Evaluate $\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} $

Question

$$\int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2} dx$$

My approach is to calc $$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx$$ and then take the limit for the answer when $X \rightarrow \infty$

However, I must do something wrong. The correct answer should be $2\ln(2)$.

$$\int_{1}^{X} \frac{\ln{(2x-1)}}{x^2} dx = \left[-\frac{1}{x} \ln (2x-1) \right]_{1}^{X} + \int_{1}^{X} \frac{1}{x} \times \frac{2}{2x-1} dx = -\frac{1}{X}\ln(2X-1) + 2\int_{1}^{X} -\frac{1}{x} + \frac{2}{2x-1} dx = -1\frac{1}{X}\ln(2X-1)-2\ln X+2\ln(2X-1) $$

Am I wrong? If I'm not, how to proceed?

=== EDIT ===

After the edit I wonder if this is the correct way to proceed:

$$ - \frac{1}{X}\ln(2X-1)-2\ln X+2\ln(2X-1) $$ The first part will do to zero because of $\frac{1}{X} $ so we ignore that one, the second and third part:

$$ -2\ln X+2\ln(2X-1) = 2\ln \left( \frac{2X-1}{X}\right) = 2\ln \left( 2-\frac{1}{X} \right) = 2\ln (2)$$

Answer 1

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

\begin{align} &\color{#66f}{\large\int_{1}^{\infty}{\ln\pars{2x - 1} \over x^{2}}\,\dd x} =-\int_{x\ =\ 1}^{x\ \to\infty}\ln\pars{2x - 1}\,\dd\pars{1 \over x} =\int_{1}^{\infty}{1 \over x}\,{2 \over 2x - 1}\,\dd x \\[3mm]& =\int_{1}^{\infty}{1 \over x\pars{x - 1/2}}\,\dd x =2\int_{1}^{\infty}\pars{{1 \over x - 1/2} - {1 \over x}}\,\dd x \\[3mm]&=\left. 2\ln\pars{\verts{x - 1/2 \over x}}\right\vert_{1}^{\infty} =\color{#66f}{\large 2\ln\pars{2}} \approx {\tt 1.3863} \end{align}

Answer 2

Your derivative is incorrect, it should be $$\frac{2}{2x-1}$$ instead of $$\frac{1}{2x-1}$$ Everything else seems correct.

Answer 3

Another approach :

Setting $x\mapsto\frac1x$, we will obtain $$ \int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2}\ dx=\int_{0}^{1} \ln\left(\frac{2-x}{x}\right) dx=\int_{0}^{1} \bigg[\ln(2-x)-\ln x\bigg]\ dx. $$ Note that $$ \int\ln y\ dy=y\ \ln y-y+C, $$ hence $$ \int_{1}^{\infty} \frac{\ln{(2x-1)}}{x^2}\ dx=2\ln 2. $$