# Logarithm integral with golden ratio $\int^1_0 \frac{\log(1+\phi x^2)}{1+x}\, dx$

How to evaluate the following integral

$$\int^1_0 \frac{\log(1+\phi x^2)}{1+x}\, dx$$

where

$$\phi =\frac{\sqrt{5}+1}{2}$$

Is there a closed form for the integral ?

My first attempt would involve relating the integral to dilogarithms but that seems complicated !

• "Solve" is the wrong word". In this kind of problem, one seeks a value, not a solution. The question is how to evaluate this integral. – Michael Hardy Aug 20 '13 at 1:57
• @MichaelHardy , I don't understand how is that relevant . I am seeking a solution for the integral not just a value . – Zaid Alyafeai Aug 20 '13 at 2:04
• You can use this method. – Mhenni Benghorbal Aug 22 '13 at 4:37

Let's replace $\phi$ with a positive parameter $a$: $$\mathcal{I}(a) = \int_0^1 \frac{\log(1+a x^2)}{1+x} \mathrm{d}x \stackrel{x^2=t}{=} \int_0^1 \frac{\log(1+ a t)}{1+\sqrt{t}} \frac{\mathrm{d}t}{2 \sqrt{t}}$$ Differentiating under the integral sign: $$\begin{eqnarray} \mathcal{I}^\prime(a) &=& \int_0^1 \frac{\sqrt{t}}{2\left(1+a t\right)\left(1+ \sqrt{t}\right)} \mathrm{d}t = \left[\frac{\log \left(1+\sqrt{t}\right)}{1+a}+\frac{\log (1+a t)}{2 a \left(1+a\right)}-\frac{\arctan\left(\sqrt{a t}\right)}{\sqrt{a}\left(1+a\right)} \right]_0^1 \\ &=& \frac{\log 2}{1+a} + \frac{\log(1+a)}{2 a \left(1+a\right)} - \frac{\arctan(\sqrt{a})}{\sqrt{a} \left(1+a\right)} \\ &=& \frac{\log(1+a)}{2} \left(\frac{1}{a} - \frac{1}{1+a} \right) + \frac{\mathrm{d}}{\mathrm{d}a}\left( \log 2 \cdot \log\left(1+a\right) - \arctan^2\left(\sqrt{a}\right)\right) \\ &=& \frac{\log(1+a)}{2 a} + \frac{\mathrm{d}}{\mathrm{d}a}\left( \log 2 \cdot \log\left(1+a\right) + \frac{\log^2\left(1+a\right)}{4} - \arctan^2\left(\sqrt{a}\right)\right) \end{eqnarray}$$ Hence, for $\alpha>0$ $$\begin{eqnarray} \mathcal{I}\left(\alpha\right) &=& \int_0^\alpha \mathcal{I}^\prime(a) \mathrm{d}a = \left[ \log 2 \cdot \log\left(1+a\right) + \frac{\log^2\left(1+a\right)}{4} - \arctan^2\left(\sqrt{a}\right) \right]_{0}^{\alpha} + \frac{1}{2} \underbrace{\int_0^\alpha \frac{\log(1+a)}{a} \mathrm{d}a }_{-\operatorname{Li}_2\left(-\alpha\right)} \\ &=& \frac{\log^2\left(1+\alpha\right)}{4} + \log(2)\log\left(1+\alpha\right) - \arctan^2\left(\sqrt{\alpha}\right) - \frac{1}{2} \operatorname{Li}_2\left(-\alpha\right) \end{eqnarray}$$ Quadrature confirmation:

In[45]:= N[
With[{a = GoldenRatio}, -ArcTan[Sqrt[a]]^2 + Log[2] Log[1 + a] -
1/4 Log[1 + a]^2 - 1/2 PolyLog[2, -a]], 30]

Out[45]= 0.226575874320240735184008669784

In[46]:= NIntegrate[Log[1 + GoldenRatio x^2]/(1 + x), {x, 0, 1},
WorkingPrecision -> 70] - %

Out[46]= 0.*10^-31

• It should be noted that $\operatorname{Li}_2(-\phi) = \frac{\pi^2}{10} - \ln^2 \phi = \frac{\pi^2}{10} - \left(\operatorname{csch}^{-1} 2\right)^2$, hence the exact value of the integral can be expressed using elementary functions. – Jon Claus Aug 20 '13 at 4:44

\begin{align} \int_{0}^{1}{\ln\left(1 + \phi x^{2}\right) \over 1+x}\,{\rm d}x &= \sum_{\sigma = \pm} \int_{0}^{1}{\ln\left(1 + {\rm i}\,\sigma\,\phi^{1/2}x\right) \over 1 + x}\,{\rm d}x = \sum_{\sigma = \pm} \int_{1}^{2}{\ln\left(1 - {\rm i}\,\sigma\,\phi^{1/2} + {\rm i}\,\sigma\,\phi^{1/2}x\right) \over x}\,{\rm d}x \\[2mm]&= \sum_{\sigma = \pm}\left\lbrack\ln\left(1 - {\rm i}\,\sigma\,\phi^{1/2}\right)\ln\left(2\right) + \int_{1}^{2}{\ln\left(1 - z_{\sigma}x\right) \over x}\,{\rm d}x\right\rbrack \end{align}

where $\displaystyle{z_{\sigma} \equiv -\,{{\rm i}\,\sigma\,\phi^{1/2} \over 1 - {\rm i}\,\sigma\,\phi^{1/2}}}$

\begin{align} \int_{0}^{1}{\ln\left(1 + \phi x^{2}\right) \over 1+x}\,{\rm d}x &= \ln\left(1 + \phi\right)\ln\left(2\right) + \sum_{\sigma = \pm}\left\lbrack% \int_{0}^{1}{\ln\left(1 - 2z_{\sigma}x\right) \over x}\,{\rm d}x - \int_{0}^{1}{\ln\left(1 - z_{\sigma}x\right) \over x}\,{\rm d}x \right\rbrack \\[2mm]&= \ln\left(1 + \phi\right)\ln\left(2\right) + \sum_{\sigma = \pm}\left\lbrack\vphantom{\Large A}% -{\rm Li_{2}}\left(2z_{\sigma}\right) + {\rm Li_{2}}\left(z_{\sigma}\right) \right\rbrack \\[1cm]& \end{align}

\begin{align} \int_{0}^{1}{\ln\left(1 + \phi x^{2}\right) \over 1+x}\,{\rm d}x &= \ln\left(1 + \phi\right)\ln\left(2\right) + 2\Re\left\lbrack\vphantom{\Large A}% {\rm Li_{2}}\left(z\right) - {\rm Li_{2}}\left(2z\right) \right\rbrack \\[2mm]& \mbox{with}\quad z = -\,{{\rm i}\,\phi^{1/2} \over 1 - {\rm i}\,\phi^{1/2}} \end{align}

• Note that $1+\phi=\phi^2\to\log(1+\phi)=2\log(\phi)$ – dot dot Aug 31 '13 at 23:06