application of logarithm series

The integrand of the improper integral

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

is not defined at the lower limit 0.  However, from the Taylor series expansion

of the natural logarithm we obtain the expansion of the integrand

whence

$\underset{x\to 0}{lim}{\displaystyle \frac{\ln (1+x)}{x}}=\mathrm{\hspace{0.33em}1}.$(2)

This implies that the integrand of (1) is bounded on the interval  $[0,\mathrm{\hspace{0.17em}1}]$ 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:

$I$$\mathrm{\hspace{0.33em}}={\displaystyle {\int }_0^1}(1-{\displaystyle \frac{x}{2}}+{\displaystyle \frac{x^2}{3}}-{\displaystyle \frac{x^3}{4}}+-\mathrm{\dots })dx$

$\mathrm{\hspace{0.33em}}=\underset{0}{\overset{1}{/}}(x-{\displaystyle \frac{x^2}{2^2}}+{\displaystyle \frac{x^3}{3^2}}-{\displaystyle \frac{x^4}{4^2}}+-\mathrm{\dots })$

$\mathrm{\hspace{0.33em}}=\mathrm{\hspace{0.33em}1}-{\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{3^2}}-{\displaystyle \frac{1}{4^2}}+-\mathrm{\dots }$

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

${\displaystyle {\int }_0^1}{\displaystyle \frac{\ln (1+x)}{x}}𝑑x={\displaystyle \frac{{\pi }^2}{12}}.$(3)

Similarly, using the series

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

${\displaystyle {\int }_0^1}{\displaystyle \frac{\ln (1-x)}{x}}𝑑x=-{\displaystyle \frac{{\pi }^2}{6}}.$(4)