2
$\begingroup$

The parameter $\lambda$ is complex and it's not on the real axis.

There are some similar cases:

Help me evaluate $\int_0^1 \frac{\log(x+1)}{1+x^2} dx$

Evaluate $\int_0^1 \frac{\ln(1+bx)}{1+x} dx $

Evaluation of the integral $\int_0^1 \frac{\ln(1 - x)}{1 + x}dx$

Supplement: How to figure out the two integal $I_1$ and $I_2$

This integral should be separated into two parts for the convergence condition of the complex parameter $\lambda$, let us consider the two integrals with a complex parameter : $\mathrm{Im}\{\lambda\} \ne 0$ , and its real part is limited by integral:

$$ I_1 = \int_0^1 \frac{\ln(1 - u)}{\lambda + u}\, d u $$

Mathematica gave the result $I_1 = \mathrm{Li}_2 \big(\frac{\lambda}{1 + \lambda}\big)- \ln \left( \frac{\lambda}{1+\lambda} \right)\, \ln(1 + \lambda) - \frac{\pi^2}{6}$

$$ I_2 = \int_0^1 \frac{\ln(1 + u)}{\lambda + u}\, du = - \mathrm{Li}_2\big(\frac{1}{1 - \lambda}\big) + \mathrm{Li}_2 \big(\frac{2}{1 - \lambda}\big) + \ln 2 \, \ln \left(\frac{\lambda +1 }{\lambda - 1} \right) $$

$\endgroup$
6
  • 1
    $\begingroup$ What about $|Re(\lambda)|$? $\endgroup$
    – Yuriy S
    Jun 20, 2016 at 9:48
  • 1
    $\begingroup$ Mathematica gave me an answer involving several polylogarithm functions $Li_2(f(\lambda))$ which implies you can't do it with elementary functions. $\endgroup$
    – Ian Miller
    Jun 20, 2016 at 9:56
  • $\begingroup$ @YuriyS : $|\mathrm{Re}(\lambda)| <1$ $\endgroup$
    – Zoe Rowa
    Jun 20, 2016 at 10:06
  • 1
    $\begingroup$ For real numbers, if $\lambda \to 1$, I suppose that the limit is $\frac{\pi ^2}{12}$ $\endgroup$
    – Claude Leibovici
    Jun 20, 2016 at 10:11
  • $\begingroup$ I'd say there is a problem of convergence for all $\lambda\in(-1,0)$. $\endgroup$
    – mickep
    Jun 20, 2016 at 11:10

3 Answers 3

Reset to default
2
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{align} &\color{#f00}{% \int_{0}^{1}{1 \over u + \lambda}\, \ln\pars{1 + u \over 1 - u}\,\dd u} = -\int_{1}^{2}{\ln\pars{u} \over \pars{1 - \lambda} - u}\,\dd u - \int_{0}^{1}{\ln\pars{u} \over \pars{1 + \lambda} - u}\,\dd u \end{align}

Then, integrals are of the form \begin{align} \int_{a}^{b}{\ln\pars{u} \over c - u}\,\dd u & = \int_{a/c}^{b/c}{\ln\pars{cu} \over 1 - u}\,\dd u = \left.\vphantom{\Large A}-\ln\pars{1 - u}\ln\pars{cu}\right\vert_{\ a/c}^{\ b/c} + \int_{a/c}^{b/c}\ \overbrace{{\ln\pars{1 - u} \over u}}^{\ds{-\,\mathrm{Li}_{2}'\pars{u}}}\ \,\dd u \\[3mm] & = -\ln\pars{c - b \over c}\ln\pars{b} + \ln\pars{c - a \over c}\ln\pars{a} - \mathrm{Li}_{2}\pars{b \over c} + \mathrm{Li}_{2}\pars{a \over c} \end{align}
  • $\ds{\large a = 1\,,\ b = 2\,,\ c = 1 - \lambda}$: \begin{align} \int_{1}^{2}{\ln\pars{u} \over \pars{1 - \lambda} - u}\,\dd u & = -\ln\pars{\lambda + 1 \over \lambda - 1}\ln\pars{2} - \mathrm{Li}_{2}\pars{2 \over 1 - \lambda} + \mathrm{Li}_{2}\pars{1 \over 1 - \lambda} \end{align}
  • $\ds{\large a = 0\,,\ b = 1\,,\ c = 1 + \lambda}$: \begin{align} \int_{0}^{1}{\ln\pars{u} \over \pars{1 + \lambda} - u}\,\dd u & = -\mathrm{Li}_{2}\pars{1 \over 1 + \lambda} \end{align}

    • \begin{align} &\color{#f00}{% \int_{0}^{1}{1 \over u + \lambda}\, \ln\pars{1 + u \over 1 - u}\,\dd u} \\[3mm] = &\ \color{#f00}{\ln\pars{\lambda + 1 \over \lambda - 1}\ln\pars{2} + \mathrm{Li}_{2}\pars{2 \over 1 - \lambda} - \mathrm{Li}_{2}\pars{1 \over 1 - \lambda} + \mathrm{Li}_{2}\pars{1 \over 1 + \lambda}} \end{align}

    $\endgroup$
    1
    $\begingroup$

    \begin{align} \int_0^1 \frac{1}{u + \lambda} \ln \frac{1 + u}{1 - u} du \overset{u=\frac{1-t}{1+t} }=& -\int_0^1 \frac{\ln t}{[(1+ \lambda)-(1-\lambda)t](1+t)}dt\\ =& -\int_0^1 \bigg(\frac{\frac{1-\lambda}{1+\lambda}\ln t}{1-\frac{1-\lambda}{1+ \lambda}t} +\frac{\ln t}{1+t}\bigg)\ dt\\ =&\ \text{Li}_2(\frac{1-\lambda}{1+\lambda} ) +\frac{\pi^2}{12}\\ \end{align}

    $\endgroup$
    0
    $\begingroup$

    Take the second integral as a demonstration.

    \begin{aligned} I_2 &= \int_0^1 \frac{\ln (1 + u)}{\lambda + u} \\ & \Downarrow \quad x:= 1 + u \\ &= \int_1^2 \frac{\ln x}{\underbrace{\lambda - 1}_{z} + x} d x= \int_1^2 \frac{\ln x}{z +x} dx \\ &= \ln x \, \ln\big(1 + \frac{x}{z} \big) \bigg|_{x=1}^{x=2} - \int_1^2 \frac{d x}{x} \ln \big(1 + \frac{x}{z} \big) \\ &= \ln 2\, \ln \left(\frac{\lambda + 1}{\lambda -1} \right) - \int_0^1 dt \,\int_1^2 \frac{dx}{z + tx} \end{aligned}

    Focus the last integral \begin{aligned} - \int_0^1 dt \,\int_1^2 \frac{dx}{z + tx} &= - \int_0^1 \frac{dt}{t} \,\int_1^2 \frac{t \, dx}{z + tx} \\ &= -\int_0^1 \frac{dt}{t} \, \ln \big(z + tx \big) \bigg|_{x=1}^{x=2} \\ &= - \int_0^1 \frac{dt}{t} \, \ln\big[ 1 - \big(-\frac{2}{z}\big) \big] + \int_0^1 \frac{dt}{t} \, \ln \big[ 1 - \big( - \frac{1}{z}\big) \big] \\ &= \mathrm{Li}_2 \big(\frac{2}{1 - \lambda }\big) - \mathrm{Li}_2\big(\frac{1}{1 - \lambda} \big) \end{aligned}

    Hence: $\displaystyle I_2 = \mathrm{Li}_2 \big(\frac{2}{1 - \lambda }\big) - \mathrm{Li}_2\big(\frac{1}{1 - \lambda} \big) + \ln2\, \ln\left(\frac{\lambda+1}{\lambda-1}\right)$

    Similarly, $I_1$ can be figured out.

    $\endgroup$

    You must log in to answer this question.

    Not the answer you're looking for? Browse other questions tagged .