Solving $\int_0^{\infty}\frac{\ln(2e^x-1)}{e^x-1}dx$

Question

$$I=\int_0^{\infty}\frac{\ln(2e^x-1)}{e^x-1}dx$$ I have tried some substitution to clean up this integral but end up getting stuck. Here is my work:

Let $u=e^x$ $$I=\int_1^{\infty}\frac{\ln(2u-1)}{u(u-1)}du$$ Now let $t=\frac{1}{u}$ $$I=\int_0^1\frac{\ln(\frac{2-t}{t})}{1-t}dt$$ $$I=\int_0^1\frac{\ln(2-t)}{1-t}dt-\int_0^1\frac{\ln t}{1-t}dt$$ Let $k=1-t$ $$I=-\int_1^0\frac{\ln(k+1)}{k}dk+\int_1^0\frac{\ln(1-k)}{k}dk$$ This is where I get stuck. I have made it look much better than before but I have no clue how to continue from here. Any help or guidance would be appreciated

Answer 1

Assuming what you did so far is correct. Consider the Taylor expansion for $\ln(1-x)$, which is $x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...$. Then you divide by x gives you $\frac{\ln(1-x)}{x} = 1+\frac{x}{2}+\frac{x^2}{3}+...$ Then you take the integral of this which becomes $\int\frac{\ln(1-x)}{x}dx = x + \frac{x^2}{4} + \frac{x^3}{9}+ \frac{x^4}{16}+...$

If you evaluate this at $x=1$ gives you the famous Basel problem, namely the sum of reciprocals of perfect squares is $\frac{\pi^2}{6}$. For the other half you repeat exactly the same procedure. It's an alternating sum of reciprocals of squares. If you factor out a 1/4 the even part is 1/4 the original part, so the difference is $\frac{\pi^2}{12}$

Answer 2

OP’s approach is correct. Let’s start with the last step.

Let $k=1-t$, then $$ I = \underbrace{ \int_0^1 \frac{\ln (1+k)}{k} d k}_{J} - \underbrace{\int_0^1 \frac{\ln (1-k)}{k} d k}_{K} $$ For the integral $J$, we integrate the series for $|k|<1,$

$$ \frac{1}{1+k}=\sum_{n=0}^{\infty}(-1)^n k^n \Rightarrow \quad \ln (1+k)= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} k^n}{n} $$

$$ \begin{aligned} J & =\int_0^1 \sum_{n=1}^{\infty} \frac{(-1)^{n-1} k^n}{n k} d k \\ & =\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_0^1 k^{n-1} d k \\ & =\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}\\&=\sum_{n=1}^{\infty} \frac{1}{n^2}-2 \sum_{n=1}^{\infty} \frac{1}{(2 n)^2} \\ & =\left(1-\frac{1}{2}\right) \zeta(2) \\ & =\frac{\pi^2}{12} \end{aligned} $$ For the integral $K$, we integrate the series for $|k|<1,$

$$ \frac{1}{1-k}=\sum_{n=0}^{\infty} k^n \Rightarrow \ln (1-k)=-\sum_{n=1}^{\infty} \frac{k^n}{n} $$

$$ \begin{aligned} K= & -\int_0^1 \sum_{n=1}^{\infty} \frac{k^n}{n k} d k \\ & =-\sum_{n=1}^{\infty} \frac{1}{n} \int_0^1 k^{n-1} d k \\ & =-\sum_{n=1}^{\infty} \frac{1}{n^2} \\ & =-\frac{\pi^2}{6} \end{aligned} $$

$$\boxed{\int_0^{\infty} \frac{\ln \left(2 e^x-1\right)}{e^x-1} d x =\frac{\pi^2}{12}-\left(-\frac{\pi^2}{6}\right)=\frac{\pi^2}{4}}$$

Wish it helps!

Answer 3

\begin{array}{l} \displaystyle I = \int\limits_0^1 {\frac{{\ln \left( {\frac{{2 - t}}{t}} \right)}}{{1 - t}}dt} = \int\limits_0^1 {\frac{{\ln \left( {\frac{{2 - \left( {1 - t} \right)}}{{1 - t}}} \right)}}{t}dt} = \int\limits_0^1 {\frac{{\ln \left( {\frac{{1 + t}}{{1 - t}}} \right)}}{t}dt} = - \int\limits_0^1 {\frac{{\ln \left( {\frac{{1 - t}}{{1 + t}}} \right)}}{t}dt} \\ \displaystyle\frac{{1 - t}}{{1 + t}} \to t \Rightarrow I = - 2\int\limits_0^1 {\frac{{\ln \left( t \right)}}{{1 - {t^2}}}dt} = - 2\sum\limits_{n = 0}^\infty {\int\limits_0^1 {{t^{2n}}\ln \left( t \right)dt} } = 2\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {2n + 1} \right)}^2}}}} = 2.\frac{{{\pi ^2}}}{8} = \frac{{{\pi ^2}}}{4} \end{array}

Answer 4

Let $2e^x-1=e^t$. Then,

$$I=\int_0^{\infty}\frac{\ln(2e^x-1)}{e^x-1}dx=\int_0^\infty\frac{2te^t}{e^{2t}-1}dt\\\stackrel{2t=u}{=}\frac12\int_0^{\infty}\frac{ue^{\frac12 u}}{e^u-1}du\\ =\frac12\int_0^{\infty}\frac{ue^{-\frac12 u}}{1-e^{-u}}du \\=\frac12\zeta(2,\frac12)=\frac12(\tfrac{\pi^2}2)=\frac{\pi^2}4.$$ where $\zeta(s,a)=\sum_{n=1}^\infty\frac1{(n+a)^s}$, $\Re s>1$ is the Hurwitz zeta function.

Answer 5

$$I=\int_0^{\infty}\frac{\ln(2e^x-1)}{e^x-1}dx$$

Substitute $e^{-x}=t\implies (0,\infty)\implies(1,0)$

$$I=\int_0^1\frac{ln(2-t)}{(1-t)}\,dt-\int_0^1\frac{ln(t)}{(1-t)}\,dt$$

$$I=I_1-I_2$$

$$I_1=\int_0^1\frac{ln(2-t)}{(1-t)}\,dt=\frac{\zeta(2)}{2}$$

$$I_2=\int_0^1\frac{ln(t)}{(1-t)}\,dt=-\zeta(2)$$

$$\boxed{I=\frac{3\zeta(2)}{2}=\frac{\pi^2}{4}}$$

Answer 6

Continue with \begin{align} I=& -\int_1^0\frac{\ln(k+1)}{k}dk+\int_1^0\frac{\ln(1-k)}{k}dk\\ =& \int_0^1 \frac{\ln \frac{1+k}{1-k}}kdk \overset{\frac{1-k}{1+k}\to k}=\int_0^1 \frac{2\ln k}{k^2-1}dk =\int_0^1 \int_0^\infty \frac{2t}{(1+t^2)(1+k^2t^2)}dt \ dk\\ =&\int_0^\infty \frac{2\arctan t}{1+t^2}dt = \int_0^\infty d(\arctan^2t)=\frac{\pi^2}4 \end{align}