∫−11x1−ln(1+x)dx is undefined on the interval [-1,1].
Mainly because
∫−11x1dx is undefined on the interval [-1,1].
See the ln(x) in Soroban's series?. It is undefined on the given interval.
But, if we wanted
∫−11xln(1+x)dx, then we could hammer out a solution using series and integration.
It works out to be
4π2
But, then
∫−11x1dx=−πi. In other words, undefined.
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The usefulness of ln(1+x) is limited because of its slow convergence and the restriction
−1<x≤1
But, if we replace x with -x to get the series for ln(1-x), then subtract the two series we get:
ln(1−x1+x)=2(x+3x3+5x5+7x7+⋅⋅⋅), −1≤x<1......[1]
The last series can be used to compute the natural log of any positive number y by letting:
y=1−x1+x
x=y+1y−1.............[2]
and noting that
−1<x<1.
Let compute ln(2) by letting y=2 in [2]. This gives
x=1/3.
Sub this into [1] and get
ln(2)=2(1/3+3(1/3)3+5(1/3)5+7(1/7)7+⋅⋅⋅)
This gives ln(2)=.6931 to four decimal places.
We could just let x=1 in the series for ln(1+x), but it converges too slow to be of any real computational value.