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Integrate the following using INFINITE SERIES (no binomial please)?

#\int(\ln(1+x^2))/x^wdx#

Answer 1

Isaiah Allen

#int ln(1+x^2)/x^w = C+sum_(n=0)^oo (-1)^n/(n+1)x^(2n+3-w)/(2n+3-w)#

Using the MacLaurin expansion of:

#ln(1+t) = sum_(n=0)^oo (-1)^nt^(n+1)/(n+1)#

let: #t= x^2#: to get

#ln(1+x^2) = sum_(n=0)^oo (-1)^nx^(2n+2)/(n+1)#

then divide by #x^w# and integrate term by term:

#ln(1+x^2)/x^w = sum_(n=0)^oo (-1)^nx^(2n+2-w)/(n+1)#

#int ln(1+x^2)/x^w = sum_(n=0)^oo (-1)^n/(n+1) int x^(2n+2-w)dx#

#int ln(1+x^2)/x^w = C+sum_(n=0)^oo (-1)^n/(n+1)x^(2n+3-w)/(2n+3-w)#

The series has radius of convergence at least equal to the original series #R=1#, but has sense only if for every #n#:

#2n+3-w != 0#

#2n+3 != w#

thus #w# cannot be an odd integer number except for #w=1#.

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Answer 2

William Abdalla

Certainly! Please provide the function or expression that you'd like to integrate using infinite series, and I'll help you with the solution.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.