Using the MacLaurin expansion of:
ln(1+t) = sum_(n=0)^oo (-1)^nt^(n+1)/(n+1)ln(1+t)=∞∑n=0(−1)ntn+1n+1
let: t= x^2t=x2: to get
ln(1+x^2) = sum_(n=0)^oo (-1)^nx^(2n+2)/(n+1)ln(1+x2)=∞∑n=0(−1)nx2n+2n+1
then divide by x^wxw and integrate term by term:
ln(1+x^2)/x^w = sum_(n=0)^oo (-1)^nx^(2n+2-w)/(n+1)ln(1+x2)xw=∞∑n=0(−1)nx2n+2−wn+1
int ln(1+x^2)/x^w = sum_(n=0)^oo (-1)^n/(n+1) int x^(2n+2-w)dx∫ln(1+x2)xw=∞∑n=0(−1)nn+1∫x2n+2−wdx
int ln(1+x^2)/x^w = C+sum_(n=0)^oo (-1)^n/(n+1)x^(2n+3-w)/(2n+3-w)∫ln(1+x2)xw=C+∞∑n=0(−1)nn+1x2n+3−w2n+3−w
The series has radius of convergence at least equal to the original series R=1R=1, but has sense only if for every nn:
2n+3-w != 02n+3−w≠0
2n+3 != w2n+3≠w
thus ww cannot be an odd integer number except for w=1w=1.
