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How to compute this integral $$\int_0^1\frac{\ln x}{x^2-x-1}dx$$

One method I have thought before is to solve the equation $x^2-x-1=0$ for its roots $x_{1,2}=\frac{\sqrt{5}\pm 1}{2}$, with $\phi=\frac{\sqrt{5}+ 1}{2}$. Then, use the polylogarithmic integral formula $$\int \frac{\ln x}{x-a}=\ln x\ln(1-\frac{x}{a})+\text{Li}_2(\frac{x}{a}),$$ to get $$\int_0^1\frac{\ln x}{x^2-x-1}dx =\frac1{\sqrt{5}}(\text{Li}_2(\frac1\phi)-\text{Li}_2(-\phi)).$$ In reference to the following known results $$\text{Li}_2(\frac1\phi)=\frac1{10}\pi^2-\ln ^2\phi,\>\>\>\>\> \text{Li}_2(-\phi)=-\frac1{10}\pi^2-\ln ^2\phi,$$ we can get

$$\int_0^1\frac{\ln x}{x^2-x-1}dx=\frac{\pi^2}{5\sqrt{5}}.$$ However, this method relies on the property of the bilogarithmic functions $\text{Li}_2(z) $; so do other methods found in Evaluation of $\int_{0}^{1}\frac{\ln x}{x^2-x-1}dx$. They look dull and may be unnecessary, given its clean elementary result $\frac{\pi^2}{5\sqrt{5}}$.

Can someone give a "cleverer" method which does not rely on special polylogarithmic functions? Many thanks in advance.

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1 Answers1

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Below is an elementary solution. \begin{align} I&=\int_0^1\frac{\ln x}{x^2-x-1}dx = \frac1{\sqrt5}\int_0^1\frac{\overset{1-\frac x\phi =t^2}{\ln x}}{x-\phi}-\frac{\overset{1+\phi x=\frac1{t^2}}{\ln x}}{x+\frac1\phi}{dx}\\ &= \frac4{\sqrt5} \int_1^{\frac1{\phi}} \frac{\ln (1-t)}t dt + \frac{\ln \left(1+\frac1t\right)}t \overset{t\to \frac1t}{dt}\\ & = \frac4{\sqrt5}\bigg[ \int_0^{\frac1{\phi}} \frac{\ln (1-t)}t \overset{t=\frac x\phi}{dt}-\int_0^\phi\frac{\ln(1+t)}t \overset{x=\frac t\phi}{dt} + \int_0^1\frac{\ln \frac{1+t}{1-t}}tdt\bigg]\\ & = \frac4{\sqrt5} \int_0^1 \frac{\ln\frac{\phi+\phi^2x}{\phi-x}}x \ \overset{ibp} { dx}+\frac4{\sqrt5}\cdot \frac{\pi^2}4 =- 4I + \frac{\pi^2}{\sqrt5} = \frac{\pi^2}{5\sqrt5} \end{align} where $\int_0^1\frac{\ln \frac{1+t}{1-t}}tdt=\frac{\pi^2}4$.

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