Derive $\int_0^1 \frac{\ln(\sqrt2-1)-(\sqrt2-x)\ln x}{(\sqrt2-x)^2-1}\,dx=\frac{\pi^2}6+\frac14\ln^2(\sqrt2-1) $

Question

I obtained the integral

$$\int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx=\frac{\pi^2}6+\frac14\ln^2(\sqrt2-1) $$

as a by-product while carrying out some complex analysis on an engineering problem. In short, given the real nature of the physical outcome, the imaginary part of the final expression has to vanish, which implies the result above and is numerically verifiable.

However, the complex method employed is not intended to solve integrals such as this and produces it merely by accident. So, the question is: How to derive this elementary close-form directly, with real methods?

Note that the integrand may not look as innocent as it appears. There is a hole as the denominator becomes zero at $x=\sqrt2-1$ and the term $\ln(\sqrt2-1)$ is present in the numerator to remove the singularity, thus ensuring convergence.

Answer 1

We have \begin{align} I &\equiv \int \limits_0^1 \frac{\log(\sqrt{2}-1) - (\sqrt{2}-x)\log(x)}{(\sqrt{2}-x)^2-1} \, \mathrm{d} x \\ &= \int \limits_0^1 \frac{- (\sqrt{2} - 1 -x) \log(\sqrt{2}-1) - (\sqrt{2}-x)\log\left(\frac{x}{\sqrt{2}-1}\right)}{(\sqrt{2}-x)^2-1} \, \mathrm{d} x \\ &= -\log(\sqrt{2}-1) \int \limits_0^1 \frac{\mathrm{d} x}{1 + \sqrt{2} - x} + J = \log^2(1+\sqrt{2}) - \log(1 + \sqrt{2}) \log(\sqrt{2}) + J \, , \end{align} where \begin{align} J &= \int \limits_0^1 \frac{-\log\left(\frac{x}{\sqrt{2}-1}\right) (\sqrt{2}-x) }{(\sqrt{2}-x)^2-1} \, \mathrm{d} x \\ &= \int \limits_0^{\sqrt{2}-1} \frac{-\log\left(\frac{x}{\sqrt{2}-1}\right) (\sqrt{2}-x) }{(\sqrt{2}-x)^2-1} \, \mathrm{d} x + \int \limits_{\sqrt{2}-1}^1 \frac{-\log\left(\frac{x}{\sqrt{2}-1}\right) (\sqrt{2}-x) }{(\sqrt{2}-x)^2-1} \, \mathrm{d} x \\ &\!\!\stackrel{\text{IBP}}{=}\frac{1}{2} \int \limits_0^{\sqrt{2}-1} \frac{-\log[(\sqrt{2} - x)^2 - 1]}{x} \, \mathrm{d} x - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\ &\phantom{=} ~ + \frac{1}{2} \int \limits_{\sqrt{2}-1}^1 \frac{-\log[1 - (\sqrt{2} - x)^2]}{x} \, \mathrm{d} x \\ &= \frac{1}{2} \int \limits_0^1 \frac{-\log(\lvert (\sqrt{2}-x)^2 - 1\rvert)}{x} \, \mathrm{d} x - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\ &= \frac{1}{2} \int \limits_0^1 \frac{-\log\left[\left\lvert 1 - \frac{x}{\sqrt{2}-1}\right\rvert \left(1 - \frac{x}{1 + \sqrt{2}}\right)\right]}{x} \, \mathrm{d} x - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\ &= \frac{1}{2} \int \limits_0^1 \frac{-\log\left(1 - \frac{x}{1 + \sqrt{2}}\right)}{x} \, \mathrm{d} x + \frac{1}{2} \int \limits_{\sqrt{2}-1}^1 \frac{-\log\left(\frac{x}{\sqrt{2} - 1} - 1\right)}{x} \, \mathrm{d} x \\ &\phantom{=} ~ + \frac{1}{2} \int \limits_0^{\sqrt{2}-1} \frac{-\log\left(1 - \frac{x}{\sqrt{2} - 1}\right)}{x} \, \mathrm{d} x - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \, . \end{align} Now let $x = (1 + \sqrt{2}) t$ in the first, $x = \frac{\sqrt{2} - 1}{t}$ in the second and $x = (\sqrt{2} - 1) t$ in the third integral. This yields \begin{align} J &= \frac{1}{2} \int \limits_0^{\sqrt{2} - 1} \frac{-\log(1-t)}{t} \, \mathrm{d} t + \frac{1}{2} \int \limits_{\sqrt{2}-1}^1 \frac{-\log\left(\frac{1}{t} - 1\right)}{t} \, \mathrm{d} t + \frac{1}{2} \int \limits_0^1 \frac{-\log(1-t)}{t} \, \mathrm{d} t \\ &\phantom{=} ~ - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\ &= \int \limits_0^1 \frac{-\log(1-t)}{t} \, \mathrm{d} t - \frac{1}{4} \log^2(1 + \sqrt{2}) - \frac{1}{2} \log(1+\sqrt{2}) \log \left(\frac{1+\sqrt{2}}{2}\right) \\ &= \frac{\pi^2}{6} - \frac{3}{4} \log^2(1+\sqrt{2}) + \log(1+\sqrt{2}) \log(\sqrt{2}) \, , \end{align} so $$ I = \frac{\pi^2}{6} + \frac{1}{4} \log^2 (1 + \sqrt{2}) \, . $$

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A similar calculation yields the more general integral \begin{align} f(a) &\equiv \int \limits_0^1 \frac{\log(a) - \log(x) (1 + a -x)}{(1 + a - x)^2 - 1} \, \mathrm{d} x \\ &= \frac{\pi^2}{6} - \log(\sqrt{a}) \log \left(\frac{(1-a)(2+a)}{(1+a)\sqrt{a}}\right) + \frac{1}{2} \left[\operatorname{Li}_2 \left(\frac{1}{2+a}\right) - \operatorname{Li}_2(a)\right] \end{align} for $a \in (0,1]$. While special values like $f(1) = \frac{\pi^2}{12} + \frac{1}{2} \operatorname{Li}_2 \left(\frac{1}{3}\right)$ are reasonably nice, the polylogarithms only cancel for $a = \sqrt{2} - 1$, leading to the particularly simple expression for $I = f(\sqrt{2} - 1)$.

Answer 2

This will be a bit complicated, but I will try to explain clearly as I go. We will first find a general antiderivative, and then use it to compute the desired result. To make things a little easier to TeX, I'm going to write $a = \sqrt{2}$, so we're trying to find $$\int_0^1 \frac{\ln(a-1)-\ln(x)(a-x)}{(a-x)^2-1}\,dx.$$ I'm going to do this by computing the antiderivative \begin{equation} \int \frac{\ln(a-1)-\ln(x)(a-x)}{(a-x)^2-1}\,dx = \int\frac{\ln(a-1)}{(a-x)^2-1}\,dx - \int\frac{\ln{x}(a-x)}{(a-x)^2-1}\,dx.\tag{1} \end{equation} The first integral on the right of $(1)$ is relatively straightforward via partial fractions, so I will omit the computation: $$\int\frac{\ln(a-1)}{(a-x)^2-1}\,dx = \frac{\ln(a-1)}{2}\ln\left(\left|\frac{a-x+1}{x-a+1}\right|\right) + C.$$ For the second integral in $(1)$, we will do some manipulation first. Since there is a $\ln(x)$ in the numerator, and the denominator is not obviously the derivative of the numerator, we're aiming for a polylogarithm. We will also use a partial fraction decomposition, but I will not show the actual steps for that part. \begin{align*} \frac{\ln{x}(a-x)}{(a-x)^2-1} &= \ln(x)\frac{a-x}{a^{2} - 2ax + x^2 - 1}\\ &=\ln(x)\frac{a-x}{(a-x-1)(a-x+1)}\\ &=-\frac{\ln(x)}{2}\left(\frac{1}{a-x-1} + \frac{1}{a-x+1}\right).\tag{2} \end{align*} At this point, we're going to do something a bit non-obvious. Note that $$\frac{1}{a-x-1} = \frac{a-1}{a-x-1}(a-1) = \frac{d}{dx}\ln\left(\left|\frac{a-x-1}{a-1}\right|\right).$$ We're going to do something akin to using an integrating factor, in the sense that we're going to extract a product rule derivative. Continuing from $(2)$: \begin{align*} \frac{\ln{x}(a-x)}{(a-x)^2-1} &= -\frac{\ln(x)}{2}\left(\frac{1}{a-x-1} + \frac{1}{a-x+1}\right)\\[10pt] &\qquad-\frac{1}{2x}\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\\[10pt] &\qquad+\frac{1}{2x}\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\\[10pt] &=-\frac{1}{2}\frac{d}{dx}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\right\rbrace\\[10pt] &\qquad+\frac{1}{2}\left(\frac{\ln\left(\left|1-\frac{x}{a-1}\right|\right)}{x} + \frac{\ln\left(\left|1-\frac{x}{a+1}\right|\right)}{x}\right)\\ &=-\frac{1}{2}\frac{d}{dx}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\right\rbrace\\[10pt] &\qquad-\frac{1}{2}\frac{d}{dx}\left\lbrace\Phi\left(\frac{x}{a-1}\right) + \Phi\left(\frac{x}{a+1}\right)\right\rbrace. \end{align*} Note that here $\Phi$ is a version of Spence's Function given by $$\Phi(x) = -\int_0^x \frac{\ln|1-u|}{u} \, du = \begin{cases} \operatorname{Li}_2(x), & x \leq 1; \\ \frac{\pi^2}{3} - \frac{1}{2} \ln^2(x) - \operatorname{Li}_2(\frac{1}{x}), & x > 1. \end{cases}$$ and $\operatorname{Li_2}$ is the polylogarithm $$\operatorname{Li_2}(x) = -\int_0^x \frac{\ln(1-u)}{u} \, du.$$ Therefore, we have that \begin{align*} \int\frac{\ln{x}(a-x)}{(a-x)^2-1}\,dx &= -\frac{1}{2}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right) + \ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right)\right.\\[10pt] &\qquad\quad + \left.\Phi\left(\frac{x}{a-1}\right) + \Phi\left(\frac{x}{a+1}\right)\right\rbrace + C. \end{align*} All in all, then, going back to $(1)$, we have that \begin{equation*} \int \frac{\ln(a-1)-\ln(x)(a-x)}{(a-x)^2-1}\,dx \end{equation*} is given by \begin{align*} \frac{\ln(a-1)}{2}&\ln\left(\left|\frac{a-x+1}{x-a+1}\right|\right)+\frac{1}{2}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{a-x-1}{a-1}\right|\right)\right.\right.\\[10pt] &+ \left.\ln\left(\left|\frac{a-x+1}{a+1}\right|\right)\right) + \left.\Phi\left(\frac{x}{a-1}\right) + \Phi\left(\frac{x}{a+1}\right)\right\rbrace + C. \end{align*} Replacing $a$ with $\sqrt{2}$, we get your original integral: \begin{equation*} \int \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx \end{equation*} is given by \begin{align*} \frac{\ln(\sqrt2-1)}{2}&\ln\left(\left|\frac{\sqrt2-x+1}{x-\sqrt2+1}\right|\right)+\frac{1}{2}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{\sqrt2-x-1}{\sqrt2-1}\right|\right)\right.\right.\\[10pt] &+ \left.\ln\left(\left|\frac{\sqrt2-x+1}{\sqrt2+1}\right|\right)\right) + \left.\Phi\left(\frac{x}{\sqrt2-1}\right) + \Phi\left(\frac{x}{\sqrt2+1}\right)\right\rbrace + C. \end{align*}

Now, this function does not have any singularities on $(0,1)$, so we can use the Fundamental Theorem of Calculus to evaluate the original integral \begin{equation*} \int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx. \end{equation*} We have the following: \begin{equation*} \frac{\ln(\sqrt2-1)}{2}\ln\left(\left|\frac{\sqrt2-x+1}{x-\sqrt2+1}\right|\right)\Bigg|_{0}^{1}=\frac{1}{2}\ln(\sqrt{2}-1)\left( \ln\left(\frac{\sqrt2}{2-\sqrt2}\right)-\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\right)\tag{3} \end{equation*} \begin{equation*} \frac{1}{2}\left\lbrace\ln(x)\left(\ln\left(\left|\frac{\sqrt2-x-1}{\sqrt2-1}\right|\right)+\ln\left(\left|\frac{\sqrt2-x+1}{\sqrt2+1}\right|\right)\right)\right\rbrace\Bigg|_{0}^{1} = 0\tag{4} \end{equation*} \begin{align*} \frac{1}{2}\left\lbrace\Phi\left(\frac{x}{\sqrt2-1}\right) + \Phi\left(\frac{x}{\sqrt2+1}\right)\right\rbrace\Bigg|_{0}^{1} &= \frac{1}{2}\left\lbrace\Phi\left(\frac{1}{\sqrt2-1}\right) + \Phi\left(\frac{1}{\sqrt2+1}\right)\right\rbrace\\ &=\frac{1}{2}\left\lbrace\frac{\pi^{2}}{3} - \frac{1}{2}\ln^{2}\left(\frac{1}{\sqrt2-1}\right)\right.\\ &\qquad\left.- \operatorname{Li_{2}}\left(\sqrt2-1\right) + \operatorname{Li_{2}}\left(\frac{1}{\sqrt2+1}\right)\right\rbrace\\ &=\frac{\pi^{2}}{6}-\frac{1}{4}\ln^{2}(\sqrt{2} - 1).\tag{5} \end{align*}

Adding $(1)$, $(2)$, and $(3)$, we get \begin{align*} \int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx &=\frac{1}{2}\ln(\sqrt{2}-1)\left( \ln\left(\frac{\sqrt2}{2-\sqrt2}\right)-\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)\right)\\ &\qquad+ \frac{\pi^{2}}{6}-\frac{1}{4}\ln^{2}(\sqrt{2} - 1)\\ &=\frac{1}{2}\ln(\sqrt2-1)\left(\ln(\sqrt2)-\ln(\sqrt2(\sqrt2-1))\right.\\&\qquad\left.-\ln(\sqrt2+1)+\ln(\sqrt2-1)\right)\\ &\qquad+\frac{\pi^{2}}{6}-\frac{1}{4}\ln^{2}(\sqrt{2} - 1)\\ &=-\frac{1}{2}\ln(\sqrt2-1)\ln(\sqrt2 +1)+\frac{\pi^{2}}{6}-\frac{1}{4}\ln^{2}(\sqrt{2} - 1)\\ &=\frac{\pi^{2}}{6} + \frac{1}{2}\ln^{2}(\sqrt2 - 1) - \frac{1}{4}\ln^{2}(\sqrt2 - 1)\\ &=\frac{\pi^{2}}{6} + \frac{1}{4}\ln^{2}(\sqrt2 - 1). \end{align*} Note that we used the fact that \begin{align*} \ln(\sqrt2 + 1) + \ln(\sqrt2 - 1) &= \ln((\sqrt2 + 1)(\sqrt2 - 1)) = \ln(1) =0\\ &\implies -\ln(\sqrt2 - 1)\ln(\sqrt2 + 1) = \ln^{2}(\sqrt2 - 1). \end{align*} Therefore, we have arrived at the desired result: $$\boxed{\int_0^1 \frac{\ln(\sqrt2-1)-\ln(x)(\sqrt2-x)}{(\sqrt2-x)^2-1}\,dx = \frac{\pi^{2}}{6} + \frac{1}{4}\ln^{2}(\sqrt2 - 1)}$$