The problem is to evaluate the following integral:

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

where $a \gt 1$ and $|b| \lt 1$.

It should be obvious to those who spend time around these integrals that this integral does not converge as stated. However, we only have a simple pole at $x=-b$ so that we can compute the Cauchy principal value of the integral. This is what we will do.

However, the computation is a little tricky. We actually have two branch cuts: one along the integration interval $(-1,1)$; the other along $(-\infty,-a)$ due to the log. How we define the values of the complex-valued version of the integrand just above and below one branch cuts will depend on how we define such values just above and below the other branch cut.

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

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

Your skill in evaluating these residue theorem integrals is world class, dude. I hope it doesn’t come off as blowing smoke up your ass, but I plan to work through every problem on this site for fun after I finish my phd.

I admire your computational skills. I’m currently studying complex analysis, and as Bennett, I also plan to work through all the problems on your blog some day.

Hi Ron,

Can you tell me where I can find

properties of Li2 functions,used in the

evaluation of integrals.

example:

integral from 0 to1 log x/(x-b)dx=

Li2(1/b)

Thank you for your help.

My e-mail

gil.decraemer@skynet.be