I want to find the value of
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$
At first, I solved this elementary integral:
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}} \ \mathrm dx$
Using the same method, I couldn't find my asked integral. Are there any ways to connect them? Any help would be appreciated.
Best Answer
$$ I = \displaystyle \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx = \frac{1}{2}\int_{0}^{\infty} \frac{\ln(x)}{\cosh x +\frac{1}{2}} dx $$
Consider the "discrete" Laplace transform
$$ \sum_{k=1}^{\infty}e^{-kt}\sin(kz) = \frac{1}{2}\frac{\sin z}{\cosh t -\cos z} \quad t>0 $$
If we put $z = \frac{2\pi}{3}$
$$ \sum_{k=1}^{\infty}e^{-kt}\sin\left(\frac{2\pi k}{3}\right) = \frac{1}{2}\frac{\frac{\sqrt{3}}{2}}{\cosh t + \frac{1}{2}} $$
Therefore
\begin{align*} I =& \frac{2}{\sqrt{3}}\int_{0}^{\infty} \frac{1}{2}\frac{\frac{\sqrt{3}}{2}}{\cosh x +\frac{1}{2}} \ln(x) dx\\ =& \frac{2}{\sqrt{3}}\int_{0}^{\infty} \sum_{k=1}^{\infty}e^{-kx} \sin\left(\frac{2\pi k}{3}\right)\ln(x) dx\\ =& \frac{2}{\sqrt{3}}\sum_{k=1}^{\infty}\sin\left(\frac{2\pi k}{3}\right)\int_{0}^{\infty} e^{-kx} \ln(x) dx \end{align*}
This last integral is the Laplace transform of the $\ln(x)$ function
$$ \mathcal{L}\left\{\ln(x) \right\} = \int_{0}^{\infty} e^{-st} \ln(t)dt = -\frac{\ln(s)+\gamma}{s} \quad \Re(s)> 0 $$
where $\gamma$ is the Euler constant.
Hence \begin{align*} I =& \frac{2}{\sqrt{3}}\sum_{k=1}^{\infty}\sin\left(\frac{2\pi k}{3}\right)\int_{0}^{\infty} e^{-kx} \ln(x) dx \\ =& -\frac{2}{\sqrt{3}}\sum_{k=1}^{\infty}\frac{\sin\left(\frac{2\pi k}{3}\right)\left(\ln(k)+\gamma\right)}{k}\\ =& -\frac{2}{\sqrt{3}}\underbrace{\sum_{k=1}^{\infty}\frac{\sin\left(\frac{2\pi k}{3}\right)\ln(k)}{k}}_{S_{1}} - \frac{2\gamma}{\sqrt{3}}\underbrace{\sum_{k=1}^{\infty}\frac{\sin\left(\frac{2\pi k}{3}\right)}{k}}_{S_{2}} \end{align*}
Recall the Kummer's Fourier series for the log-gamma function:
$$ \ln \Gamma(t) = \frac{1}{2}\ln \left(\frac{\pi}{\sin(\pi t)} \right)+ \left(\gamma + \ln(2\pi) \right)\left(\frac{1}{2}-t\right) + \frac{1}{\pi} \sum_{k=1}^{\infty} \frac{\sin(2\pi k t) \ln(k) }{k} $$
If we put $\displaystyle t=\frac{1}{3}$
$$ S_{1}= \sum_{k=1}^{\infty} \frac{\sin(\frac{2\pi k}{3}) \ln(k) }{k} = \pi\ln\Gamma\left(\frac{1}{3}\right) -\frac{\pi}{2}\ln\left(\frac{2\pi}{\sqrt{3}}\right)-\frac{\pi\ln(2\pi)}{6} -\frac{\gamma\pi}{6} $$
For $S_{2}$ take the principal branch of the $\ln(z)$ function
$$ S_{2} = \sum_{k=1}^{\infty}\frac{\sin\left(\frac{2\pi k}{3}\right)}{k} = \Im\left(\sum_{k=1}^{\infty} \frac{e^{\frac{2\pi ik}{3}}}{k} \right) = \Im\left(-\ln(1-e^{\frac{2\pi i}{3}})\right) = \Im \left(-\frac{\ln(3)}{2}+\frac{i\pi}{6}\right) = \frac{\pi}{6}$$
Therefore
$$I = -\frac{2}{\sqrt{3}}S_{1}-\frac{2\gamma}{\sqrt{3}}S_{2} =-\frac{2\pi}{\sqrt{3}}\ln\Gamma\left(\frac{1}{3}\right)+\frac{\pi}{\sqrt{3}}\ln\left(\frac{2\pi}{\sqrt{3}}\right)+\frac{\pi}{3\sqrt{3}}\ln(2\pi) $$
Therefore, we can conclude
$$\boxed{ \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx = \frac{\pi}{\sqrt{3}}\ln\left(\frac{2\pi}{\sqrt{3}}\right)+\frac{\pi}{3\sqrt{3}}\ln(2\pi)-\frac{2\pi}{\sqrt{3}}\ln\Gamma\left(\frac{1}{3}\right)} $$