Integral of the day XXXIV

Integral involving the golden mean $\phi$

Today we show the proof of the following integral posted by @integralsbot \[ \int_0^\pi \frac{x^2}{\sqrt{5}-2\cos x} dx = 2\pi\ln^2(\phi) + \frac{\pi^3}{15} \] The proof relies on a "discrete" Laplace transform a technique that we have used in the past here and here

Proof:

Consider the following discrete Laplace transform (proof in Appendix 1): \[ 1+ 2\sum_{k=1}^{\infty} e^{-kt}\cos(kx) = \frac{\sinh t}{\cosh t -\cos x } \quad t>0\] If $\cosh t = \frac{\sqrt{5}}{2} $ then \[ t = \ln\left(\frac{1+\sqrt{5}}{2} \right) = \ln(\phi) \] \[ \sinh\left(\frac{1+\sqrt{5}}{2} \right) = \frac{1}{2} \] therefore \[ \frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}-\cos x} = 1+ 2\sum_{k=1}^{\infty} \phi^{-k}\cos(kx) \] where $\displaystyle \phi = \frac{1+\sqrt{5}}{2}$

Hence \begin{align*} I = \int_0^\pi \frac{x^2}{\sqrt{5}-2\cos x} dx =& \frac{1}{2} \int_0^\pi \frac{x^2}{\frac{\sqrt{5}}{2}-\cos x} dx \\ =& \int_0^\pi \left(\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}-\cos x}\right) x^2 dx\\ =& \int_0^\pi \left(1+ 2\sum_{k=1}^{\infty} e^{-k\ln(\phi)}\cos(kx)\right) x^2 dx\\ =& \int_{0}^{\pi }\left[ x^2 + 2\sum_{k=1}^{\infty} \phi^{-k} \cos(kx)x^2 \right] dx \\ =& \int_{0}^{\pi } \left[x^2 dx + 2\sum_{k=1}^{\infty} \phi^{-k} \int _{0}^{\pi} \cos(kx)x^2 \right] dx \\ =& \int_{0}^{\pi } \left[x^2 dx + 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3} \int _{0}^{k\pi} \cos(w)w^2\right] dx \\ \stackrel{IBP}{=} &\frac{\pi^3}{3}+ 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3}\left[ w^2\sin(w)\Big|_{0}^{k\pi} -2\int_{0}^{k\pi} \sin(w)w dw \right] \\ \stackrel{IBP}{=} &\frac{\pi^3}{3}+ 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3}\left[ (k\pi)^2\sin(k \pi ) +2\cos(w)w\Big|_{0}^{k\pi} +2 \int_{0}^{\pi k} \cos(w) dw \right] \\ = &\frac{\pi^3}{3}+ 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3}\left[ (k\pi)^2\sin(k \pi ) +2\cos(k \pi )k\pi -2\sin(\pi k)\right] \\ =& \frac{\pi^3}{3} + \underbrace{2\sum_{k=1}^{\infty} \phi^{-k}\frac{(k^2\pi^2-2)\sin(k\pi)}{k^3}}_{\sin(\pi k) = 0 } + \underbrace{4\sum_{k=1}^{\infty} \phi^{-k}\pi \frac{\cos(k \pi)}{k^2} }_{\cos(\pi k) =(-1)^k}\\ =& \frac{\pi^3}{3} + 4\pi \sum_{k=1}^{\infty} \frac{(-\phi^{-1})^k}{k^2} \\ =& \frac{\pi^3}{3} + 4\pi \operatorname{Li}_{2}(-\phi^{-1}) \\ \end{align*} where we used the series expansion for the dilogarithm function $\operatorname{Li}_{2}(z)$: \[ \operatorname{Li}_{2}(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2} \] Landen and Euler proved (proof in the Appendix 2): \[ \operatorname{Li}_{2}\left(-\phi^{-1} \right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2(\phi) \] Therefore \[ \boxed{ \int_0^\pi \frac{x^2}{\sqrt{5}-2\cos x} dx = 2\pi\ln^2(\phi) + \frac{\pi^3}{15} }\]

Appendix 1

\[ 1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx = \frac{\sinh t}{\cosh t - \cos x} \quad { t>0 }\] Proof: \begin{align*} 1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{-kt}e^{ikx} \right)\\ =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{k(ix-t)} \right)\\ =& 1+2\Re\left(\frac{1}{e^{t-ix}-1} \right)\\ =& 1+2\Re\left(\frac{e^{-t}}{e^{-ix}-e^{-t}} \right)\\ =& 1+2e^{-t}\Re\left(\frac{1}{\cos x-i\sin x- e^{-t}}\right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{(\cos x-e^{-t})^2 + \sin^2 x} \right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{e^{-2t}-2e^{t}\cos x + 1} \right)\\ =& 1+\Re\left(\frac{\cos x -e^{-t}+ i\sin x}{\cosh t-\cos x} \right)\\ =& \frac{\cosh t- \cos x}{\cosh t- \cos x}+\frac{\cos x-e^{-t}}{(\cosh t- \cos x )} \\ =& \frac{\sinh t}{\cosh t - \cos x} \end{align*}

Appendix 2

\[ \operatorname{Li}_{2}\left(-\phi^{-1} \right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2(\phi) \]

Proof:

Consider the following derivative: \[ \frac{d}{dx} \operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) = = -\left[ \ln\left(1+ \frac{x}{1-x}\right)\right]\left(\frac{1}{x} + \frac{1}{1-x} \right)= \ln(1-x)\left(\frac{1}{x} + \frac{1}{1-x} \right) \] Integrating \[ \operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) = -\operatorname{Li}_{2}(x) - \frac{1}{2} \ln^2(1-x) \] \[ \Longrightarrow \operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) + \operatorname{Li}_{2}(x) = - \frac{1}{2} \ln^2(1-x) \tag{1} \] If we put $\displaystyle x = \frac{3-\sqrt{5}}{2}$ \[ \Longrightarrow \operatorname{Li}_{2}\left(\frac{1-\sqrt{5}}{2} \right) + \operatorname{Li}_{2}\left(\frac{3-\sqrt{5}}{2}\right) = - \frac{1}{2} \ln^2\left(\frac{1+\sqrt{5}}{2}\right) \tag{2} \] Now we will show that \[ \frac{1}{2} \operatorname{Li}_{2}(x^2) = \operatorname{Li}_{2}(x) +\operatorname{Li}_{2}(-x) \tag{3} \] From the factorization \[ (1-x^r) = (1-x)(1-wx)(1-w^2x)\cdots(1-w^{r-1}x)\] where $\displaystyle w = e^{\frac{i2\pi}{r}}$ Taking logarithms and dividing by $x$ both sides \[ \frac{\ln(1-x^r)}{x} = \frac{\ln(1-x)}{x} + \frac{\ln(1-wx)}{x} + \cdots +\frac{\ln(1-w^{r-1}x)}{x} \] Integrating \[ \frac{1}{r}\operatorname{Li}_{2}(x^r) = \operatorname{Li}_{2}(x) + \operatorname{Li}_{2}(xw) + \cdots + \operatorname{Li}_{2}(w^{r-1}x) + C\] If $x \to 0$ then $C\to 0$ \[ \frac{1}{r}\operatorname{Li}_{2}(x^r) = \operatorname{Li}_{2}(x) + \operatorname{Li}_{2}(xw) + \cdots + \operatorname{Li}_{2}(w^{r-1}x) \] If we put $ r=2$ \[ \frac{1}{2} \operatorname{Li}_{2}(x^2) = \operatorname{Li}_{2}(x) +\operatorname{Li}_{2}(-x) \tag{3} \] From (1) and (3) we have \[ \operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) + \frac{1}{2}\operatorname{Li}_{2}(x) - \operatorname{Li}_2(-x) = -\frac{1}{2}\ln^2(1-x) \tag{4} \] If we solve the equation \[ \frac{-x}{1-x} = x^2 \Longrightarrow x^2-x+1=0 \Longrightarrow x = \frac{1-\sqrt{5}}{2} \] Hence from (4) we have \[ \frac{3}{2}\operatorname{Li}_{2} \left(\frac{3-\sqrt{5}}{2}\right)- \operatorname{Li}_{2}\left(\frac{\sqrt{5}-1}{2}\right) = -\frac{1}{2}\ln^2\left(\frac{1+\sqrt{5}}{2}\right) \tag{5} \] If we put $\displaystyle x = \frac{3-\sqrt{5}}{2}$ in the relation (Proof in the Appendix 2 here) \[ \operatorname{Li}_{2}(x) + \operatorname{Li}_{2}(1-x) = \frac{\pi^2}{6} -\ln(x)\ln(1-x) \tag{6} \] we have \[ \operatorname{Li}_{2}\left(\frac{3-\sqrt{5}}{2}\right) + \operatorname{Li}_{2}\left(\frac{\sqrt{5}-1}{2}\right) = \frac{\pi^2}{6} -\ln\left(\frac{3-\sqrt{5}}{2}\right)\ln\left(\frac{\sqrt{5}-1}{2}\right) \tag{7}\] Solving the system given by (5) and (7) we have \[ \operatorname{Li}_{2}\left(\frac{3-\sqrt{5}}{2}\right) = \frac{\pi^2}{15} - \frac{1}{4}\ln^2\left(\frac{3-\sqrt{5}}{2}\right) \tag{8} \] Then from (2) and (8) we get \[ \operatorname{Li}_{2}\left(\frac{1-\sqrt{5}}{2}\right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2\left(\frac{\sqrt{5}-1}{2}\right) \] which is basically the same as \[ \operatorname{Li}_{2}\left(-\phi^{-1}\right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2\left(\phi\right) \]