Evaluation of $\int_{0}^{1}\frac{\ln x}{x^2-x-1}dx$

Question

Evaluation of $\displaystyle \int_{0}^{1}\frac{\ln x}{x^2-x-1}dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2-x-1}dx=\int_{0}^{\infty}\frac{\ln(x)}{\left(x-\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2}dx$

Now Put $\displaystyle \left(x-\frac{1}{2}\right)=\frac{\sqrt{5}}{2}\sec \theta\;,$ Then $\displaystyle dx = \frac{\sqrt{5}}{2}\sec \theta \tan \theta$

So $$ I= -\frac{2}{\sqrt{5}}\int_{-\frac{1}{2}}^{\infty}\frac{\ln(\sqrt{5}\sec \theta+1)-\ln(2)}{\tan \theta}\cdot \sec \theta d\theta$$

So $$I = \frac{2}{\sqrt{5}}\int_{-\frac{1}{2}}^{\infty}\frac{\ln(2)+\ln(\cos \theta)-\ln(\sqrt{5}+\cos \theta)}{\sin \theta} d\theta$$

Now How can I solve after that, Help Required, Thanks

Answer 1

So, by the comment of the OP, the integral we have to study is $$\int_{0}^{1}\frac{\log\left(x\right)}{x^{2}-x-1}dx.$$ We note that $$\begin{align} I= & \int_{0}^{1}\frac{\log\left(x\right)}{x^{2}-x-1}dx \\ = & \int_{0}^{1}\frac{\log\left(x\right)}{\left(x-\frac{1-\sqrt{5}}{2}\right)\left(x-\frac{1+\sqrt{5}}{2}\right)}dx \\ = & -\frac{2}{\sqrt{5}}\left(\int_{0}^{1}\frac{\log\left(x\right)}{2x+\sqrt{5}-1}dx-\int_{0}^{1}\frac{\log\left(x\right)}{2x-\sqrt{5}-1}dx\right) \\ = & -\frac{2}{\sqrt{5}}\left(I_{1}-I_{2}\right), \end{align}$$ say. Let us analyze $I_{1}$. Integrating by part we get $$I_{1}=\int_{0}^{1}\frac{\log\left(x\right)}{2x+\sqrt{5}-1}dx=\frac{1}{\sqrt{5}-1}\int_{0}^{1}\frac{\log\left(x\right)}{\frac{2x}{\sqrt{5}-1}+1}dx=-\frac{1}{2}\int_{0}^{1}\frac{\log\left(\frac{2x}{\sqrt{5}-1}+1\right)}{x}dx $$ $$=\frac{1}{2}\textrm{Li}_{2}\left(-\frac{2}{\sqrt{5}-1}\right) $$ where $\textrm{Li}_{2}(x)$ is the Dilogarithm function. In a similar way we can find that $$I_{2}=\int_{0}^{1}\frac{\log\left(x\right)}{2x-\sqrt{5}-1}dx=\frac{1}{2}\textrm{Li}_{2}\left(\frac{2}{\sqrt{5}+1}\right) $$ so $$I=\frac{\textrm{Li}_{2}\left(\frac{2}{\sqrt{5}+1}\right)-\textrm{Li}_{2}\left(-\frac{2}{\sqrt{5}-1}\right)}{\sqrt{5}} $$ and since holds $$\textrm{Li}_{2}\left(\phi^{-1}\right)=\frac{1}{10}\pi^{2}-\log^{2}\left(\phi\right),\,\textrm{Li}_{2}\left(-\left(\phi-1\right)^{-1}\right)=-\frac{1}{10}\pi^{2}-\log^{2}\left(\phi\right) $$ where $\phi=\frac{\sqrt{5}+1}{2} $ is the golden ratio we have $$I=\frac{\pi^{2}}{5\sqrt{5}}.$$

Answer 2

$$\begin{align*} I &= \int_0^1 \frac{\ln x}{x^2 - x - 1} \, dx \\ &= \int_0^1 \left[-\int_x^1 \frac{dy}y\right] \, \frac{dx}{x^2-x-1} \\ &= - \int_0^1 \left[\int_0^y \frac{dx}{x^2-x-1}\right] \, \frac{dy}y \\ &= -\frac1{\sqrt5} \int_0^1 \ln\left(\frac{2+(1-\sqrt5)y}{2+(1+\sqrt5)y}\right) \, \frac{dy}y \\ &= \frac1{\sqrt5} \int_0^1 \frac{\ln\left(1+\phi y\right) - \ln\left(1+(1-\phi)y\right)}{y} \, dy \\ &= \frac{J(\phi) - J(1-\phi)}{\sqrt5} \\ &= \frac{\operatorname{Li}_2(\phi-1) - \operatorname{Li}_2(-\phi)}{\sqrt5} \\ &= \boxed{\frac{\pi^2}{5\sqrt5}} \end{align*}$$

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Evaluating the parameterized integral:

$$\begin{align*} J(a) &= \int_0^1 \frac{\ln(1+ay)}y \, dy \\[1ex] J'(a) &= \int_0^1 \frac{dy}{1+ay} \\ &= \frac{\ln(1+a)}a \\[1ex] \implies J(a) &= J(0) + \int_0^a \frac{\ln(1+b)}b \, db \\ &= \int_0^1 \frac{\ln(1+ab)}b \, db \\ &= - \sum_{n=1}^\infty \frac{(-a)^n}n \int_0^1 b^{n-1} \, db \\ &= - \sum_{n=1}^\infty \frac{(-a)^n}{n^2} \\ &= -\operatorname{Li}_2(-a) \end{align*}$$

Answer 3

\begin{align*} \text{equation} \quad & x^2 - x - 1 = 0 \quad \text{gives solutions } r_1 = \varphi = \frac{1 + \sqrt{5}}{2} \text{ and } r_2 = \frac{1 - \sqrt{5}}{2} = 1 - \varphi \text{, where } r_1 - r_2 = \sqrt{5}, \\ & \varphi^2 = \varphi + 1, \quad \frac{\varphi - 1}{\varphi} = \frac{1}{\varphi^2} \text{, etc.} \\ \text{Breakdown into simple fractions} \quad & \dots \\ \Rightarrow \boxed{\text{I}} \quad & = \frac{\pi^2}{3\sqrt{5}} + \frac{2}{\sqrt{5}}\int\limits_0^{1/\varphi^2} \frac{\ln(1 - u)}{u} du - \frac{2}{\sqrt{5}}\ln^2(\varphi) \\ \text{Using the bilogarithmic function} \quad & f_2(z) = \int\limits_z^0 \frac{\ln(1 - x)}{x} dx \text{ and given that } f_2(\varphi^{-2}) = \frac{\pi^2}{15} - \ln^2(\varphi) \\ \text{We have} \quad & I = \frac{\pi^2}{3\sqrt{5}} - \frac{2}{\sqrt{5}}f_2\left(\frac{1}{\varphi^2}\right) - \frac{2}{\sqrt{5}}\ln^2(\varphi) \\ & = \frac{\pi^2}{3\sqrt{5}} - \frac{2}{\sqrt{5}}\left(\frac{\pi^2}{15} - \ln^2(\varphi)\right) - \frac{2}{\sqrt{5}}\ln^2(\varphi) \\ & = \frac{\pi^2}{3\sqrt{5}} - \frac{2\pi^2}{15\sqrt{5}} \\ \Rightarrow \boxed{\int\limits_0^1 \frac{\ln(x)}{x^2 - x - 1}dx} \quad & = \frac{\pi^2}{5\sqrt{5}}. \end{align*} \text{We also studied the bilogarithmic function here: \url{http://mathworld.wolfram.com/Dilogarithm.html}.} \end{align*}