Integral of the day XXXIII

An integral representation of the dilogarithm function

Today we show the proof of this integral posted by @Ali39342137 \[\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{2+\cos x}\right) dx = \frac{\pi^3}{12} -\frac{\pi}{2}\ln^2\left(2\right)\] In general we found \[\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{a+\cos x}\right) dx = \pi \operatorname{Li}_{2}\left(\frac{1}{a}\right) \quad |a|\geq1 \] In the proof we make use of Fourier series and some properties of the dilogarithm function $\operatorname{Li}_{2}(z)$.

Proof:

Consider the following Fourier series (Proof in the Appendix 1): \[ \sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} = \arctan\left(\frac{p\sin x}{1-p\cos x}\right) \quad 0\lt x\lt 2\pi, \quad p^2\leq 1 \] Then, if $p = -q$ \[ \sum_{k=1}^{\infty } \frac{(-1)^kq^{k}\sin(kx)}{k} = \arctan\left(\frac{-q\sin x}{1+q\cos x}\right) = - \arctan\left(\frac{q\sin x}{1+q\cos x}\right) \] Therefore \[ \arctan\left(\frac{\sin x}{\frac{1}{q}+\cos x}\right) = \sum_{k=1}^{\infty } \frac{(-1)^{k+1}q^{k}\sin(kx)}{k} \quad \frac{1}{q}\geq 1 \] Hence \begin{align*} I = \int_{0}^{\pi} x\arctan\left(\frac{\sin x}{a+\cos x}\right) dx =& \int_{0}^{\pi} x \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}\sin(kx)}{k}dx\\ =& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k} \int_{0}^{\pi} x\sin(kx)dx\\ =& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k^3} \int_{0}^{k\pi} w\sin(w)dx \quad (w \mapsto kx)\\ \stackrel{IBP}{=}& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k^3} \left[-\cos(x)x\Big|_{0}^{k\pi}+ \int_{0}^{k\pi}\cos(w)dx \right] \\ =& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k^3} \left[-k\pi\cos(\pi k) + \sin(\pi k) \right] \\ =& \pi \underbrace{\sum_{k=1}^{\infty } \frac{(-1)^{k}\left(\frac{1}{a}\right)^{k}\cos(k\pi) }{k^2}}_{\cos(k \pi) = (-1)^k} + \underbrace{\sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}\sin(\pi k) }{k^3}}_{\sin(k\pi)=0} \\ =& \pi \sum_{k=1}^{\infty } \frac{\left(\frac{1}{a}\right)^{k} }{k^2} \\ =& \pi \operatorname{Li}_{2}\left(\frac{1}{a}\right) \end{align*} If $a = 2$ \[ \int_{0}^{\pi} x\arctan\left(\frac{\sin x}{2+\cos x}\right) dx = \pi \operatorname{Li}_{2}\left(\frac{1}{2}\right)\] From the relation (proof in the Appendix 2) \[ \operatorname{Li}_{2}(z) +\operatorname{Li}_{2}(1-z) = \frac{\pi^2}{6} -\ln(z)\ln(1-z) \] if we put $\displaystyle z = \frac{1}{2}$ \[ \operatorname{Li}_{2}\left(\frac{1}{2} \right) = \frac{\pi^2}{12} -\frac{1}{2}\ln^2\left(2\right) \] Therefore \[\boxed{\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{2+\cos x}\right) dx = \frac{\pi^3}{12} -\frac{\pi}{2}\ln^2\left(2\right)}\] \[\boxed{\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{a+\cos x}\right) dx = \pi \operatorname{Li}_{2}\left(\frac{1}{a}\right) \quad |a|\geq1 }\]

Appendix 1 \[ \sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} = \arctan\left(\frac{p\sin x}{1-p\cos x}\right) \quad 0\lt x\lt 2\pi, \quad p^2\leq 1 \]

Proof

Take the principal branch of the $\ln(z)$ function. Hence \begin{align*} \sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} =& \Im\left( \sum_{k=1}^{\infty } \frac{(pe^{ix})^k}{k} \right) = \Im\left( -\ln(1-pe^{ix})\right)\\ =& -\arg\left(1-pe^{ix}\right)\\ =& -\arg\left(1-p\cos(x)-pi\sin(x)\right)\\ \end{align*} Since $|p|\lt 1$ then $1-p\cos(x)>0$ therefore \begin{align*} \sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} =& -\arg\left(1-p\cos(x)-pi\sin(x)\right)\\ =& -\arctan\left( \frac{-p\sin x}{1-p\cos(x)}\right) \\ =& \arctan\left( \frac{p\sin x}{1-p\cos(x)}\right) \quad \left(\arctan \textrm{ is odd } \right) \end{align*}

Appendix 2 \[ \operatorname{Li}_{2}(z) +\operatorname{Li}_{2}(1-z) = \frac{\pi^2}{6} -\ln(z)\ln(1-z) \]

Proof \begin{align*} \operatorname{Li}_{2}(z) = &\int_{z}^{0} \frac{\ln(1-t)}{t} dt \stackrel{IBP}{=} -\ln(z)\ln(1-z) -\int_{0}^{z}\frac{\ln (t)}{1-t}dt\\ =& -\ln(z)\ln(1-z) + \int_{1}^{1-z}\frac{\ln (1-w)}{w}dw \quad \left( w\mapsto 1-t \right)\\ =& -\ln(z)\ln(1-z) + \int_{1}^{0}\frac{\ln (1-w)}{w}dw + \int_{0}^{1-z}\frac{\ln (1-w)}{w}dw\\ =& -\ln(z)\ln(1-z) + \int_{1}^{0}\frac{\ln (1-w)}{w}dw - \int_{1-z}^{0}\frac{\ln (1-w)}{w}dw\\ =& -\ln(z)\ln(1-z) + \operatorname{Li}_{2}(1) - \operatorname{Li}_{2}(1-z)\\ \end{align*} Therefore \[ \operatorname{Li}_{2}(z) + \operatorname{Li}_{2}(1-z) = -\ln(z)\ln(1-z) + \zeta(2) = -\ln(z)\ln(1-z) + \frac{\pi^2}{6} \]