Integral of the day VII

The strong connection between the $\operatorname{Li}_{2}$ function and $\phi$ vol. II

Today we show the proof of this pair of integrals posted by @infseriesbot \[\int_{0}^{\infty} \frac{\arctan\left(\frac{x}{\phi}\right)}{1+x^2}dx = \frac{\pi^2}{12} + \frac{3}{4} \ln^2(\phi) \] \[\int_{0}^{\infty} \frac{\arctan\left(\phi x\right)}{1+x^2}dx= \frac{\pi^2}{6} - \frac{3}{4} \ln^2(\phi) \] In general, we found that \[ \int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx = \frac{1}{4}\ln(1-a)\ln(a^2)+\frac{1}{2}\operatorname{Li}_{2}(a) - \frac{1}{4}\ln(a+1)\ln(a^2) - \frac{1}{2}\operatorname{Li}_{2}(-a) \quad a\neq 1,-1\]

Proof:

Consider the integral \[ \Phi(a) = \int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx \] Differentiating under the integral sign: \[ \Phi'(a) = \int_{0}^{\infty} \frac{x}{(1+x^2)(1+a^2x^2)}dx \] Applying partial fraction decomposition \[ \Phi'(a) = \int_{0}^{\infty} \frac{x}{(1+x^2)(1+a^2x^2)}dx = \frac{1}{2(a^2-1)}\int_{0}^{\infty}\left[ \frac{2a^2x}{1+a^2x^2} -\frac{2x}{1+x^2}\right]dx \] Note that \[\int\left[ \frac{2a^2x}{1+a^2x^2} -\frac{2x}{1+x^2}\right]dx = \int \frac{2a^2x}{1+a^2x^2} dx - \int \frac{2x}{1+x^2} dx = \ln(1+a^2x^2)-\ln(1+x^2) + C \] Therefore \[ \Phi'(a) = \frac{1}{2(a^2-1)}\int_{0}^{\infty}\left[ \frac{2a^2x}{1+a^2x^2} -\frac{2x}{1+x^2}\right]dx = \frac{1}{2(a^2-1)} \left[\ln\left(\frac{1+a^2x^2}{1+x^2}\right)\right]_{0}^{\infty} \] Note that \[ \lim_{x\to \infty} \ln\left(\frac{1+a^2x^2}{1+x^2}\right) = \ln\left(\lim_{x \to \infty} \frac{\frac{1}{x^2}+a^2}{\frac{1}{x^2}+1}\right) = \ln(a^2) \] Hence \[ \Phi'(a) = \frac{1}{2(a^2-1)} \left[\ln\left(\frac{1+a^2x^2}{1+x^2}\right)\right]_{0}^{\infty} = \frac{\ln(a^2)}{2(a^2-1)}\] Given that for $a\in \mathbb{C}$ in general \[ \ln(a^n) \neq n\ln(a)\] we have to take care when evaluating the right hand side.

Integrating

\[ \Phi(a) = \int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx = \int\frac{\ln(a^2)}{2(a^2-1)}da + C\] \[\frac{1}{2}\int \frac{\ln(a^2)}{a^2-1}da = \frac{1}{4}\underbrace{\int\frac{\ln(a^2)}{a-1}da }_{J} -\frac{1}{4} \underbrace{\int \frac{\ln(a^2)}{a+1}da}_{K} \] For $J$: \begin{align*} J = \int\frac{\ln(a^2)}{a-1}da \stackrel{IBP}{=}& -\ln(1-a)\ln(a^2)+2\int \frac{\ln(1-a)}{a} +C \\ =&-\ln(1-a)\ln(a^2)-2\operatorname{Li}_{2}(a)+C \end{align*} \[ \therefore J = -\ln(1-a)\ln(a^2)-2\operatorname{Li}_{2}(a)+C\] Now, for $K$: \[ K = \int \frac{\ln(a^2)}{a+1}da \stackrel{IBP}{=} \ln(1+a)\ln(a^2)-2\int \frac{\ln(1+a)}{a}da\] \[\int\frac{\ln(a+1)}{a}da \stackrel{a=-w}{=} \int \frac{\ln(1-w)}{w}dw = -\operatorname{Li}_{2}(w)+ C = -\operatorname{Li}(-a) + C \] \[ \therefore K = \int \frac{\ln(a^2)}{a+1}da= \ln(a+1)\ln(a^2) + 2\operatorname{Li}_{2}(-a)+ C\] Hence \[ \Phi(a) = \int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx =\frac{1}{2}\int \frac{\ln(a)}{a^2-1}da = \frac{1}{4}\ln(1-a)\ln(a^2)+\frac{1}{2}\operatorname{Li}_{2}(a) - \frac{1}{4}\ln(a+1)\ln(a^2) - \frac{1}{2}\operatorname{Li}_{2}(-a)+C\] To find $C$ note that if $a \to 0 \Longrightarrow C \to 0 $: Hence \[ \Phi(a) = \int_{0}^{\infty} \frac{\arctan(\phi x)}{1+x^2}dx =\frac{1}{4}\ln(1-a)\ln(a^2)+\frac{1}{2}\operatorname{Li}_{2}(a) - \frac{1}{4}\ln(a+1)\ln(a^2) - \frac{1}{2}\operatorname{Li}_{2}(-a) \] If we put $ a = \phi$

and using the following identities and properties of the dilogarithm function and the golden ratio: \[ \ln(\phi)\ln(\phi+1)= 2\ln^2(\phi) \] \[ \ln(1-\phi)\ln(\phi) = -\frac{\ln^2(\phi)}{2}+\frac{i\pi \ln(\phi)}{2}\] \[\operatorname{Li}_{2}(-\phi) = -\frac{\pi^2}{10}-\ln^2(\phi)\] \[\operatorname{Li}_{2}(\phi^{-1}) = \frac{\pi^2}{10} -\ln^2(\phi) \] \[ \operatorname{Li}_{2}(\phi) +\operatorname{Li}_{2}(\phi^{-1}) = -\frac{\pi^2}{6}-\frac{\ln^2(-\phi)}{2} \] \[ \operatorname{Li}_{2}(\phi) = \frac{7\pi^2}{30} + \frac{\ln^2(\phi)}{2} -i\pi \ln(\phi) \] Hence \[ \int_{0}^{\infty} \frac{\arctan(\phi x)}{1+x^2}dx = \underbrace{\frac{1}{2}\ln(1-\phi)\ln(\phi)}_{-\frac{\ln^2(\phi)}{2}+\frac{i\pi\ln(\phi)}{2}}+\underbrace{\frac{1}{2}\operatorname{Li}_{2}(\phi)}_{ \frac{7\pi^2}{60} + \frac{\ln^2(\phi)}{4} -\frac{i\pi \ln(\phi)}{2}} \underbrace{- \frac{1}{2}\ln(\phi+1)\ln(\phi)}_{-\ln^2(\phi)} \underbrace{- \frac{1}{2}\operatorname{Li}_{2}(-\phi)}_{\frac{\pi^2}{20}+\frac{\ln^2(\phi)}{2}}\] \[ \boxed{\therefore \int_{0}^{\infty} \frac{\arctan(ax)}{1+x^2}dx = \frac{\pi^2}{6} - \frac{3}{4}\ln^2(\phi)}\] If now we put $a = \phi^{-1}$

and using th following identities in addition to the others: \[ \ln(1-\phi^{-1})\ln(\phi^{-1}) = \ln^2(\phi) \] \[ \ln(\phi^{-1}+1)\ln(\phi^{-1}) = -2\ln^2(\phi) \] \[ \operatorname{Li}_{2}(-\phi^{-1}) = -\frac{\pi^2}{15} + \frac{1}{2}\ln^2(\phi) \] we have \[ \Phi( \phi^{-1}) = \int_{0}^{\infty} \frac{\arctan\left(\frac{x}{\phi}\right)}{1+x^2}dx =\underbrace{\frac{1}{2}\ln(1-\phi^{-1})\ln(\phi^{-1})}_{\frac{1}{2}\ln^2(\phi)}+\underbrace{\frac{1}{2}\operatorname{Li}_{2}(\phi^{-1})}_{\frac{\pi^2}{20}-\ln^2(\phi)}\underbrace{ - \frac{1}{2}\ln(\phi^{-1}+1)\ln(\phi^{-1})}_{\ln^2(\phi)} \underbrace{- \frac{1}{2}\operatorname{Li}_{2}(-\phi^{-1}) }_{\frac{\pi^2}{30}-\frac{1}{4}\ln^2(\phi)}\] \[ \boxed{\therefore \int_{0}^{\infty} \frac{\arctan\left(\frac{x}{\phi}\right)}{1+x^2}dx = \frac{\pi^2}{12} +\frac{3}{4}\ln^2(\phi)} \]