Integral Involving Harmonic Numbers: $\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx$

Question

(Motivation) In an attempt to answer this question, I got stuck on evaluating a certain integral. I have made up the following conjecture:

$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right).$$

(Attempt) Let $H_n$ denote the n-th harmonic number $\displaystyle H_{n}=\sum_{k=1}^{n}\frac{1}{k}$. I will warn this process gets ugly, but it is the best I have so far. Expanding the integrand as a series, we get:

$$ \eqalign{ \int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+2}}dx+\sum_{n=0}^{\infty}\int_{\sqrt{3}}^{\infty}\frac{4\ln\left(x\right)-H_{n}}{x^{4n+4}}dx, \cr } $$

which simplifies down to

$$\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-1}}{-4n-1}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-1}}{-4n-1}+\frac{4\sqrt{3}^{-4n-1}}{\left(-4n-1\right)^{2}}\right)+\sum_{n=0}^{\infty}\left(\frac{H_{n}\sqrt{3}^{-4n-3}}{-4n-3}-\frac{4\ln\left(\sqrt{3}\right)\sqrt{3}^{-4n-3}}{-4n-3}+\frac{4\sqrt{3}^{-4n-3}}{\left(-4n-3\right)^{2}}\right).$$

Since both series converge, we can split up the first series like

$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-1}}{4n+1}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{\left(4n+1\right)^{2}}$$

and the second series like

$$-\sum_{n=0}^{\infty}\frac{H_{n}\sqrt{3}^{-4n-3}}{4n+3}+4\ln\left(\sqrt{3}\right)\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}+4\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{\left(4n+3\right)^{2}}.$$

Next, I found that

$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-1}}{4n+1}=\frac{\pi}{12}+\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)$$

and

$$\sum_{n=0}^{\infty}\frac{\sqrt{3}^{-4n-3}}{4n+3}=\frac{1}{2}\operatorname{arctanh}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{12}.$$

However, after trying for a while, I am out of ideas for evaluating the other sums.

I realize I skipped a lot of steps, but that is because I don't want this question to be too long. So for your convenience, I put all of these into Desmos, so I believe my process is correct so far based on numerical approximations.

(Question) Does anyone have an idea of how to evaluate the integral in question, or how to evaluate the sums I am stuck on? Any hints and ideas are appreciated.

(Miscellaneous) Here are some other ideas I have:

$$ \eqalign{ \int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{\operatorname{arcsec}\left(\sqrt{3}\right)}^{\frac{\pi}{2}}\frac{\ln\left(\left(\sec x\right)^{4}-1\right)}{\sec^{2}x-1}\sec\left(x\right)\tan\left(x\right)dx \cr \int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx &= \int_{0}^{\infty}\frac{\ln\left(\left(x+\sqrt{3}\right)^{4}-1\right)}{\left(x+\sqrt{3}\right)^{2}-1}dx. } $$

Maybe I could construct a keyhole contour for the last integral?

Answer 1

I will answer my own question by proving

$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(2+\sqrt{3}\right).$$

Let the integral in question equal $I$. Letting $x \to \dfrac{1-x}{1+x}$, we rewrite the integral as

$$I = \int_{\sqrt{3}-2}^{-1}\frac{\ln\left(\left(\frac{1-x}{1+x}\right)^{4}-1\right)}{\left(\frac{1-x}{1+x}\right)^{2}-1}\left(\frac{-2}{\left(1+x\right)^{2}}\right)dx.$$

Doing some simplifications, we get

$$I = \frac{3\ln\left(2\right)}{2}\int_{\sqrt{3}-2}^{-1}\frac{1}{x}dx+\frac{1}{2}\int_{\sqrt{3}-2}^{-1}\frac{\ln\left(-x\right)}{x}dx+\frac{1}{2}\int_{\sqrt{3}-2}^{-1}\frac{\ln\left(1+x^{2}\right)}{x}dx-\frac{1}{2}\int_{\sqrt{3}-2}^{-1}\frac{\ln\left(\left(x+1\right)^{4}\right)}{x}dx.$$

Trivially, we can solve the first two integrals. For the last two, we can use the dilogarithm definition. Thus,

$$I = \frac{3\ln\left(2\right)}{2}\left(-\ln\left(2-\sqrt{3}\right)\right)+\frac{1}{2}\left(-\frac{1}{2}\ln^{2}\left(2-\sqrt{3}\right)\right)+\frac{1}{2}\left(\frac{1}{24}\left(12\operatorname{Li}_2 \left(4\sqrt{3}-7\right)+\pi^{2}\right)\right)-\frac{1}{2}\left(4\left(\operatorname{Li}_2 \left(2-\sqrt{3}\right)-\frac{\pi^{2}}{6}\right)\right)$$

which simplifies down to

$$I = \ln\left(\frac{1}{2-\sqrt{3}}\right)\left(\frac{1}{4}\ln\left(2-\sqrt{3}\right)+\frac{3}{2}\ln\left(2\right)\right)+\frac{17\pi^{2}}{48}+\frac{1}{4}\operatorname{Li}_2 \left(-\left(2-\sqrt{3}\right)^{2}\right)-2\operatorname{Li}_2 \left(2-\sqrt{3}\right)$$

From this answer, we can prove that

$$\frac{1}{4}\operatorname{Li}_2 \left(-\left(2-\sqrt{3}\right)^{2}\right)-2\operatorname{Li}_2 \left(2-\sqrt{3}\right) = \frac{\ln^{2}\left(2-\sqrt{3}\right)}{4}-\frac{5\pi^{2}}{48}.$$

Combining the results, we conclude that the integral $I$ is

$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(2+\sqrt{3}\right).$$

Answer 2

A possible path and a partial answer.

I would split the (indefinite) integral, first, and then take the limit. \begin{eqnarray} \mathcal I &=& \int \frac{\log(x^4-1)}{x^2-1}dx=\\ &=&\underbrace{\int\frac{\log(x-1)}{x^2-1}dx}_{\mathcal I_1} + \underbrace{\int \frac{\log(x+1)}{x^2-1}dx}_{\mathcal I_2} +\\ & &+\int \frac{\log(x+i)}{x^2-1}dx +\int \frac{\log(x-i)}{x^2-1}dx=\\ &=& \mathcal I_1 + \mathcal I_2 + \underbrace{2\mbox{Re}\left\{\int\frac{\log(x+i)}{x^2-1}dx\right\}}_{\mathcal I_3}. \end{eqnarray}

Then we have \begin{eqnarray} \mathcal I_1 &\stackrel{x-1 \mapsto t}{=}& \int \frac{\log t}{t(t+2)}dt=\\ &=&\frac12 \int \frac{\log t}t dt -\underbrace{\frac12\int \frac{\log t}{t+2}dt}_{IBP}=\\ &=&\frac14\log^2(x-1)-\frac12 \log t\log\left(\frac{t+2}2\right)-\frac12\int\frac{\log\left(1+\frac{t}2\right)}{-\frac{t}2}dt=\\ &=&\frac14 \log^2(x-1)-\frac12 \log(x-1)\log\left(\frac{x+1}2\right)-\frac12 \mbox{Li}_2 \left(-\frac{t}2\right)+C=\\ &=&\underbrace{\frac14 \log^2(x-1)-\frac12 \log(x-1)\log\left(\frac{x+1}2\right)-\frac12 \mbox{Li}_2 \left(-\frac{x-1}2\right)}_{F_1(x)}+C \end{eqnarray} Thus the definite integral gives \begin{eqnarray} \int_\sqrt 3^\infty \frac{\log(x-1)}{x^2-1}dx &=&-\frac14\log^2(\sqrt 3-1) +\frac12 \log(\sqrt 3-1)\log\left(\frac{\sqrt 3+1}2\right)+\\ & &+\frac12\mbox{Li}_2\left(-\frac{\sqrt 3-1}2\right)+\\ & &+\lim_{x\to \infty}F_1(x). \end{eqnarray} Let us first rewrite $F_1(x)$ using the reflection formula $$\mbox{Li}_2(z) = -\mbox{Li}_2\left(\frac1{z}\right)-\frac{\pi^2}6-\frac12 \log^2(-z),$$ in order to have \begin{eqnarray} F_1(x)&=& \frac14\log^2(x-1)-\frac12\log(x-1)\log\left(\frac{x+1}2\right)+\frac14\log^2\left(\frac{x-1}2\right)+\\& &+\frac{\pi^2}{12}+\frac12\mbox{Li}_2\left(-\frac2{x-1}\right)=\\ &=&\frac12\log(x-1)\log\left(\frac{x-1}{x+1}\right)+ \frac{\log^22}4+\frac{\pi^2}{12}+\frac12\mbox{Li}_2\left(-\frac2{x-1}\right). \end{eqnarray} Thus $$\lim_{x\to \infty} F_1(x) = \frac{\log^22}4 + \frac{\pi^2}{12},$$ and the first (definite) integral becomes \begin{eqnarray} \int_\sqrt 3^\infty \frac{\log(x-1)}{x^2-1}dx &=&-\frac14\log^2(\sqrt 3-1) +\frac12 \log(\sqrt 3-1)\log\left(\frac{\sqrt 3+1}2\right)+\\ & &+\frac12\mbox{Li}_2\left(-\frac{\sqrt 3-1}2\right)+\frac{\log^22}4 + \frac{\pi^2}{12}. \end{eqnarray}

An identical approach can be adopted for $\mathcal I_2$ yielding \begin{eqnarray} \int_\sqrt 3^\infty \frac{\log(x+1)}{x^2-1} dx &=&\frac14\log^2(\sqrt 3+1)-\frac12 \log 2\log(\sqrt 3-1)+\\ & &+\frac12 \mbox{Li}_2\left(-\frac{\sqrt 3-1}2\right)+\frac{\log^22}4 + \frac{\pi^2}{12}, \end{eqnarray} and therefore \begin{eqnarray} \int_\sqrt 3^\infty \frac{\log(x^2-1)}{x^2-1}dx &=&\frac14\log^2(\sqrt 3+1)-\frac14\log^2(\sqrt 3-1)- \log 2\log(\sqrt 3-1)+\\ & &+\frac{\log2}2 + \frac{\log^22}2+\mbox{Li}_2\left(-\frac{\sqrt 3-1}2\right)+\frac{\pi^2}6. \end{eqnarray}

What I have done so far for $\mathcal I_3$: \begin{eqnarray} \mathcal I_3 &=&\mbox{Re}\left\{\underbrace{\int\frac{\log(x+i)}{x-1}dx}_{IBP}-\underbrace{\int\frac{\log(x+i)}{x+1}dx}_{IBP}\right\}=\\ &=& \mbox{Re}\left\{\log\left[\left(-\frac12+\frac{i}2\right)(x-1)\right]\log(x+i)-\int\frac{\log\left[1+\left(\frac12-\frac{i}2\right)(x+i)\right]}{(x+i)}dx\right.+\\ & &\left.-\log\left[\left(\frac12+\frac{i}2\right)(x+1)\right]\log(x+i)+\int\frac{\log\left[1+\left(-\frac12-\frac{i}2\right)(x+i)\right]}{(x+i)}dx\right\}=\\ &=&\mbox{Re}\left\{\log\left[\left(-\frac12+\frac{i}2\right)(x-1)\right]\log(x+i)+\mbox{Li}_2\left[\left(\frac12-\frac{i}2\right)(x+i)\right]\right.+\\ & &\left.-\log\left[\left(\frac12+\frac{i}2\right)(x+1)\right]\log(x+i)-\mbox{Li}_2\left[\left(-\frac12-\frac{i}2\right)(x+i)\right]+C\right\} \end{eqnarray}

Answer 3

Mathematica 13.1 produces

Integrate[Log[x^4 - 1]/(x^2 - 1), {x, Sqrt[3], Infinity}]//FullSimplify

$$\frac{1}{32} \left(16 \left(\text{Li}_2\left(\frac{1}{2} \left(1-\sqrt{3}\right)\right)+\text{Li}_2\left(\sqrt{3}-1\right)+\text{Li}_2\left(\left(-\frac{1}{2}+\frac{i}{2}\right) \left(\sqrt{3}-1\right)\right)+\text{Li}_2\left(\left(-\frac{1}{2}-\frac{i}{2}\right) \left(\sqrt{3}-1\right)\right)+\text{Li}_2\left(\frac{1-i}{\sqrt{3}+1}\right)+\text{Li}_2\left(\frac{1+i}{\sqrt{3}+1}\right)+\log ^2\left(\sqrt{3}+1\right)\right)+7 \pi ^2+8 \log ^2\left(\sqrt{3}-1\right)+4 \log (2) \left(\log (8)+4 \log \left(2 \sqrt{3}+\frac{7}{2}\right)\right)\right) $$

Answer 4

This is not a complete solution but an extended comment.

I was particularly interested to find out what makes the lower integration border $\sqrt{3}$ so special as a prerequisite for an appreciable simplification of the final expression of the integral.

Hence I started with the same decomposition of our integral into the sum of three integrals as @dfnu, but now with a general lower integration border $a$, where $a>1$ to ensure convergence:

$$i_1(a) =\int_a^{\infty } \frac{\log (x-1)}{x^2-1} \, dx$$ $$i_2(a) =\int_a^{\infty } \frac{\log (x+1)}{x^2-1} \, dx$$

and looked particularly at the integral

$$i_3(a) = \int_{a}^{\infty}\frac{\log(x^2+1)}{x^2-1}$$

After a lengthy journey I found the following expression

$$\begin {align} i_3(a) = \frac{\pi ^2}{4}-2 \sum _{k=1}^{\infty } \frac{\left(\frac{\sqrt{a^2+1}}{\sqrt{2}}\right)^k \sin \left(\frac{\pi k}{4}\right) \sin \left(k \tan ^{-1}(a)\right)}{k^2}\\+\frac{1}{2} \log \left(\frac{a+1}{a-1}\right) \log \left(a^2+1\right)+\frac{1}{2} \pi \tan ^{-1}\left(\frac{1}{a}\right)\end{align}$$

Now to the special role of $\sqrt{3}$. The only value of $a=\sqrt{m}$ with positive integer $m$ making $\tan ^{-1}\left(\frac{1}{a}\right)$ a rational multiple of $\pi$ is $\sqrt{3}$, giving $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi }{6}.$

The complete integral evaluates to

$$\begin{align} i(a) & = \int_{a}^{\infty}\frac {\log(x^4-1)}{x^2-1} \\ & = \frac{1}{2} \pi \tan ^{-1}\left(\frac{1}{a}\right)+\frac{\pi ^2}{3} \text{(*1*)} \\ & -\frac{1}{4} \log ^2(a-1)+\frac{1}{2} \log ^2(a+1) \\ & -\frac{1}{2} \log (2) \log (a-1)+\frac{\log ^2(2)}{4} \\ & + \frac{1}{2} \log \left(\frac{a+1}{a-1}\right) \log \left(a^2+1\right) \text{(*2*)} \\ & +\frac{1}{2} \text{Li}_2\left(\frac{1-a}{2}\right)+\frac{1}{2} \text{Li}_2\left(\frac{2}{a+1}\right) \\ & +\frac{1}{2} \left(-\text{Li}_2\left(\left(\frac{1}{2}+\frac{i}{2}\right) (a-i)\right) \\ + \text{Li}_2\left(\left(-\frac{1}{2}+\frac{i}{2}\right) (a-i)\right) \\ -\text{Li}_2\left(\left(\frac{1}{2}-\frac{i}{2}\right) (a+i)\right) \\ +\text{Li}_2\left(\left(-\frac{1}{2}-\frac{i}{2}\right) (a+i)\right)\right) \end{align}$$

As the two lines marked $(*1*)$ and $(*2*)$ are already similar to the final result for $a=\sqrt {3}$ there should be a lot of cancellation from the polylog functions.

(to be completed)