How do you evaluate: $\int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$ 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.
 A: $$ I = \displaystyle  \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx = \frac{1}{2}\int_{0}^{\infty} \frac{\ln(x)}{\cosh x +\frac{1}{2}} dx $$
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
$$\boxed{ \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx = \frac{\pi}{\sqrt{3}}\ln\left(\frac{2\pi}{\sqrt{3}}\right)+\frac{\pi}{3\sqrt{3}}\ln(2\pi)-\frac{2\pi}{\sqrt{3}}\ln\Gamma\left(\frac{1}{3}\right)}  $$
A: The similar problem was posted on AoPS the other day. There are nice answers posted. For the sake of completeness, I would like to add the solution, based on approach developed by Yaroslav Blagouchine; it is convenient for solving the problems with a specific symmetry by means of the integration along a rectangular contour in the complex plane.
Let
$$I=\int_0^\infty \frac{\ln{x}}{e^{x}+e^{-x}+1}dx=\int_0^\infty \frac{\ln{x}}{2\cosh x+1}dx$$
$$=\lim_{a\to0}\frac{1}{2}\int_0^\infty\frac{\ln(x^2+a^2)}{2\cosh x+1}dx=\frac{1}{4}\lim_{a\to0}\Re\int_{-\infty}^\infty\frac{\ln(a-ix)}{\cosh x+\frac{1}{2}}dx$$
Let's consider
$$I(a)=\frac{1}{4}\int_{-\infty}^\infty\frac{\ln(a-ix)}{\cosh x+\frac{1}{2}}dx=\frac{\pi}{2}\int_{-\infty}^\infty\frac{\ln(a-2\pi i t)}{\cosh 2\pi t+\frac{1}{2}}dt$$
$$=\frac{\pi}{2}\ln2\pi\int_{-\infty}^\infty\frac{dt}{\cosh 2\pi t+\frac{1}{2}}+\frac{\pi}{2}\int_{-\infty}^\infty\frac{\ln\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t+\frac{1}{2}}dt=I_1+I_2(a)$$
Now, we move to the complex plane and consider the closed rectangular contour $-R\,\to R\,\to (R+i)\,\to (-R+i)\,\to -R;\,\,R\to \infty$; counter clockwise.
Let's also consider the following integral along this contour. The integrand has simple poles at the points $z=\frac{i}{3}$ and $\frac{2i}{3}$.
$$\frac{\pi}{2}\ln2\pi\oint\frac{e^{i\beta z}}{\cosh 2\pi z-\frac{1}{2}}dz=I_1(\beta)\big(1-e^{-\beta}\big)=\frac{\pi}{2}\ln2\pi \,2\pi i\operatorname{Res}_{\binom{\frac{i}{3}}{\frac{2i}{3}}}\frac{e^{i\beta z}}{\cosh 2\pi z-\frac{1}{2}}$$
$$I_1(\beta)\big(1-e^{-\beta}\big)=\frac{\pi}{2}\ln2\pi\frac{2\pi i}{\sqrt 3\pi i}\Big(e^{-\frac{\beta}{3}}-e^{-\frac{2\beta}{3}}\Big)$$
(We also have to add side integrals - along $R\to R+i$ and $-R+i\to-R$, but these integrals $\to 0$ at $R\to\infty$). Taking the limit $\beta\to 0$, we find
$$\boxed{\,\,I_1(0)=I_1=\frac{\pi}{3\sqrt 3}\ln2\pi\,\,}$$
To evaluate $I_2(a)$, we notice that
$$\frac{\ln\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t-\frac{1}{2}}=\frac{\ln\Gamma\big(\frac{a}{2\pi}-it+1\big)-\ln\Gamma\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t-\frac{1}{2}}$$
$$=\frac{\ln\Gamma\big(\frac{a}{2\pi}-i(t+i)\big)}{\cosh 2\pi (t+i)-\frac{1}{2}}\,-\,\frac{\ln\Gamma\big(\frac{a}{2\pi}-it\big)}{\cosh 2\pi t-\frac{1}{2}}$$
(We used the fact that $\cosh (x+2\pi i)=\cosh x$). Adding two integrals (along $R\to R+i$ and $-R+i\to-R$ - these integrals $\to 0$ at $R\to\infty$), we can present $I_2(a)$ in the form of the integral along the same rectangular contour
$$I_2(a)=-\frac{\pi}{2}\oint\frac{\ln\Gamma\big(\frac{a}{2\pi}-iz\big)}{\cosh 2\pi z-\frac{1}{2}}dz=-\frac{\pi}{2}\,2\pi i\operatorname{Res}_{\binom{\frac{i}{3}}{\frac{2i}{3}}}\frac{\ln\Gamma\big(\frac{a}{2\pi}-iz\big)}{\cosh 2\pi z-\frac{1}{2}}=-\frac{\pi}{\sqrt 3}\Big(\ln\Gamma\big(\frac{a}{2\pi}+\frac{1}{3}\big)-\ln\Gamma\big(\frac{a}{2\pi}+\frac{2}{3}\big)\Big)$$
$$\boxed{\,\,I_2(a)=\frac{\pi}{\sqrt 3}\ln\frac{\Gamma\big(\frac{a}{2\pi}+\frac{2}{3}\big)}{\Gamma\big(\frac{a}{2\pi}+\frac{1}{3}\big)}\,\,}$$
Coming back to our initial integral
$$I=I_1+\Re\,I_2(0)=\frac{\pi}{3\sqrt 3}\ln2\pi+\frac{\pi}{\sqrt 3}\ln\frac{\Gamma\big(\frac{2}{3}\big)}{\Gamma\big(\frac{1}{3}\big)}$$
Using the reflection formula for gamma-function $\Gamma\big(\frac{2}{3}\big)\Gamma\big(\frac{1}{3}\big)=\frac{\pi}{\sin\frac{\pi}{3}}=\frac{2\pi}{\sqrt3}$
$$\boxed{\,\,I=\frac{\pi}{3\sqrt 3}\ln2\pi+\frac{\pi}{\sqrt 3}\ln\frac{2\pi}{\sqrt3}-\frac{2\pi}{\sqrt 3}\ln\Gamma\Big(\frac{1}{3}\Big)=-0.126321...\,\,}$$
A: This is not an answer.
I do not know if, $\forall a >0$, it could be possible to compute
$$I(a)=\int_0^\infty \frac{\log (x)}{e^x+e^{-x}+a}\,dx$$ But for small values of $a$ we could write
$$I(a)=\sum_{n=0}^\infty (-1)^n a^n \int_0^\infty \left(\frac{e^x}{e^{2 x}+1}\right)^{n+1} \log (x)\,dx$$ and we know the exact values of
$$J_n=\int_0^\infty \left(\frac{e^x}{e^{2 x}+1}\right)^{n+1} \log (x)\,dx$$ The first ones are
$$J_0=\frac{1}{4} \pi  \log \left(\frac{4 \pi ^3}{\Gamma
   \left(\frac{1}{4}\right)^4}\right)\qquad \qquad J_1=\frac{1}{4} \left(\gamma +\log \left(\frac{4}{\pi }\right)\right)$$
$$J_2=\frac{1}{32} \pi  \log \left(\frac{4 \pi ^3}{\Gamma
   \left(\frac{1}{4}\right)^4}\right)-\frac{C}{4 \pi }$$
$$J_3=\frac{1}{48} \left(\frac{7 \zeta (3)}{\pi ^2}+2 \gamma +\log (16)-2 \log (\pi
   )\right)$$
$$J_4=-\frac{320 \pi ^2 C+\psi ^{(3)}\left(\frac{1}{4}\right)-4 \pi ^4 \left(2+9
   \log (4)+27 \log (\pi )-36 \log \left(\Gamma
   \left(\frac{1}{4}\right)\right)\right)}{6144 \pi ^3}$$
$$J_5=\frac{35 \pi ^2 \zeta (3)+93 \zeta (5)+8 \pi ^4 \left(\gamma +\log
   \left(\frac{4}{\pi }\right)\right)}{960 \pi ^4}$$
The next do not fit on a line but they do not make any problem (at least up to $n=7$; for $n>7$ start appearing derivatives of the zeta function).
For $a=1$ and the summation up to $n=7$, the decimal value is $-0.125365$ while the exact value is $-0.126321$.
For $a=0.5$, the decimal value is $-0.179238$ while the exact value is $-0.179243$.
