Integral with polylogarithm

Question

I have to calculate integral: $$\int \frac{1}{x}\ln \left( \frac{x-a}{b}\right) \,dx.$$ I tried by substitution, but I did not receive such a result as wolfram.

Answer 1

According to Wolfram Alpha, the integral is:

$$\int \frac{\ln( \frac{x-a}{b}) }{x}dx = \ln\left(\frac{x}{a}\right)\ln\left(\frac{x-a}{b}\right)+Li_2\left(1-\frac{x}{a}\right)+c$$ where $Li_2(x)$ is the offset logarithmic integral: $$Li_2(x) = \int_2^x \frac{1}{\ln(t)}dt$$ You can find more information about the logarithmic integral if you click on the above link, there are some ways to calculate it for certain values, but unfortunately it cannot be expressed with known functions in most cases (without using $\sum$'s).

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{t \equiv {x - a \over b} \implies x = bt + a}$:

\begin{align} \int{1 \over x}\ln\pars{x - a \over b}\,\dd x & = \int{1 \over bt + a}\ln\pars{t}\,b\,\dd t = -\int{\ln\pars{\bracks{-a/b}\bracks{-bt/a}} \over 1 - \pars{-bt/a}} \,\pars{-\,{b \over a}}\dd t \end{align}

Lets $\ds{y \equiv -\,{b \over a}\,t \implies t = -\,{a \over b}\,y}$:

\begin{align} \int{1 \over x}\ln\pars{x - a \over b}\,\dd x & = -\int{\ln\pars{-ay/b} \over 1 - y}\,\dd y \,\,\,\stackrel{\mrm{IBP}}{=}\,\,\, \ln\pars{1 - y}\ln\pars{-\,{a \over b}\,y} - \int{\ln\pars{1 - y} \over y}\,\dd y \\[5mm] & = \ln\pars{1 - y}\ln\pars{-\,{a \over b}\,y} + \mrm{Li}_{2}\pars{y} = \ln\pars{1 + {b \over a}\,t}\ln\pars{t} + \mrm{Li}_{2}\pars{-\,{b \over a}\,t} \\[5mm] & = \ln\pars{1 + {b \over a}\,{x - a \over b}}\ln\pars{x - a \over b} + \mrm{Li}_{2}\pars{-\,{b \over a}\,{x - a \over b}} \\[5mm] & = \bbx{\ln\pars{x \over a}\ln\pars{x - a \over b} + \mrm{Li}_{2}\pars{1 - {x \over a}} + \pars{~\mbox{a constant}~}} \end{align}