# Integral $\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx$

Regarding this problem, I conjectured that

$$I(r, s) = \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx = 4 \pi \operatorname{arccot} \sqrt{ \frac{2r + 2\sqrt{r^{2} - s^{2}}}{s^{2}} - 1}.$$

Though we may try the same technique as in the previous problem, now I'm curious if this generality leads us to a different (and possible a more elegant) proof.

Indeed, I observed that $I(r, 0) = 0$ and

$$\frac{\partial I}{\partial s}(r, s) = \int_{0}^{\infty} \left\{ \frac{2\sqrt{y}}{(r-s)y^{2} + 2(2-r)y + (r+s)}+\frac{2\sqrt{y}}{(r+s)y^{2}+ 2(2-r)y + (r-s)} \right\} \,\mathrm dy,$$

which can be evaluated using standard contour integration technique. But simplifying the residue and integrating them seems still daunting.

EDIT. By applying a series of change of variables, I noticed that the problem is equivalent to prove that

$$\tilde{I}(\alpha, s) := \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{ 1 + 2sx \sin\alpha + (s^{2} - \cos^{2}\alpha) x^{2}}{ 1 - 2sx \sin\alpha + (s^{2} - \cos^{2}\alpha) x^{2}} \right) \, \mathrm dx = 4\pi \alpha$$

for $-\frac{\pi}{2} < \alpha < \frac{\pi}{2}$ and $s > 1$. (This is equivalent to the condition that the expression inside the logarithm is positive for all $x \in \Bbb{R}$.)

Another simple observation. once you prove that $\tilde{I}(\alpha, s)$ does not depend on the variable $s$ for $s > 1$, then by suitable limiting process it follows that

$$\tilde{I}(\alpha, s) = \int_{-\infty}^{\infty} \log \left( \frac{ 1 + 2x \sin\alpha + x^{2}}{ 1 - 2x \sin\alpha + x^{2}} \right) \, \frac{\mathrm dx}{x},$$

which (I guess) can be calculated by hand. The following graph may also help us understand the behavior of this integral.

• I believe that the method I outlined in that solution is not the best possible, as I ended up exploiting much symmetry in deriving the roots and getting an incredibly simple value for the residues. What it is...someone will likely uncover it and make what I did seem like child's play. That said, I can derive an integral much like the one I did for your expression and thereby prove your conjecture, but it will be about as pretty. – Ron Gordon Nov 16 '13 at 3:08

So, following the procedure I outlined here, I get for the transformed integral:

$$I(r,s) = \int_0^{\infty} dv \frac{4 s \left(v^2-1\right) \left(v^4-(4 r-6) v^2+1\right)}{v^8+4 \left(2 r-s^2-1\right) v^6 +2 \left(8 r^2-8 r-4 s^2+3\right) v^4 +4 \left(2 r-s^2-1\right) v^2 +1} \log{v}$$

Note that this reduces to the integral in the original problem when $r=3$ and $s=2$. Then we see that the roots of the denominator satisfy the same symmetries as before, so we need only find one root of the form $\rho e^{i \theta}$ where

$$\rho = \sqrt{\frac{r+\sqrt{r^2-s^2}}{2}} + \sqrt{\frac{r+\sqrt{r^2-s^2}}{2}-1}$$

and

$$\theta = \arctan{\sqrt{\frac{2 \left (r+\sqrt{r^2-s^2}\right )}{s^2}-1}}$$

Using the same methodology I derived, I am able to confirm your conjecture.

Just for references, I remark that the following proposition was proved in my answer:

Proposition. If $0 < r < 1$ and $r < s$, then $$I(r, s) := \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx = 4\pi \arcsin r. \tag{*}$$

Recently, I found an alternate proof which is much simpler and does not use contour integration technique.

Lemma 1. For any $k = 0, 1, 2, \cdots$ we have $$\int_{0}^{1} \frac{x^{2k}}{\sqrt{1-x^{2}}} \, dx = (-1)^{k} \frac{\pi}{2} \binom{-1/2}{k}.$$

Since this is so famous, we skip the proof.

Lemma 2. For any $z \in \Bbb{C}$ with $|z| \leq 1$, we have $$f(z) := - \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log(1 - zx) \, dz= \pi \sin^{-1} z - \pi \log \left( \tfrac{1}{2}+\tfrac{1}{2}\sqrt{1-z^{2}} \right) . \tag{1}$$

Proof of Lemma. Expand $-\log(1-zx)$ using the MacLaurin series. Then we have

$$f(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n} \int_{-1}^{1} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx. \tag{2}$$

To identify the coefficient, we observe that

\begin{align*} \int_{-1}^{1} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx &= \int_{0}^{1} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx + \int_{-1}^{0} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx \\ &= \int_{0}^{1} x^{n-1} \frac{(1+x) + (-1)^{n-1}(1-x)}{\sqrt{1-x^{2}}} \, dx \end{align*}

Dividing the cases based on the parity of $n$, it follows that

$$\int_{-1}^{1} x^{n-1} \sqrt{\frac{1+x}{1-x}} \, dx = \begin{cases} \displaystyle 2\int_{0}^{1} \frac{x^{n}}{\sqrt{1-x^{2}}} \, dx, & n \text{ even} \\ \displaystyle 2\int_{0}^{1} \frac{x^{n-1}}{\sqrt{1-x^{2}}} \, dx, & n \text{ odd}. \end{cases}.$$

Thus by Lemma 1 we know an exact formula for the coefficients of $f(z)$ in $\text{(2)}$, and we obtain

\begin{align*} f(z) &= \pi \sum_{k=0}^{\infty} \binom{-1/2}{k} \frac{(-1)^{k} z^{2k+1}}{2k+1} + \pi \sum_{k=1}^{\infty} \binom{-1/2}{k} \frac{(-1)^{k} z^{2k}}{2k} \\ &= \pi \int_{0}^{z} \frac{dw}{\sqrt{1- w^{2}}} + \pi \int_{0}^{z} \left( \frac{1}{\sqrt{1- w^{2}}} - 1 \right) \, \frac{dw}{w}. \end{align*}

Therefore evaluating the last integral yields $\text{(1)}$ as desired. ////

Proof of Proposition. Now let us return to the proof of our proposition. Let $r = \cos\alpha$ and $s = \cos\beta$ for any $\alpha, \beta \in \Bbb{R}$. Then by a simple application of trigonometry, we find that

$$1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} = (1 \pm x \cos(\alpha+\beta))(1 \pm x \cos(\alpha-\beta)).$$ So it follows that

\begin{align*} I(r, s) &= f(\cos(\alpha+\beta)) + f(\cos(\alpha-\beta)) - f(-\cos(\alpha+\beta)) - f(\cos(\alpha-\beta)) \\ &= 2\pi \sin^{-1}\cos(\alpha+\beta) + 2\pi \sin^{-1}\cos(\alpha-\beta). \end{align*}

If we restrict our attention to the case $0 < \alpha < \beta < \pi/2$, then it follows that we have

\begin{align*} I(r, s) &= 2\pi \sin^{-1}\cos(\alpha+\beta) + 2\pi \sin^{-1}\cos(\alpha-\beta) \\ &= 4\pi ( \tfrac{\pi}{2} - \alpha ) \\ &= 4\pi \arcsin r. \end{align*}

This completes the proof.

• Thank you for sharing this method using real analysis only. (+1) – Venus Dec 16 '14 at 13:45