Converting Power Series to Integrals: How to Handle Constants of Integration?

[nameVoid]

<br /> \int \frac{x-arctanx}{x^3}dx<br />
<br /> \frac{d}{dx}( x-arctanx ) = 1-\frac{1}{1+x^2}=\frac{x^2}{x^2+1}<br />
<br /> = x^2 \sum_{n=0}^{\infty}(-1)^nx^{2n} = \sum_{n=0}^{\infty}(-1)^nx^{2n+2} <br />
<br /> \int \sum_{n=0}^{\infty}(-1)^nx^{2n+2} dx = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}+C<br />
<br /> C=0?<br />
<br /> \int \frac{\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}}{x^3} dx<br />
<br /> \int \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{2n+3}}dx<br />
<br /> \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+3)(2n+1)}}<br />
<br /> \sum_{n=1}^{\infty}(-1)^n\frac{x^{2n}}{(2n+2)(2n)}}+C<br />
here i took d/dx converted to geomtric then integrated divided by x^3 then integrated again not sure how to deal with the ontants of integration in this case took c=0 on the first integral

 

[vela]

<br /> \int \frac{x-arctanx}{x^3}dx<br />
<br /> \frac{d}{dx}( x-arctanx ) = 1-\frac{1}{1+x^2}=\frac{x^2}{x^2+1}<br />
<br /> = x^2 \sum_{n=0}^{\infty}(-1)^nx^{2n} = \sum_{n=0}^{\infty}(-1)^nx^{2n+2} <br />
<br /> \int \sum_{n=0}^{\infty}(-1)^nx^{2n+2} dx = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}+C<br />
<br /> C=0?<br />

If you set x=0, you can show that C is, in fact, 0 as you assumed.

<br /> \int \frac{\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{2n+3}}{x^3} dx<br />
<br /> \int \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{2n+3}}dx<br />
<br /> \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+3)(2n+1)}}<br />
<br /> \sum_{n=1}^{\infty}(-1)^n\frac{x^{2n}}{(2n+2)(2n)}}+C<br />
here i took d/dx converted to geomtric then integrated divided by x^3 then integrated again not sure how to deal with the ontants of integration in this case took c=0 on the first integral

You're fine up until the last step. Note that in the next-to-the-last line, you have odd powers of x, and in the final line you have only even powers of x. The two expressions aren't equal to each other.

 

[nameVoid]

right..