# Need help evaluating an improper integral as a power series.

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1. Apr 8, 2017

### uchuu-man chi

1. The problem statement, all variables and given/known data

Evaluate the indefinite integral as a power series. What is the radius of convergence (R)?
$\int x^2ln(1+x) \, dx$

Book's answer: $\int x^2ln(1+x) dx = C + \sum_{n=1}^\infty (-1)^n \frac {x^{n+3}} {n(n+3)}; R = 1$
2. Relevant equations
Geometric series
$\frac {1} {1-x} = \sum_{n=0}^\infty x^n ; |x|<1$

3. The attempt at a solution
$\frac {1} {1-x} = \sum_{n=0}^\infty x^n ; |x|<1$

-Substitute -x in for x
$\frac {1} {1+x} = \sum_{n=0}^\infty (-1)^n x^n ; |x|<1$

-Integrate
$\int \frac {1} {1+x} dx = \int \sum_{n=0}^\infty (-1)^n x^n dx ; |x|<1 \\ = ln(1+x) = C + \sum_{n=0}^\infty (-1)^n \frac {x^{n+1}}{n+1} ; |x|<1$

$\text {When x=0, C = ln(1) = 0} \\ = ln(1+x) = \sum_{n=0}^\infty (-1)^n \frac {x^{n+1}}{n+1} ; |x|<1$

-Multiply by x2
$x^2 ln(1+x) = x^2 \sum_{n=0}^\infty (-1)^n \frac {x^{n+1}} {n+1} ; |x|<1 \\ = x^2ln(1+x) = \sum_{n=0}^\infty (-1)^n \frac {x^{n+3}} {(n+1)}; |x|<1$

-Integrate

$\int x^2 ln(1+x) dx = \int \sum_{n=0}^\infty (-1)^n \frac {x^{n+3}} {n+1} dx ; |x|<1 \\ = \int x^2 ln(1+x) dx = C + \sum_{n=0}^\infty (-1)^n \frac {x^{n+4}} {(n+1)(n+4)} , R = 1$

$C + \sum_{n=0}^\infty (-1)^n \frac {x^{n+4}} {(n+1)(n+4)} , R = 1$

I've been trying to get the same answer as the book, but even if I shifted the index to start at 1, it would be $\sum_{n=1}^\infty (-1)^{n+1} \frac {x^{n+3}} {n(n+3)} \\ \text {or} \\ \sum_{n=1}^\infty (-1)^{n-1} \frac {x^{n+3}} {n(n+3)}$

Last edited: Apr 8, 2017
2. Apr 9, 2017

### Staff: Mentor

I can't find a sign error and Wikipedia is also on our side: $ln(1+x)=-\sum_{n\geq 0}\frac{(-x)^{n+1}}{n+1}$ which I took for comparison to what the differentiation of the book's answer got me. So as long as I didn't fall into the same pit as you, the sign error is in the book. You could differentiate, too, just to be sure.