigorous we also need to show continuity across the two overlapping cuts on $(-\infty, -1)$ as shown in this MSE link.

Remark. It really helps to think of the map from $z$ to $-z$ as a $180$ degree rotation when one tries to visualize what is happening here.