Evaluate $\int\limits^{\infty}_{0} \frac{1}{x} \, \ln\left(\frac{x^{3} - x^{2} - x + 1}{x^{3} + x^{2} + x + 1}\right) \, \mathrm{d}x$

Question

Question: Evaluate $$\int\limits^{\infty}_{0} \frac{1}{x} \, \ln\left(\frac{x^{3} - x^{2} - x + 1}{x^{3} + x^{2} + x + 1}\right) \, \mathrm{d}x$$

Disclaimer: I really enjoyed solving this problem and it felt very satisfying. I was wondering if there are alternative methods that might be more elegant, insightful, or efficient. I'd love to see how others might approach this!

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My approach:

$$I = \int_{0}\limits^{\infty}\frac{1}{x}\ln\left(\frac{(x-1)^2(x+1)}{(x+1)(x^2+1)}\right) \, \mathrm{d}x=\int_{0}\limits^{\infty}\frac{1}{x}\ln\left(\frac{(x-1)^2}{x^2+1}\right) \, \mathrm{d}x$$

$$I = \int_{0}\limits^{\infty}\frac{\ln\left(x^2+1-2x\right)-\ln\left(x^2+1\right)}{x} \, \mathrm{d}x$$

$$I = \int_{0}\limits^{\infty}\int_{0}\limits^{-2}\frac{1}{x^2+1+ax} \,\mathrm{d}a\mathrm{d}x = \int_{0}\limits^{-2}\int_{0}\limits^{\infty}\frac{1}{x^2+1+ax} \,\mathrm{d}x\mathrm{d}a$$

$$I = \int_{0}\limits^{-2}\left(\frac{\pi}{2}-\arctan\left(\frac{a}{\sqrt{4-a^{2}}}\right)\right)\cdot\frac{2}{\sqrt{4-a^{2}}} \, \mathrm{d}a =\int_{0}\limits^{-2}\frac{2}{\sqrt{4-a^2}}\arccos\left(\frac{a}{2}\right) \, \mathrm{d}a $$

$$I = \left[-\left(\cos^{-1}\frac{a}{2}\right)^2\right]_{0}^{-2} = \fbox{$-\frac{3\pi^2}{4}$}$$

Answer 1

Continue with \begin{align} I = & \int_{0}^{\infty}\frac{1}{x}\ln\frac{(x-1)^2}{x^2+1}dx =\int_{0}^{\infty}\frac{1}{x}\ln(x^2-2x\cos \theta +1)\bigg|_{\theta= \frac\pi2}^{\theta=0} dx\\ =& \int_{0}^{\infty}\int_ {\frac\pi2}^{0} \frac{2\sin\theta}{x^2-2x\cos \theta +1}d\theta \ dx =2 \int_ {\frac\pi2}^{0} (\pi-\theta)d\theta=-\frac{3\pi^2}4 \end{align}

Answer 2

Let $$A= \int\limits^{\infty}_{0} \frac{1}{x} \, \ln\left(\frac{x^{3} - x^{2} - x + 1}{x^{3} + x^{2} + x + 1}\right) \, \mathrm{d}x $$

As you have already seen, the integrand simplifies to

$$ \frac{1}{x} \, \ln\left(\frac{(x - 1)^2}{x^2 + 1}\right) $$

and by using logarithm properties, this becomes

$$ \frac{2 \ln |x - 1|}{x} - \frac{\ln(x^2 + 1)}{x} $$

We can now split the integral into two parts

$$ A = 2 \int\limits^{\infty}_{0} \frac{\ln |x - 1|}{x} \, \mathrm{d}x - \int\limits^{\infty}_{0} \frac{\ln(x^2 + 1)}{x} \, \mathrm{d}x $$

We will evaluate each part seperately

Evaluate $I_1 = \int\limits^{\infty}_{0} \frac{\ln |x - 1|}{x} \, \mathrm{d}x$

Splitting the integral at $x = 1$ yeilds

$$ I_1 = \int\limits^{1}_{0} \frac{\ln(1 - x)}{x} \, \mathrm{d}x + \int\limits^{\infty}_{1} \frac{\ln(x - 1)}{x} \, \mathrm{d}x $$

For the first part, use the known result

$$ \int\limits^{1}_{0} \frac{\ln(1 - x)}{x} \, \mathrm{d}x = -\frac{\pi^2}{6} $$

For the second part, substitute $u = \frac{1}{x}$

$$ \int\limits^{\infty}_{1} \frac{\ln(x - 1)}{x} \, \mathrm{d}x = \int\limits^{0}_{1} \frac{\ln\left(\frac{1}{u} - 1\right)}{\frac{1}{u}} \left(-\frac{1}{u^2}\right) \, \mathrm{d}u = \int\limits^{1}_{0} \frac{\ln(1 - u)}{u} \, \mathrm{d}u = -\frac{\pi^2}{6} $$

Thus

$$ I_1 = -\frac{\pi^2}{6} - \frac{\pi^2}{6} = -\frac{\pi^2}{3} $$

Evaluate $I_2 = \int\limits^{\infty}_{0} \frac{\ln(x^2 + 1)}{x} \, \mathrm{d}x$

Splitting the integral at $x = 1$ yeilds

$$ I_2 = \int\limits^{1}_{0} \frac{\ln(x^2 + 1)}{x} \, \mathrm{d}x + \int\limits^{\infty}_{1} \frac{\ln(x^2 + 1)}{x} \, \mathrm{d}x $$

For the first part, expand $\ln(x^2 + 1)$ as a series

$$ \ln(x^2 + 1) = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \cdots $$

Integrate term by term

$$ \int\limits^{1}_{0} \frac{\ln(x^2 + 1)}{x} \, \mathrm{d}x = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n^2} = \frac{\pi^2}{24} $$

For the second part, substitute $u = \frac{1}{x}$

$$ \int\limits^{\infty}_{1} \frac{\ln(x^2 + 1)}{x} \, \mathrm{d}x = \int\limits^{0}_{1} \frac{\ln\left(\frac{1}{u^2} + 1\right)}{\frac{1}{u}} \left(-\frac{1}{u^2}\right) \, \mathrm{d}u = \int\limits^{1}_{0} \frac{\ln(u^2 + 1)}{u} \, \mathrm{d}u = \frac{\pi^2}{24} $$

Thus

$$ I_2 = \frac{\pi^2}{24} + \frac{\pi^2}{24} = \frac{\pi^2}{12} $$

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$$A=I_1+I_2$$

$$A=\frac{-\pi^2}{3}+\frac{\pi^2}{12}$$

$$ \boxed{A=-\dfrac{3\pi^2}{4}} $$

Answer 3

$$\begin{align}\int_0^\infty\ln\left(\frac{x^3-x^2-x+1}{x^3+x^2+x+1}\right)\frac{\mathrm dx}x&=2\int_1^\infty\ln\left(\frac{x^\frac32+x^\frac{-3}2-x^\frac12-x^\frac{-1}2}{x^\frac32+x^\frac{-3}2+x^\frac12+x^\frac{-1}2}\right)\frac{\mathrm dx}x\\&\overset{(1)}{=}4\int_0^\infty\ln\left(\frac{\cosh3v-\cosh v}{\cosh2v+\cosh v}\right)\mathrm dv\\&\overset{(2)}{=}4\int_0^\infty\ln(\tanh v)\mathrm dv+4\int_0^\infty\ln(\tanh 2v)\mathrm dv\\&\overset{(3)}{=}4\left(\frac{-\pi^2}{8}+\frac{-\pi^2}{16}\right)\\&=\frac{-3\pi^2}4\end{align}$$

Explanation of steps:

$(1):$ $x=e^{2v}$

$(2):$ $\cosh x+\cosh y=2\cosh\left(\frac{x+y}2\right)\cosh\left(\frac{x-y}2\right)$ and $\cosh x-\cosh y=2\sinh\left(\frac{x+y}2\right)\sinh\left(\frac{x-y}2\right)$

$(3):$ $\displaystyle\int_0^\infty\ln\left(\tanh(ax)\right)\mathrm dx=\frac{-\pi^2}{8a}$

Answer 4

$$I_1=\int \frac{\log \left((x-1)^2\right)}{x}\,dx=2 \text{Li}_2(1-x)+\log \left((x-1)^2\right) \log (x)$$ $$I_2=\int \frac{\log \left(x^2+1\right)}{x} \,dx=-\frac{\text{Li}_2\left(-x^2\right)}{2}$$ Computing $(I_1-I_2)$ and using the bounds $(0,t)$ $$J(t)=\frac{\text{Li}_2\left(-t^2\right)}{2}+ 2 \text{Li}_2(1-t)+\log\left((t-1)^2\right) \log (t)-\frac {\pi^2}3$$ Using asymptotics $$J(t)=-\frac{3 \pi^2}{4}+\frac{2}{t}+\frac{1}{t^2} +\frac{2}{9 t^3}+O\left(\frac{1}{t^5}\right)$$

Answer 5

\begin{eqnarray} I &= & \int_{0}^{\infty}\frac{1}{x}\ln\frac{(x-1)^2}{x^2+1}dx\\ &=& \int_{0}^{1}\frac{1}{x}\ln\frac{(x-1)^2}{x^2+1}dx+\int_{1}^{\infty}\frac{1}{x}\ln\frac{(x-1)^2}{x^2+1}\overset{x\to1/x}{dx}\\ &=& 2\int_{0}^{1}\frac{1}{x}\ln\frac{(x-1)^2}{x^2+1}dx \\ &=& 2\ln x\ln\frac{(x-1)^2}{x^2+1}\bigg|_0^1-2\int_0^1\ln x\bigg(-\frac{2}{1-x}-\frac{2x}{x^2+1}\bigg)dx\\ &=&4\int_0^1\frac{\ln x}{1-x}dx+4\int_0^1\frac{x\ln x}{x^2+1}dx\\ &=&-4\cdot\frac{\pi^2}{6}-\frac{\pi^2}{12}=-\frac{3\pi^2}{4}. \end{eqnarray}