# How can we show that $\int_{-1}^{1}{1\over x}\sqrt{1+x\over 1-x}\ln\left({1-x+2x^3\over 1+x-2x^3}\right)dx=2\pi arccot\left(2\sqrt{\phi^3}\right)?$

Motivated by this question

$$\int_{-1}^{1}{1\over x}\sqrt{1+x\over 1-x}\ln\left({1-x+2x^3\over 1+x-2x^3}\right)\mathrm dx=\color{blue}{2\pi \operatorname*{arccot}\left(2\sqrt{\phi^3}\right)}\tag1$$

My try:

Not sure how to go about to tackle $(1)$ but I try

$u={1-x+2x^3\over 1+x-2x^3}\implies du={12x^2-2\over (1+x-2x^3)^2}~dx$

How may we prove $(1)?$

• What is the point to build from other computations some new integral and hide (the transformation you have performed) how you get it? – FDP May 7 '17 at 10:50

$\displaystyle J=\int_{-1}^{1}{1\over x}\sqrt{1+x\over 1-x}\ln\left({1-x+2x^3\over 1+x-2x^3}\right)\mathrm dx$

Perform the change of variable $y=\sqrt{\dfrac{1-x}{1+x}}$,

\begin{align} J&=-8\int_0^{+\infty}\dfrac{\ln x}{1-x^4}dx+4\int_0^{+\infty}\dfrac{\ln\left(\dfrac{5x^4-2x^2+1}{x^4-2x^2+5}\right)}{1-x^4}dx\\ &=-8\int_0^{+\infty}\dfrac{\ln x}{1-x^4}dx+2\int_0^{+\infty}\dfrac{\ln\left(\dfrac{5x^4-2x^2+1}{x^4-2x^2+5}\right)}{1+x^2}dx+2\int_0^{+\infty}\dfrac{\ln\left(\dfrac{5x^4-2x^2+1}{x^4-2x^2+5}\right)}{1-x^2}dx\\ &=-8\int_0^{1}\dfrac{\ln x}{1-x^4}dx-8\int_1^{+\infty}\dfrac{\ln x}{1-x^4}dx+2\int_0^{1}\dfrac{\ln\left(\dfrac{5x^4-2x^2+1}{x^4-2x^2+5}\right)}{1+x^2}dx+\\&2\int_1^{+\infty}\dfrac{\ln\left(\dfrac{5x^4-2x^2+1}{x^4-2x^2+5}\right)}{1+x^2}dx+2\int_0^{1}\dfrac{\ln\left(\dfrac{5x^4-2x^2+1}{x^4-2x^2+5}\right)}{1-x^2}dx\\ &+2\int_1^{+\infty}\dfrac{\ln\left(\dfrac{5x^4-2x^2+1}{x^4-2x^2+5}\right)}{1-x^2}dx\end{align}

In the second, in the fourth, in the sixth integrals perform the change of variable $y=\dfrac{1}{x}$

\begin{align} J&=-8\int_0^1 \dfrac{\ln x}{1-x^2}dx+4\int_0^1 \dfrac{\ln\left(\dfrac{5x^4-2x^2+1}{x^4-2x^2+5}\right)}{1-x^2}dx \end{align}

The latter integral have been already computed, (see Arcturus's answer, https://math.stackexchange.com/q/1886298 )

therefore,

\begin{align}J&=\pi^2-4\pi\arctan\left(\sqrt{\dfrac{\sqrt{5}-1}{2}}\right)\\ &=4\pi \left(\dfrac{\pi}{4}-\arctan\left(\sqrt{\dfrac{\sqrt{5}-1}{2}}\right) \right)\\ &=4\pi \left(\arctan(1)-\arctan\left(\sqrt{\dfrac{\sqrt{5}-1}{2}}\right) \right)\\ &=4\pi\arctan\left(\dfrac{1-\sqrt{\tfrac{\sqrt{5}-1}{2}}}{1+\sqrt{\tfrac{\sqrt{5}-1}{2}}}\right)\\ &=4\pi\arctan\left(\dfrac{1-\sqrt{\phi-1}}{1+\sqrt{\phi-1}}\right)\\ &=2\pi \arctan\left(\dfrac{2\left(\tfrac{1-\sqrt{\phi-1}}{1+\sqrt{\phi-1}}\right)}{1-\left(\tfrac{1-\sqrt{\phi-1}}{1+\sqrt{\phi-1}}\right)^2}\right)\\ &=2\pi \arctan\left(\dfrac{2(1-\sqrt{\phi-1})(1+\sqrt{\phi-1})}{(1+\sqrt{\phi-1})^2-(1-\sqrt{\phi-1})^2}\right)\\ &=2\pi\arctan\left(\dfrac{(1-\sqrt{\phi-1})(1+\sqrt{\phi-1})}{2\sqrt{\phi-1}}\right)\\ &=\boxed{2\pi\arctan\left(\dfrac{2-\phi}{2\sqrt{\phi-1}}\right)}\\ \end{align}

Since $\phi^2=\phi+1$

therefore,

$\phi^2-\phi=1$

therefore,

$\phi-1=\dfrac{1}{\phi}$

Therefore,

\begin{align} J&=2\pi\arctan\left(\dfrac{(1-\tfrac{1}{\phi})\sqrt{\phi}}{2}\right)\\ &=2\pi\arctan\left(\dfrac{(\phi-1)\sqrt{\phi}}{2\phi}\right)\\ &=2\pi\arctan\left(\dfrac{\sqrt{\phi}}{2\phi^2}\right)\\ &=2\pi\arctan\left(\dfrac{1}{2\phi^{\tfrac{3}{2}}}\right)\\ &=\boxed{2\pi\text{arccot}\left(2\sqrt{\phi^3}\right)}\\ \end{align}

• Thank you @FDP, fascinating proof. Do hold a maths PhD? Anyway promised to reward 100 rep. – gymbvghjkgkjkhgfkl May 7 '17 at 14:46
• No PhD here. My tools: Maxima and WA and my brain, my instinct. – FDP May 7 '17 at 16:42