Easiest way to calculate $\int_{0}^{1} \frac{ \log (1+x)}{x} dx$

What is the easiest way to calculate

$$\int_{0}^{1} \frac{ \log (1+x)}{x}\, dx$$

?

Need a hint.

• I would simply suggest integration by parts. Careful how you assign $u, du$ and $v, dv$. – Autolatry Jan 6 '15 at 15:45
• @Autolatry Okay,I'll try that. – I am not a robot Jan 6 '15 at 15:48
• What should I take as u and v? @Autolatry – I am not a robot Jan 6 '15 at 15:50
• My thought would be to (possibly) try to avoid having to integrate $\log(1+x)$ but differentiating it would be ok. So choose $u=\log(1+x)$. – Autolatry Jan 6 '15 at 15:53
• @Nivedita: Jack d'Aurizio has given an answer. It "only" works for the limits $0$ and $1$, in the sense that we will not find an elementary expression for $\int_0^w$. – André Nicolas Jan 6 '15 at 16:13

$$I=\int_{0}^{1}\frac{\log(1+x)}{x}\,dx = \int_{0}^{1}\sum_{n\geq 1}\frac{(-1)^{n+1}x^{n-1}}{n}\,dx=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}=\frac{1}{2}\zeta(2)=\color{red}{\frac{\pi^2}{12}}.$$
• in addition to this fine answer and with regards to the beta..Let $$I_a = \int_0^1(1+x)^ax^{-1}dx$$ your integral then is simply $$\lim_{a\rightarrow 0}\frac{\partial}{\partial a}\int_0^1(1+x)^ax^{-1}dx = \int_0^1\frac{\ln(1+x)}{x}dx$$ now $$\int_0^1(1+x)^ax^{-1}dx = \int_0^1 u^a(1-u)^{-1}du = B(a+1,0)$$ now we all know that the beta function is only defined for arguments $>0$ – Chinny84 Jan 6 '15 at 16:11
We have an indefinite integral $$\int\frac{\ln(1+x)}{x } dx=-\operatorname{Li}_2(-x).$$ Therefore $$\int_ 0^1 \frac{\ln(1+x)}{x } dx=-\operatorname{Li}_2(-1) = -\frac 1 2 \zeta(2)=- \frac{\pi^2}{12}.$$
Of course this is overkill for this integral, but this is the method of choice if the upper limit is $1/2$ or $\phi$.