Integral of the day XI

Integral involving the trigamma constant $\psi^{(1)}\left(\frac{1}{3}\right)$

Today we show the proof of the following result proposed by @Ali3934213 \[ \int_{0}^{\infty} \frac{\ln(x+1)}{x^3+1} dx = \frac{1}{9} \left[-\pi^2+\pi\sqrt{3}\ln(3) +\psi^{(1)}\left(\frac{1}{3}\right) \right]\] The proof relies on some properties of the polylogarithm and the trigamma function.

Proof

\[ I = \int_{0}^{\infty} \frac{\ln(x+1)}{x^3+1} dx = \int_{0}^{1} \frac{\ln(x+1)}{x^3+1} dx+ \underbrace{\int_{1}^{\infty} \frac{\ln(x+1)}{x^3+1} dx}_{J}\] \[ J = \int_{1}^{\infty} \frac{\ln(x+1)}{x^3+1} dx \stackrel{w \mapsto \frac{1}{x}}{=} \int_{0}^{1} \frac{\ln\left(\frac{w+1}{w}\right)w}{w^3+1} dw= \int_{0}^{1} \frac{\ln\left(w+1\right)w}{w^3+1} dw - \int_{0}^{1} \frac{\ln\left(w\right)w}{w^3+1} dw\] Hence \begin{align*} I =& \int_{0}^{1} \frac{\ln(x+1)}{x^3+1} dx + \int_{0}^{1} \frac{\ln\left(w+1\right)w}{w^3+1} dw - \int_{0}^{1} \frac{\ln\left(w\right)w}{w^3+1} dw\\ =& \int_{0}^{1} \frac{\ln(w+1)(w+1)}{1+w^3}dw - \int_{0}^{1} \frac{\ln\left(w\right)w}{w^3+1} dw \\ =& \underbrace{\int_{0}^{1} \frac{\ln(w+1)}{w^2-w+1}dw}_{K} - \underbrace{\int_{0}^{1} \frac{\ln\left(w\right)w}{w^3+1} dw}_{L}\\ \end{align*} Note that \[ \Im\left(\frac{2}{\sqrt{3}}\frac{1}{w-\frac{1}{2}-i\frac{\sqrt{3}}{2}}\right) = \frac{1}{w^2-w+1}\] Hence \begin{align*} K = \int_{0}^{1} \frac{\ln(w+1)}{w^2-w+1}dw =& \Im \left( \frac{2}{\sqrt{3}}\int_{0}^{1} \frac{\ln(w+1)}{w-\frac{1}{2}-i\frac{\sqrt{3}}{2}} dw \right)\\ =&\Im \left( \frac{2}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{\ln\left(t+\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}{t} dt \right)\\ =& \Im \left( \frac{2}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{\ln\left(\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)\left(\frac{t}{\frac{3}{2}+i\frac{\sqrt{3}}{2}}+1\right)\right)}{t} dt \right)\\ =& \Im \left( \frac{2\ln\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{1}{t} dt + \frac{2}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{\ln\left(\frac{t}{\frac{3}{2}+i\frac{\sqrt{3}}{2}}+1\right)}{t} dt\right)\\ =& \Im \left( \frac{2\ln\left(\frac{3}{2}+i\frac{\sqrt{3}}{2}\right)}{\sqrt{3}}\left(\ln\left(\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\right)-\ln\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)\right) + \Im\left(\frac{2}{\sqrt{3}}\int_{-\left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)}^{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \frac{\ln\left(\frac{t}{\frac{3}{2}+i\frac{\sqrt{3}}{2}}+1\right)}{t} dt\right) \\ =& \frac{\pi\ln(3)}{3\sqrt{3}}+ \Im \left( \frac{2}{\sqrt{3}}\int_{\frac{1}{2}+\frac{i}{2\sqrt{3}}}^{\frac{i}{\sqrt{3}}} \frac{\ln\left(1-s\right)}{s} ds\right) \quad \left( s \mapsto -\frac{t}{\frac{3}{2}+i\frac{\sqrt{3}}{2}}\right)\\ =& \frac{\pi\ln(3)}{3\sqrt{3}} - \frac{2}{\sqrt{3}}\Im\left(\operatorname{Li}_{2}\left(\frac{i}{\sqrt{3}}\right)-\operatorname{Li}_{2}\left(\frac{1}{2}+\frac{i}{2\sqrt{3}}\right)\right) \end{align*} The polylogarithm function satisfies: \[ \Im\left(\operatorname{Li}_{2}\left(\frac{i}{\sqrt{3}}\right)-\operatorname{Li}_{2}\left(\frac{1}{2}+\frac{i}{2\sqrt{3}}\right)\right) = -\frac{\pi^2}{18\sqrt{3}} + \frac{1}{12\sqrt{3}}\psi^{(1)}\left(\frac{1}{3}\right)\] Hence \[ K = \frac{\pi\ln(3)}{3\sqrt{3}}+\frac{\pi^2}{27} -\frac{1}{18}\psi^{(1)}\left(\frac{1}{3}\right) \] Now, for $L$; \begin{align*} L= \int_{0}^{1} \frac{\ln(w)w}{1+w^3}dw = \int_{0}^{1} \frac{\left(\frac{d}{dt}\Big|_{t=0+} w^t \right)w}{1+w^3}dw = &\frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} \frac{w^{t+1}}{1+w^3} dw \\ =& \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} \sum_{n=0}^{\infty} (-1)^nw^{3n}w^{t+1} dw \\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} (-1)^n\int_{0}^{1} w^{t+1+3n} dw \\ =& \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} (-1)^n\frac{1}{(3n+2+t)} \\ =& \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(3n+2)^2} \\ =& \sum_{n=0}^{\infty} \frac{1}{(6n+5)^2} - \sum_{n=0}^{\infty} \frac{1}{(6n+2)^2}\\ =& \frac{1}{36}\sum_{n=0}^{\infty} \frac{1}{\left(n+\frac{5}{6}\right)^2} - \frac{1}{36} \sum_{n=0}^{\infty} \frac{1}{\left(n+\frac{1}{3}\right)^2}\\ \end{align*} Recall the series representation of the polygamma function: \[ \psi^{(m)}(z) = (-1)^{m+1} m! \sum_{k=0}^{\infty} \frac{1}{(z+k)^{m+1}}\] Hence \[ L = \frac{1}{36} \left[ \psi^{(1)}\left(\frac{5}{6}\right) -\psi^{(1)}\left(\frac{1}{3}\right)\right]\] Therefore \[ I = K-L = \frac{\pi\ln(3)}{3\sqrt{3}}+\frac{\pi^2}{27} -\frac{1}{36}\psi^{(1)}\left(\frac{1}{3}\right) -\frac{1}{36}\psi^{(1)}\left(\frac{5}{6}\right)\] Now using the identity \[ \psi^{(1)}\left(\frac{1}{3}\right) + \psi^{(1)}\left(\frac{5}{6}\right) = 4\psi^{(1)}\left(\frac{2}{3}\right)\] and the reflection formula \[ \psi^{(1)}(1-z) + \psi^{(1)}(z) = \frac{\pi^2}{\sin^2(\pi z)} \] with $\displaystyle z = \frac{2}{3}$ we have \[ \psi^{(1)}\left(\frac{1}{3}\right) + \psi^{(1)}\left(\frac{2}{3}\right) = \frac{4}{3}\pi^2\] Therefore \[ \boxed{\int_{0}^{\infty} \frac{\ln(x+1)}{x^3+1} dx = \frac{1}{9} \left[-\pi^2+\pi\sqrt{3}\ln(3) +\psi^{(1)}\left(\frac{1}{3}\right) \right]}\]