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Motivated by the fact that $$\int_{-1}^1\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\frac{\mathrm dx}x=4\int_\infty^\frac14\int_0^\infty\frac1{\cosh^4x+y}-\frac1{\sinh^4x+y}\mathrm dx\,\mathrm dy$$

I tried to think of $\displaystyle\int_1^\infty\ln\left(\frac{5x^2-1}{x^2+3}\right)\frac{\mathrm dx}{x^2-1}$ which can be converted into the following double integral by the substitution $\dfrac{x^2-1}{x^2+1}\to\operatorname{sech}2x$:

$$\begin{align}\int_0^\infty\ln\left(\frac{\cosh2x+1+\frac12}{\cosh2x-1+\frac12}\right)\mathrm dx&=\frac{\pi^2}4+\int_0^\frac122y\int_0^\infty\frac1{\cosh^2x+y^2}-\frac1{\sinh^2x+y^2}\mathrm dx\,\mathrm dy\\&=\frac{\pi^2}4+\int_0^\frac122y\int_0^\infty\frac{\operatorname{sech}^2x}{y^2+1-y^2\tanh^2x}-\frac{\operatorname{sech}^2x}{y^2+(1-y^2)\tanh^2x}\mathrm dx\,\mathrm dy\\&\overset{\tanh x\to x}=\frac{\pi^2}4+\int_0^\frac12\frac2y\int_0^1\frac1{1+\frac1{y^2}-x^2}-\frac1{1-y^2}\frac1{\frac{y^2}{1-y^2}+x^2}\mathrm dx\,\mathrm dy\\&=\frac{\pi^2}4+\int_0^\frac12\frac2{\sqrt{y^2+1}}\sinh^{-1}y-\frac2{\sqrt{1-y^2}}\cos^{-1}y\,\mathrm dy\\&=\ln^2\phi+\frac{\pi^2}9\end{align}$$

where $\phi=\frac{\sqrt5+1}2$, the golden ratio.

Is there any other method to evaluate this interesting integral? Any comments and alternative methods are highly appreciated.

Nic
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Integreek
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    $$I=\displaystyle\int_1^\infty\ln\left(\frac{5x^2-1}{x^2+3}\right)\frac{\mathrm dx}{x^2-1}$$

    $$I=\displaystyle\int_1^0\ln\left(\frac{\frac{5}{t^2}-1}{\frac{1}{t^2}+3}\right)\frac{1}{\frac{1}{t^2}-1}-\frac{1}{t^2},dt$$

    $$I=\displaystyle\int_0^1\ln\left(\frac{5-t^2}{1+3t^2}\right)\frac{1}{1-t^2},dt$$

    $$I=\int_0^1\frac{\ln(5-t^2)}{1-t^2},dt-\int_0^1\frac{\ln(1+3t^2)}{1-t^2},dt=\cdots$$ Then comes up Claude's idea

     – Amrut Ayan Apr 03 '25 at 13:26
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    @AmrutAyan. Good idea to make this change of variable. – Claude Leibovici Apr 03 '25 at 13:32
  • Would the downvoter mind to comment? – Integreek Apr 03 '25 at 14:29

4 Answers4

7

Decompose the logarithm and consider $$\frac{\log \left(a x^2+b\right)}{x^2-1}=\frac{\log \left(a x^2+b\right)}{(x+1)(x-1)}=\frac 1 2 \Bigg(\frac{\log \left(a x^2+b\right)}{x-1}-\frac{\log\left(a x^2+b\right)}{x+1} \Bigg)$$ and $$a x^2+b=a \left(x-\frac{i\sqrt{b}}{\sqrt{a}}\right) \left(x+\frac{i\sqrt{b}}{\sqrt{a}}\right)$$ So, decomposing again the logarithms, four integrals looking like $$\int \frac{\log (x+\alpha )}{x+\beta }\,dx=\text{Li}_2\left(-\frac{x+\alpha }{\beta -\alpha }\right)+\log (\alpha +x) \log \left(\frac{\beta +x}{\beta -\alpha }\right)$$

Claude Leibovici
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4

Using $$ \int_0^1\frac{\ln(1-at)}t\mathrm dt=-\text{Li}_2(a),\int_0^1\frac{\ln(1-t+t^2)}{t}\mathrm dt=-\frac{\pi^2}{18} $$ and $$ \text{Li}_2(a)+\text{Li}_2(1/a)=-\frac{\pi^2}{6}-\frac12\ln^2(-a) $$ for $a\notin\in[0,1]$ from https://en.wikipedia.org/wiki/Polylogarithm, one has \begin{eqnarray} &&\int_1^\infty\ln\left(\frac{5x^2-1}{x^2+3}\right)\frac{\mathrm dx}{x^2-1}\\ &\overset{x\to\frac{1+t}{1-t}}=&\frac12\int_0^1\frac1t\ln\bigg(\frac{1+3t+t^2}{1-t+t^2}\bigg)\mathrm dt\\ &=&\frac12\bigg[\int_0^1\frac{\ln(1-at)}{t}\mathrm dt+\int_0^1\frac{\ln(1-\frac1at)}{t}\mathrm dt-\int_0^1\frac{\ln(1-t+t^2)}{t}\mathrm dt\bigg]\\ &=&\frac12\bigg[\frac{\pi^2}{18}-\text{Li}_2(a)-\text{Li}_2(\frac1a)\bigg]\\ &=&\frac12\bigg[\frac{\pi^2}{18}+\frac{\pi^2}6+\frac12\ln^2(-a)\bigg]\\ &=&\frac{\pi^2}9+\ln^2\phi \end{eqnarray} where $$ a=-\frac{3-\sqrt 5}{2}=-\phi^2. $$

xpaul
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4

\begin{align} &\int_1^\infty\ln\left(\frac{5x^2-1}{x^2+3}\right)\frac{1}{x^2-1}dx\\ =& \int_1^\infty \int_{-\frac12}^{\frac32}\frac{1}{(x^2-1) y+(x^2+1)} dy\ dx\\ =& \ \frac12\int_{-\frac12}^{\frac32} \frac{\cosh^{-1}y}{\sqrt{y^2-1}} dy= \frac14(\cosh^{-1}y)^2\bigg|_{-\frac12}^{\frac32}= \ln^2\phi+ \frac{\pi^2}9 \end{align}

Quanto
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  • Wouldn't it be more appropriate to write $\displaystyle\frac12\int_{\frac{-1}2}^1\frac{\cos^{-1}(-y)}{\sqrt{1-y^2}}\mathrm dy+\frac12\int_1^\frac32\frac{\cosh^{-1}y}{\sqrt{y^2-1}}\mathrm dy$? – Integreek Apr 07 '25 at 10:04
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    @Integreek You could. But the implicit complex notation is more concise. – Quanto Apr 07 '25 at 13:44
3

Perform the change of variable $\dfrac{x^2-1}{x^2+1}\to\sin x$:

$$I=\frac12\int_0^{\pi/2}\csc x\ln\left(\frac{2+3\sin x}{2-\sin x}\right)\mathrm dx$$

Consider $\mathcal I(a)=\displaystyle\int_0^{\pi/2}\csc x\ln\left(\frac{1+(a+1)\sin x}{1+(a-1)\sin x}\right)\mathrm dx$, $0\le a\le\frac12$. Then, differentiating under the integral sign,

$$\begin{align}\mathcal I'(a)&=\int_0^{\pi/2}\frac{\mathrm dx}{(a+1)\sin x+1}+\int_0^{\pi/2}\frac{\mathrm dx}{(1-a)\sin x-1}\\&=\left.\frac{2\coth^{-1}\left(\dfrac{\tan\frac{x}2+a+1}{\sqrt{a^2+2a}}\right)}{\sqrt{a^2+2a}}+\frac{2\tan^{-1}\left(\dfrac{-\tan\frac{x}2-a+1}{\sqrt{2a-a^2}}\right)}{\sqrt{2a-a^2}}\right|_0^{\pi/2}\\&=\frac{\cosh^{-1}(a+1)}{\sqrt{(a+1)^2-1}}-\frac{\cos^{-1}(1-a)-\pi}{\sqrt{1-(1-a)^2}}\\\implies2I&=\mathcal I(0)+\Big[\left(\cosh^{-1}(a+1)\right)^2+\left(\cos^{-1}(1-a)-\pi\right)^2\Big]_0^{1/2}\\&=\frac{\pi^2}2+\frac12\ln^2(\phi^2)-\frac{5\pi^2}{18}\\\therefore I&=\ln^2\phi+\frac{\pi^2}9\end{align}$$

Integreek
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