While calculating a specific problem, I succeeded in proving a more general problem.

Proposition. For $0 < r < 1$ and $r < s$, the following holds:[1] \begin{equation} \label{eq_wts} \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. \end{equation}

Proof. We divide the proof into several steps.

Step 1. (Case reduction by analytic continuation)

We first remark that given $r$ and $s$, we always have $$ \min_{-1 \leq x \leq 1} \{ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} \} > 0. \tag{2} $$ Indeed, it is not hard to check if we utilize the following equality $$ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} = (1 \pm rsx)^{2} - (1 - r^{2})(1 - s^{2}) x^{2}. $$ Then $\text{(2)}$ allows us to extend $s \mapsto I(r, s)$ as a holomorphic function on some open set containing the line segment $(r, \infty) \subset \Bbb{C}$. Then by the principle of analytic continuation, it is sufficient to prove that $\text{(1)}$ holds for $r < s < 1$.

Step 2. (Integral representation of $I$)

Put $r = \sin \alpha$ and $s = \sin \beta$, where $ 0 < \alpha < \beta < \frac{\pi}{2}$. Then \begin{align*} I(r, s) &= \int_{-1}^{1} \frac{1+x}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \\ &= \int_{0}^{1} \frac{2}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \qquad (\because \text{ parity}) \\ &= \int_{1}^{\infty} \frac{2}{\sqrt{x^{2}-1}} \log \left( \frac{x^{2} + 2rsx + (r^{2} + s^{2} - 1)}{x^{2} - 2rsx + (r^{2} + s^{2} - 1)} \right) \, dx \qquad (x \mapsto x^{-1}) \\ &= \int_{0}^{1} \frac{2}{t} \log \left( \frac{\left(t+t^{-1}\right)^{2} + 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)}{\left(t+t^{-1}\right)^{2} - 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)} \right) \, dt, \end{align*} where in the last line we utilized the substitution $x = \frac{1}{2}(t + t^{-1})$. If we introduce the quartic polynomial \begin{align*} p(t) = t^{4} + 4rst^{3} + (4r^{2}+4s^{2}-2)t^{2} + 4rst + 1, \end{align*} then by the property $p(1/t) = t^{-4}p(t)$, we can simplify \begin{align*} I(r, s) &= 2 \int_{0}^{1} \frac{\log p(t) - \log p(-t)}{t} \, dt = \int_{0}^{\infty} \frac{\log p(t) - \log p(-t)}{t} \, dt \\ &= - \int_{0}^{\infty} \left( \frac{p'(t)}{p(t)} + \frac{p'(-t)}{p(-t)} \right) \log t \, dt = - \frac{1}{2} \Re \int_{-\infty}^{\infty} \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z \, dz, \end{align*} where we choose the branch cut of $\log$ in such a way that it avoids the upper-half plane $$\Bbb{H} = \{ z \in \Bbb{C} : \Im z > 0 \}.$$

Step 3. (Residue calculation)

Now let $Z_{+}$ be the set of zeros of $p(z)$ in $\Bbb{H}$ and $Z_{-}$ be the set of zeros of $p(z)$ in $-\Bbb{H}$. Since $$ f(z) := \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z = O\left(\frac{\log z}{z^{2}} \right) \quad \text{as } z \to \infty, $$ by replacing the contour of integration by a semicircle of sufficiently large radius, it follows that \begin{align*} I(r, s) = - \frac{1}{2} \Re \left[ 2 \pi i \sum_{z_{0} \in Z_{+}} \operatorname{Res}_{z = z_{0}} f(z) + 2 \pi i \sum_{z_{0} \in -Z_{-}} \operatorname{Res}_{z = z_{0}} f(z) \right]. \end{align*} But by a simple calculation, together with the condition $ 0 < \alpha < \beta < \frac{\pi}{2}$, we easily notice that the zeros of $p(z)$ are exactly $$ e^{\pm i(\alpha + \beta)} \quad \text{and} \quad -e^{\pm i(\alpha - \beta)}. $$ So we have $$ Z_{+} = \{ e^{i(\beta+\alpha)}, -e^{-i(\beta - \alpha)} \} \quad \text{and} \quad Z_{-} = \{ e^{-i(\beta+\alpha)}, -e^{i(\beta- \alpha)} \} $$ and therefore \begin{align*} I(r, s) &= - \frac{1}{2} \Re 2 \pi i \left[ \sum_{z_{0} \in Z_{+}} \log z_{0} + \sum_{z_{0} \in -Z_{-}} \log z_{0} \right] \\ &= -\frac{1}{2} \Re 2 \pi i \left\{ \log e^{i(\beta+\alpha)} + \log e^{i(\pi-\beta+\alpha)} - \log e^{i(\pi-\beta-\alpha)} - \log e^{i(\beta-\alpha)} \right\} \\ &= -\frac{1}{2} \Re 2 \pi i \left\{ i(\beta+\alpha) + i(\pi-\beta+\alpha) - i(\pi-\beta-\alpha) - i(\beta-\alpha) \right\} \\ &= \pi \left\{ (\beta+\alpha) + (\pi-\beta+\alpha) - (\pi-\beta-\alpha) - (\beta-\alpha) \right\} \\ &= 4\pi \alpha = 4\pi \arcsin r. \end{align*} This completes the proof. ////

References

  1. Laila Podlesny, Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)dx$ - Math StackExchange
Posted by aficionado

댓글을 달아 주세요

  1. Leun Kim 2013.11.18 22:37  댓글주소  수정/삭제  댓글쓰기

    역시 후덜덜 하십니다..

  2. XRumerTest 2014.06.01 18:16  댓글주소  수정/삭제  댓글쓰기

    Hello. And Bye.