application of logarithm series

The integrand of the improper integral

 $\displaystyle I\;:=\;\int_{0}^{1}\frac{\ln(1\!+\!x)}{x}dx$ (1)
 $\ln(1\!+\!x)\;=\;x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+-\ldots% \qquad(-1
 $\frac{\ln(1\!+\!x)}{x}\;=\;1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+-% \ldots\qquad(-1

whence

 $\displaystyle\lim_{x\to 0}\frac{\ln(1\!+\!x)}{x}\;=\;1.$ (2)

This implies that the integrand of (1) is bounded  on the interval$[0,\,1]$ and also continuous  , if we think that (2) defines its value at  $x=0$.  Accordingly, the integrand is Riemann integrable  on the interval, and we can determine the improper integral by integrating termwise:

 $\displaystyle I$ $\displaystyle\;=\;\int_{0}^{1}\!\left(1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3% }}{4}+-\ldots\right)dx$ $\displaystyle\;=\;\operatornamewithlimits{\Big{/}}_{\!\!\!0}^{\,\quad 1}\!% \left(x-\frac{x^{2}}{2^{2}}+\frac{x^{3}}{3^{2}}-\frac{x^{4}}{4^{2}}+-\ldots\right)$ $\displaystyle\;=\;1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+-\ldots$

By the entry on Dirichlet eta function  at 2 (http://planetmath.org/ValueOfDirichletEtaFunctionAtS2), the sum of the obtained series is  $\eta(2)=\frac{\pi^{2}}{12}$.  Thus we have the result

 $\displaystyle\int_{0}^{1}\frac{\ln(1\!+\!x)}{x}dx\;=\;\frac{\pi^{2}}{12}.$ (3)

Similarly, using the series

 $\ln(1\!-\!x)\;=\;-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\ldots% \qquad(-1\leqq x<1)$

and the result in the entry Riemann zeta function    at 2 (http://planetmath.org/ValueOfTheRiemannZetaFunctionAtS2), one can calculate that

 $\displaystyle\int_{0}^{1}\frac{\ln(1\!-\!x)}{x}dx\;=\;-\frac{\pi^{2}}{6}.$ (4)
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