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What is the easiest way to calculate

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

?

Need a hint.

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    $\begingroup$ I would simply suggest integration by parts. Careful how you assign $u, du$ and $v, dv$. $\endgroup$ – Autolatry Jan 6 '15 at 15:45
  • $\begingroup$ @Autolatry Okay,I'll try that. $\endgroup$ – Nivedita Jan 6 '15 at 15:48
  • $\begingroup$ What should I take as u and v? @Autolatry $\endgroup$ – Nivedita Jan 6 '15 at 15:50
  • $\begingroup$ My thought would be to (possibly) try to avoid having to integrate $\log(1+x)$ but differentiating it would be ok. So choose $u=\log(1+x)$. $\endgroup$ – Autolatry Jan 6 '15 at 15:53
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    $\begingroup$ @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 $\int_0^w$. $\endgroup$ – André Nicolas Jan 6 '15 at 16:13
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$$I=\int_{0}^{1}\frac{\log(1+x)}{x}\,dx = \int_{0}^{1}\sum_{n\geq 1}\frac{(-1)^{n+1}x^{n-1}}{n}\,dx=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}=\frac{1}{2}\zeta(2)=\color{red}{\frac{\pi^2}{12}}. $$

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  • $\begingroup$ Anything other than this one? I have done it the same way. $\endgroup$ – Nivedita Jan 6 '15 at 15:46
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    $\begingroup$ @Nivedita: It looks pretty straightforward to me, honestly I don't know if there are easier methods. Differentiation under the integral sign is a chance, but the computation involving the Beta function is lengthy. $\endgroup$ – Jack D'Aurizio Jan 6 '15 at 15:48
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    $\begingroup$ 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 $>0$ $\endgroup$ – Chinny84 Jan 6 '15 at 16:11
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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 $\phi$.

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