application of logarithm series


The integrand of the improper integral

I:=∫01ln⁑(1+x)x⁒𝑑x (1)

is not defined at the lower limitMathworldPlanetmath 0.  However, from the Taylor seriesMathworldPlanetmath expansion

ln(1+x)=x-x22+x33-x44+-…  (-1<x≦1)

of the natural logarithmMathworldPlanetmathPlanetmathPlanetmath we obtain the expansion of the integrand

ln⁑(1+x)x= 1-x2+x23-x34+-…  (-1<x<0,  0<x≦1)

whence

limxβ†’0⁑ln⁑(1+x)x= 1. (2)

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

I  =∫01(1-x2+x23-x34+-…)dx
 =/01(x-x222+x332-x442+-…)
 = 1-122+132-142+-…

By the entry on Dirichlet eta functionMathworldPlanetmath at 2 (http://planetmath.org/ValueOfDirichletEtaFunctionAtS2), the sum of the obtained series is  η⁒(2)=Ο€212.  Thus we have the result

∫01ln⁑(1+x)x⁒𝑑x=Ο€212. (3)

Similarly, using the series

ln(1-x)=-x-x22-x33-x44-…  (-1≦x<1)

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

∫01ln⁑(1-x)x⁒𝑑x=-Ο€26. (4)
Title application of logarithm series
Canonical name ApplicationOfLogarithmSeries
Date of creation 2013-03-22 18:56:09
Last modified on 2013-03-22 18:56:09
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 12
Author pahio (2872)
Entry type Application
Classification msc 33B10
Related topic DilogarithmFunction
Related topic ExamplesOnHowToFindTaylorSeriesFromOtherKnownSeries
Related topic SubstitutionNotation