# Branch cuts for the contour integral $\int_{-1}^{1} \frac{\ln(x+a)}{(x+b) \, \sqrt{1-x^{2}}} \, dx$

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}

• Do you know that $b\not\in[-1,1]$? – A.Γ. Aug 11 '15 at 19:16
• @A.G. I added the ranges for $a$ and $b$ – Leucippus Aug 11 '15 at 19:32
• Does the integral really converge if $|b|\leq 1$? Do you mean $|b|>1$? – mickep Aug 11 '15 at 19:38
• @Leucippus Are you committed to a contour integral approach? If you're main goal is to evaluate the integral, there's a very nice way to solve it via elementary methods. – David H Aug 11 '15 at 20:03
• Please avoid titles like "BLAH BLAH I", "SAME BLAH II". Try to come up with descriptive and informative titles instead. – Pedro Tamaroff Aug 11 '15 at 22:33

The problem of the integrand is that, if we use the principal branches then the singular points somehow overlap, which makes it hard to deal with directly. So if you want to apply complex analysis technique, we need some modification.

Notice that the integral

$$I = \mathrm{PV}\!\int_{-1}^{1} \frac{\log(x+a)}{(x+b)\sqrt{1-x^2}} \, dx$$

exists in principal value sense, due to the presence of the pole at $x = -b$. One way to circumvent this problem is to move the pole slightly away from the line of integration. So we introduce the following function:

$$I(w) = \int_{-1}^{1} \frac{\log(x+a)}{(x+w)\sqrt{1-x^2}} \, dx.$$

Then we can express $I$ as

$$I = \lim_{s \downarrow 0} \frac{I(b+is) + I(b-is)}{2}. \tag{1}$$

So it suffices to deal with $I(w)$ when $w$ lies outside the line $[-1, 1]$. Now choose a contour $C$ that winds $[-1, 1]$ counter-clockwise:

Using the principal square root, introduce the function $f(z) = i\sqrt{z-1}\sqrt{z+1}$. It is easy to see that the branch cut of $f$ is $[-1, 1]$ and that for $-1 < r < 1$,

$$f(r + 0^+i) = -\sqrt{1-r^2}, \qquad f(r - 0^+i) = \sqrt{1-r^2}.$$

Then the integral is written as

$$I(w) = \frac{1}{2} \oint_C \frac{\log(z+a)}{(z+w)f(z)} \, dz.$$

Now deform this contour as follows:

This can be done by enlarging the contour $C$ while wrapping around the singular points. (And it is possible since the integrand grows like $\mathcal{O}(\log R/ R^2)$ as the radius $R \to \infty$.) Then it follows that

\begin{align*} I(w) &= \pi i \int_{-\infty}^{-a} \frac{dx}{(x+w)f(x)} - \pi i \underset{z=-w}{\operatorname{Res}} \frac{\log(z+a)}{(z+w)f(z)} \\ &= \pi \int_{a}^{\infty} \frac{dx}{(x-w)\sqrt{x^2 - 1}} - \frac{\pi \log(a-w)}{\sqrt{-w-1}\sqrt{-w+1}} \end{align*}

Plugging this to (1), we have

\begin{align*} I &= \pi \int_{a}^{\infty} \frac{dx}{(x-b)\sqrt{x^2 - 1}} \\ &= \frac{2\pi}{\sqrt{1-b^2}} \left( \arctan\sqrt{\frac{1+b}{1-b}} - \arctan\sqrt{\frac{1+b}{1-b}\cdot\frac{a-1}{a+1}} \right). \end{align*}

• Quite nice (+1) – tired Aug 12 '15 at 10:13
• @SangchulLee: Nice and instructive answer (+1) – Markus Scheuer Aug 12 '15 at 13:28

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

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