4

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

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

How may we prove $(1)?$

Integreek
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gymbvghjkgkjkhgfkl
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    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 07 '17 at 10:50
  • The result can be simplified as $\pi\sin^{-1}\left(4\phi^{-4.5}\right)$. – Integreek Jun 09 '25 at 16:34

4 Answers4

11

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

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,

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

FDP
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5

Note that\begin{align} I=&\int_{-1}^{1}{1\over x}\sqrt{1+x\over 1-x}\ln{1-x+2x^3\over 1+x-2x^3}\ dx\\ &\int_{-1}^{1}{1\over x}\sqrt{1+x\over 1-x}\bigg( \ln{1+x\over 1-x}-\ln{1+2x+2x^2\over 1-2x+2x^2}\bigg)dx\\ \end{align} where $\int_{-1}^{1}{1\over x}\sqrt{1+x\over 1-x} \ln{1+x\over 1-x}dx=\pi^2$ $$ \int_{-1}^{1}{1\over x}\sqrt{1+x\over 1-x}\ln{1+2x+2x^2\over 1-2x+2x^2}dx=4\pi\csc^{-1}\phi $$ Thus $$I=\pi^2-4\pi\csc^{-1}\phi=2\pi\bigg(\frac\pi2-\csc^{-1}\frac{\phi^{\frac32}}{2} \bigg)= 2\pi \sec^{-1}\frac{\phi^{\frac32}}{2} $$

Quanto
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3

Note \begin{eqnarray} I&=&\int_{-1}^{1}{1\over x}\sqrt{1+x\over 1-x}\ln{1-x+2x^3\over 1+x-2x^3}\ dx\\ &\overset{}=&\int_{-1}^{1}{1\over x}\frac1{\sqrt{1-x^2}}\ln{1-x+2x^3\over 1+x-2x^3}\ dx\\ &=&\int_{-1}^{1}{1\over x}\frac1{\sqrt{1-x^2}}\bigg( \ln{1+x\over 1-x}-\ln{1+2x+2x^2\over 1-2x+2x^2}\bigg)dx\\ &=&J(0,\frac12)-J(2,1). \end{eqnarray} From Integral $\int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{(r-1)x^{2} + sx + 1}{(r-1)x^{2} - sx + 1} \right) \, \mathrm dx$, $$ J(p, q) = \int_{-1}^{1} \frac{1}{x}\sqrt{\frac{1+x}{1-x}} \log \left( \frac{px^{2} + qx + 1}{px^{2} - qx + 1} \right) \, \mathrm dx=2\pi(\arcsin(a)+\arcsin(b)) $$ where $$ a = q+\sqrt{q^2-p},b=q-\sqrt{q^2-p}.$$ So $$ I=2\pi\arcsin(1)-2\pi(\arcsin(1+i)+\arcsin(1-i))=2\pi\text{arccot}\left(2\sqrt{\phi^3}\right) $$

xpaul
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3

\begin{align}I&=2\int_0^1\ln\left(\frac{(1+x)(2x^2-2x+1)}{(1-x)(2x^2+2x+1)}\right)\frac{\mathrm dx}{x\sqrt{1-x^2}}\\&=4\int_0^\infty2\ln\coth x+\ln\left(\frac{4\sinh^4x+1}{4\cosh^4x+1}\right)\mathrm dx&x\to\operatorname{sech}\frac{x}2\\&=\pi^2-4\pi\cot^{-1}\sqrt\phi\\&=4\pi\cot^{-1}\left(\frac{\sqrt\phi+1}{\sqrt\phi-1}\right)\\&=4\pi\cot^{-1}\left(\frac{\phi+1+2\sqrt\phi}{\phi-1}\right)\\&=4\pi\cot^{-1}\left(\phi^3+2\phi^{3/2}\right)&\phi+1=\phi^2,\phi-1=\frac1\phi\\&=2\pi\cot^{-1}\left(\sinh\left(\ln\left(\phi^3+2\phi^{3/2}\right)\right)\right)&\cot\frac\theta2=e^{\sinh^{-1}(\cot\theta)}\\&=2\pi\cot^{-1}\left(2\phi^{3/2}\right)&\phi^6-4\phi^3=1\\&=\pi\csc^{-1}\left(\cosh\left(\ln\left(2\phi^{3/2}\right)\right)\right)&\cot\frac\theta2=e^{\cosh^{-1}(\csc\theta)}\\&=\pi\sin^{-1}\left(4\phi^{-9/2}\right)&\end{align}

Results used:

$$\int_0^\infty\ln(\coth x)\mathrm dx=\frac{\pi^2}8$$ $$\int_0^\infty\ln\left(\frac{4\cosh^4x+1}{4\sinh^4x+1}\right)\mathrm dx=\pi\cot^{-1}\sqrt\phi$$

Integreek
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