# Prove that $\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\ln{3}$

We wish to prove that

$$I=\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\ln{3}\tag1$$

$$1-e^{3x}=(1-e^x)(1+e^x+e^{2x})\tag2$$

$$1+e^{3x}=(1+e^x)(1-e^x+e^{2x})\tag3$$

Sub $(2)$ and $(3)$ into $(1)\rightarrow (4)$

$$I=\int_{0}^{\infty}\left(2\cdot{1\over 1+e^x+e^{2x}}+{1\over 1-e^x+e^{2x}}\right)dx\tag5$$

Any hint, please, I am unable to continue.

$$I=\int_{0}^{1}\left({2u\over u^2+u+1}+{u\over u^2-u+1}\right)du=I_1+I_2\tag6$$

Respectively.

Apply formula (17) to $(6)$

Hence

$$I_1=\ln{3}-{2\sqrt3\over 3}\tan^{-1}\left({3\over \sqrt3}\right)$$

$$I_2={2\over \sqrt3}\tan^{-1}\left({\sqrt3 \over 3}\right)$$

Hence $I=\ln{3}$

Any hint, please, I am unable to continue.

Hint. By the change of variable $u=e^{-x}$, $dx=-\dfrac{du}u$, one gets $$\int_{0}^{\infty}\left(2\cdot{1\over 1+e^x+e^{2x}}+{1\over 1-e^x+e^{2x}}\right)dx=\int_{0}^1\left(2\cdot{u\over u^2+u+1}+{u\over u^2-u+1}\right)du$$ which is standard to evaluate.


Digamma Function $\ds{\Psi}$ with fractional arguments $\ds{0 < {p \over q} < 1}$ $\ds{\pars{~p = 1,2,3,\ldots\,,\quad p < q = 3,4,5,\ldots~}}$ are evaluated with the identity: $$\Psi\pars{p \over q} + \gamma = H_{p/q} = -\ln\pars{2q} - {\pi \over 2}\,\cot\pars{p\pi \over q} + 2\sum_{k = 1}^{\left\lfloor\pars{q + 1}/2\right\rfloor - 1} \cos\pars{2kp\pi \over q}\ln\pars{\sin\pars{k\pi \over q}}$$ which yields $$\left\lbrace\begin{array}{rcl} \ds{\Psi\pars{1 \over 3} + \gamma} & \ds{=} & \ds{-\,{\root{3} \over 6}\,\pi - {3 \over 2}\,\ln\pars{3}} \\[2mm] \ds{\Psi\pars{5 \over 6} + \gamma} & \ds{=} & \ds{{\root{3} \over 2}\,\pi - 2\ln\pars{2} - {3 \over 2}\,\ln\pars{3}} \end{array}\right.$$ and $\ds{\Psi\pars{\half} + \gamma = -2\ln\pars{2}}$.
\begin{align} \color{#f00}{I} & = \int_{0}^{\infty}\pars{2\,{1 - \expo{x} \over 1 - \expo{3x}} + {1 + \expo{x} \over 1 + \expo{3x}}}\,\dd x = \color{#f00}{\ln\pars{3}} \approx 1.0986 \end{align}

$$I=\int_{0}^{\infty}\left(2\cdot{1-e^x\over 1-e^{3x}}+{1+e^x\over 1+e^{3x}}\right)dx=\int_{0}^{\infty}\frac{1-e^x}{1-e^{3x}}dx+2\int_{0}^{\infty}\frac{1-e^{4x}}{1-e^{6x}}dx$$ $$I=\int_{0}^{\infty}\frac{1}{e^x(1+e^x+e^{2x})}e^xdx+\int_{0}^{\infty}\frac{1+e^{2x}}{e^{2x}(1+e^{2x}+e^{4x})}2e^{2x}dx$$ Set $u=e^x$ in the first integral and $u=e^{2x}$ in the second one, so we have $$I=\int_{1}^{\infty}\frac{u+2}{u(1+u+u^2)}du=\int_{1}^{\infty}\left(\frac{2}{u}-\frac{2u+1}{1+u+u^2}\right)du=\ln\left(\frac{u^2}{1+u+u^2}\right)\Big{|}_{1}^{\infty}=\color{red}{\ln 3}$$

• So Thanks egreg – Behrouz Maleki Jul 18 '16 at 13:19

By replacing $x$ with $-\log t$ we get:

$$I = \int_{0}^{1}\frac{3t-t^2+3t^3}{1+t^2+t^4}\,dt=\int_{0}^{1}\frac{3t-t^2+t^4-3t^5}{1-t^6}\,dt\tag{1}$$ and by expanding the last integrand function as a Taylor series we get: $$I = \sum_{n\geq 0}\left(\frac{3}{6n+2}-\frac{1}{6n+3}+\frac{1}{6n+5}-\frac{3}{6n+6}\right)\tag{2}$$ that can be easily computed through: $$\sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}\tag{3}$$ leading to: $$\sum_{n\geq 0}\frac{1}{(6n+2)(6n+6)}=\frac{\pi\sqrt{3}+9\log 3}{144},\quad \sum_{n\geq 0}\frac{1}{(6n+3)(6n+5)}=\frac{\pi\sqrt{3}-3\log 3}{24}. \tag{4}$$