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Find $\int_0^1 \mathrm{\frac{x-1}{ln(x)}}\,\mathrm{d}x$

I tryed this:

$\int_0^1 \mathrm{\frac{x-1}{ln(x)}} = \int_0^1 \mathrm{\frac{x}{ln(x)}} - \int_0^1 \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x$

To $\int_0^1 \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x$

Let $t=lnx $ then $\frac{dt}{dx}=\frac{1}{x}$ and $dx=e^tdt$

$\int \mathrm{\frac{1}{ln(x)}}\,\mathrm{d}x \equiv \int \mathrm{\frac{e^t}{t}}$$\mathrm{d}t$ and well

$e^t=\sum _{n=0}^\infty \frac{t^n}{n!}$

$\frac{e^t}{t}=\sum_{n=0}^\infty \frac{t^{n-1}}{n!}$

$\int \mathrm{\frac{e^t}{t}}\,\mathrm{d}t=\int \mathrm \sum_{n=0}^\infty{\frac{t^{n-1}}{n!}}\,\mathrm{d}t=\sum_{n=0}^\infty\frac{t^n}{n*(n)!}$

How can I solve $\int_0^1 \mathrm{\frac{x}{ln(x)}}$ ?

Thanks for your help :) have a nice day

Mikasa
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user63192
  • 1,081

1 Answers1

3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{\int_{0}^{1}{x - 1 \over \ln\pars{x}}\,\dd x: {\large ?}.\quad}$ Whith the change of variables $x \equiv \expo{-z}:$ \begin{align} \color{#0000ff}{\large\int_{0}^{1}{x - 1 \over \ln\pars{x}}\,\dd x} &= \int_{\infty}^{0}{\expo{-z} - 1 \over -z}\,\pars{-\expo{-z}\,\dd z} = \int^{\infty}_{0}{\expo{-z} - \expo{-2z}\over z}\,\dd z \\[3mm]&= -\int^{\infty}_{0}\ln\pars{z}\pars{-\expo{-z} +2 \expo{-2z}}\,\dd z = \int^{\infty}_{0}\ln\pars{z}\expo{-z}\,\dd z - \int^{\infty}_{0}\ln\pars{z \over 2}\expo{-z}\,\dd z \\[3mm]&= \int^{\infty}_{0}\ln\pars{z}\expo{-z}\,\dd z - \int^{\infty}_{0}\bracks{\ln\pars{z} - \ln\pars{2}}\expo{-z}\,\dd z = \ln\pars{2}\overbrace{\int^{\infty}_{0}\expo{-z}\,\dd z}^{=\ 1} \\[3mm]&= \color{#0000ff}{\large\ln\pars{2}} \end{align}

Felix Marin
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