Finding $\int_{1}^{\infty} \frac{x 3^x}{(3^x-1)^2}$

Question

How do we prove that $$\int_{1}^{\infty} \frac{x \cdot 3^x}{(3^x-1)^2} \mathrm{d}x=\dfrac{3\log{3}-2\log{2}}{2\log^2{3}}$$

I have tried some substitutions such as $3^x=t$, but it didn't work out. The denominator is already factored. I do not know what to do next. Please help me out. Thank you.

Answer 1

Integrating by parts ($u=x$, $dv=\frac{3^x}{(3^x-1)^2}dx$), $$\int x\frac{3^x}{(3^x-1)^2}dx=x\int\frac{3^x}{(3^x-1)^2}dx-\int\left[\frac{dx}{dx}\int\frac{3^x}{(3^x-1)^2}dx\right]dx$$

Now for $\displaystyle\int\dfrac{3^x}{(3^x-1)^2}dx,$ set $3^x-1=u$

Finally $\displaystyle\int\dfrac{dx}{3^x-1},$ set $3^x-1=v$

Answer 2

Another answers are nice but if you want to stick to Integration by parts, Here's how

$$I_1=\int_1^\infty\frac{x 3^x}{(3^x-1)^2}$$

$$I=\int\frac{x 3^x}{(3^x-1)^2}$$ substitute $$3^x-1=t\iff3^x\ln 3=dt$$

$$I=\int\frac{x \cdot 3^x}{(3^x-1)^2} \mathrm{d}x= \frac{1}{\ln^23}\int \frac{\ln|t+1|}{t^2} \mathrm{d}t$$

$$I(\ln^23)= \int \frac{\ln|t+1|}{t^2} \mathrm{d}t=-\frac{\ln|t+1|}{t}+\int \frac{1}{t(t+1)} \mathrm{d}t$$

$$I(\ln^23)=-\frac{\ln|t+1|}{t}-\ln|t|+\ln|t+1|$$

$$I=\frac{(3^x-1)\ln|1-3^x|-3^xx\ln(3)}{(\ln^23)(3^x-1)}$$

$$\rightarrow I_1=\frac{\ln27-\ln4}{2\ln^23}$$

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{1}^{\infty}{x\ 3^{x} \over \pars{3^{x} - 1}^{2}}\,\dd x ={3\ln\pars{3} - 2\ln\pars{2} \over 2\ln^{2}\pars{3}}:\ {\large ?}}$.

\begin{align}&\color{#c00000}{\int_{1}^{\infty}{\dd x \over \expo{\mu x} - 1}} =\int_{1}^{\infty}{\expo{-\mu x}\,\dd x \over 1 - \expo{-\mu x}} =\left.{\ln\pars{1 - \expo{-\mu x}} \over \mu} \right\vert_{x\ =\ 1}^{x\ \to\ \infty} =-\,{\ln\pars{1 - \expo{-\mu}} \over \mu} \end{align}

Derivate both members respect of $\ds{\mu}$:

\begin{align}&\color{#c00000}{\int_{1}^{\infty}% \bracks{-\,{x\expo{\mu x} \over \pars{\expo{\mu x} - 1}^{2}}}\,\dd x} =-\,{1 \over \mu\pars{\expo{\mu} - 1}} + {\ln\pars{1 - \expo{-\mu}} \over \mu^{2}} \end{align}

Multiply both members by $\ds{-1}$ and set $\ds{\mu = \ln\pars{3}}$:

\begin{align}&\color{#66f}{\large\int_{1}^{\infty}% {x\ 3^{x} \over \pars{3^{x} - 1}^{2}}\,\dd x} ={1 \over \ln\pars{3}\pars{3 - 1}} - {\ln\pars{1 - 1/3} \over \ln^{2}\pars{3}} =\color{#66f}{\large{3\ln\pars{3} - 2\ln\pars{2} \over 2\ln^{2}\pars{3}}} \\[5mm]&\approx {\tt 0.7911} \end{align}