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! :)

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ math.stackexchange.com/questions/288830 $\endgroup$ – Aforest Dec 6 '16 at 9:48

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: $$ \int_{0}^{1}\frac{\log(1+x)\,dx}{x} = 2\sum_{k\text{ odd}}\frac{1}{k^2}-\sum_{k\geq 1}\frac{1}{k^2}=\sum_{k\geq 1}\frac{1}{k^2}-2\sum_{k\text{ even}}\frac{1}{k^2}=\frac{1}{2}\sum_{k\geq 1}\frac{1}{k^2}=\color{blue}{\frac{\pi^2}{12}}. $$


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


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 $\frac{\pi^2}{12}$).



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

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

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?

  • $\begingroup$ 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
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    $\begingroup$ @Dominik Sure, that's why I call it a "hint". $\endgroup$ – Olivier Oloa Dec 6 '16 at 9:56

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