5
$\begingroup$

Find: $$ \int^1_0 \frac{\ln(1+x)}{x}dx$$

There is suppose to be a clean solution (maybe some symmetry involved?)

I have tried integration by parts as followed:

$\ln(x+1)=u$ ,$\frac{1}{x+1} dx = du$ and also $\frac{1}{x}dx=dv$, $\ln(x)=v$

which means our integral becomes

$$\int^1_0 \frac{\ln(1+x)}{x}dx=\ln(x+1)\ln(x)|^1_0 - \int^1_0 \frac{\ln(x)}{x+1}$$

which does not make this easier.

I have also tried using the identity $\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$

So let $I = \int^1_0 \frac{\ln(1+x)}{x}dx$

and also $I = \int^1_0 \frac{\ln(2-x)}{1-x} $

so $2I= \int^1_0 \ln(1+x)+\ln(2-x)$ which also doesn't make it easier?

Any ideas! :)

$\endgroup$
  • $\begingroup$ I don' t see how you get $\ln (-x)$ in the second attempt. $\endgroup$ – SchrodingersCat Dec 6 '16 at 9:39
  • $\begingroup$ $ln(1+0-(1+x))=ln(-x)$ $\endgroup$ – bigfocalchord Dec 6 '16 at 9:40
  • 1
    $\begingroup$ It should be $\ln [1+(0+1-x)]$. Check carefully. $\endgroup$ – SchrodingersCat Dec 6 '16 at 9:42
  • $\begingroup$ @dydxx You have to replace $x$ by $1+0 - x$ (as you did correctly in the denominator. So the numerator becomes $$ \log(1 + (1-x)) = \log (2-x) \ne \log (-x) $$ $\endgroup$ – martini Dec 6 '16 at 9:43
  • 3
    $\begingroup$ math.stackexchange.com/questions/288830 $\endgroup$ – Aforest Dec 6 '16 at 9:48
4
$\begingroup$

By integration by parts: $$ \int_{0}^{1}\frac{\log(1+x)\,dx}{x} = -\int_{0}^{1}\frac{\log x}{1+x}\,dx=2\int_{0}^{1}\frac{-\log x}{1-x^2}\,dx-\int_{0}^{1}\frac{-\log x}{1-x} $$ but since $\int_{0}^{1}(-\log x)x^k\,dx = \frac{1}{(k+1)^2}$, by expanding $\frac{1}{1-x^2}$ and $\frac{1}{1-x}$ as geometric series we get: $$ \int_{0}^{1}\frac{\log(1+x)\,dx}{x} = 2\sum_{k\text{ odd}}\frac{1}{k^2}-\sum_{k\geq 1}\frac{1}{k^2}=\sum_{k\geq 1}\frac{1}{k^2}-2\sum_{k\text{ even}}\frac{1}{k^2}=\frac{1}{2}\sum_{k\geq 1}\frac{1}{k^2}=\color{blue}{\frac{\pi^2}{12}}. $$

$\endgroup$
3
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{\ln\pars{1 + x} \over x}\,\dd x & \,\,\,\stackrel{x\ \mapsto\ -x}{=}\,\,\, \int_{0}^{-1}{\ln\pars{1 - x} \over x}\,\dd x = -\int_{0}^{-1}\mrm{Li}_{2}'\pars{x}\,\dd x = -\mrm{Li}_{2}\pars{-1} + \mrm{Li}_{2}\pars{0} \\[5mm] & = \bbx{\ds{\pi^{2} \over 12}} \end{align}

$\endgroup$
2
$\begingroup$

If you know the gamma function $\Gamma$, the Dirichlet function $\eta$ and the formula $$F(n):=\int_0^{\infty}\frac{u^{n-1}}{e^u+1}\; \mathrm du=\Gamma(n)\cdot\eta(n) \qquad (n\in \mathbf N)$$ you can integrate $$I:= \int_0^{1}\frac{\ln(1+x)}{x} \; \mathrm d x$$ by parts to get $$I=-\int^1_0 \frac{\log(x)}{x+1} \; \mathrm d x.$$ If you substitue $x=\mathrm e^u$ ($\mathrm d x = \mathrm e^u \mathrm du$) you will get $$ \int_0^{\infty}\frac{u}{e^u+1}\mathrm du.$$ Therefore you get $$F(2)=I\stackrel{.}{=}0.82246703342$$ (which is equal to $\frac{\pi^2}{12}$).

$\endgroup$
2
$\begingroup$

HINT:

use the expansion $$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$$ $$\frac{ln(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots$$ $$\int \frac{ln(1+x)}{x}\ dx=x-\frac{x^2}{2^2}+\frac{x^3}{3^2}-\frac{x^4}{4^2}+\ldots$$

$\endgroup$
  • $\begingroup$ why down voted? $\endgroup$ – jeanne clement Dec 6 '16 at 11:11
2
$\begingroup$

Hint. One may recall that $$ \ln (1+x)=\sum_{n=1}^\infty(-1)^{n-1}\frac{x^n}{n},\quad |x|<1, $$ one may then divide by $x$ and one is allowed to integrate termwise obtaining $$ \int_0^1\frac{\ln (1+x)}x\:dx=\sum_{n=1}^\infty(-1)^{n-1}\int_0^1\frac{x^{n-1}}{n}=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}. $$ Can you take it from here?

$\endgroup$
  • $\begingroup$ Do note that while this approach is correct, you need some further argument to interchange the sum with the integral (e.g. dominated convergence). $\endgroup$ – Dominik Dec 6 '16 at 9:55
  • 2
    $\begingroup$ @Dominik Sure, that's why I call it a "hint". $\endgroup$ – Olivier Oloa Dec 6 '16 at 9:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.