50
$\begingroup$

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.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Ron Gordon Nov 16 '13 at 3:08
32
$\begingroup$

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.

$\endgroup$
20
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.