# integral of arctan(x^2) as a power series

• Oct 20th 2010, 03:46 AM
pirateboy
[Solved] integral of arctan(x^2) as a power series
We're asked to show the improper integral
$\displaystyle \int \tan^{-1}(x^2)\,dx$
as a power series.

I start with what I know. That is
$\displaystyle \displaystyle \tan^{-1}(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} = x - \frac{1}{3}x^3 + \frac{1}{5}x^5 - \frac{1}{7}x^7 + \dots$

So...
$\displaystyle \displaystyle \tan^{-1}(x^2) = \sum_{n=0}^\infty (-1)^n \frac{x^{4n+2}}{2n+1} = x^2 - \frac{1}{3}x^6 + \frac{1}{5}x^{10} - \frac{1}{7}x^{14} + \dots$

I'm trying to see the pattern here, in order to put this in sigma notation, however I'm afraid that when I do I still won't grasp what's going on here with series.

Knowing that, it's not hard to integrate the terms and see that

$\displaystyle \displaystyle \int\tan^{-1}(x)\,dx = C + \sum_{n=0}^\infty (-1)^n \frac{x^{2n+2}}{(2n+2)(2n+1)} = \frac{1}{2}x^2 - \frac{1}{14}x^4 + \frac{1}{30}x^6 - \frac{1}{56}x^8 + \dots$

But here's where I get stuck...
$\displaystyle \displaystyle \int \tan^{-1}(x^2)\,dx = C + \sum_{n=0}^\infty\text{???} = \frac{1}{3}x^3 - \frac{1}{21}x^7 + \frac{1}{55}x^{11} - \frac{1}{105}x^{15} + \dots$

• Oct 20th 2010, 03:52 AM
Also sprach Zarathustra
Quote:

$\displaystyle \displaystyle \int \tan^{-1}(x^2)\,dx = C + \sum_{n=0}^\infty\text{???} = \frac{1}{3}x^3 - \frac{1}{21}x^7 + \frac{1}{55}x^{11} - \frac{1}{105}x^{15} + \dots$
I think that this is wrong.

Integrate the formula of sum arctan(x^2) again...
• Oct 20th 2010, 04:11 AM
Pandevil1990
$\displaystyle \displaystyle \tan^{-1}(x^2) = \sum_{n=0}^\infty (-1)^n \frac{x^{4n+\textbf{2}}}{2n+1}$
• Oct 20th 2010, 04:13 AM
pirateboy
oh, you're right. the first term is wrong in arctan(x^2). it should be x^2, not x. i'll fix it.
• Oct 20th 2010, 04:19 AM
pirateboy
oh! that wasn't a copy and paste error. that was my problem. it was missing that 4n+2.

thanks pandevil. i thinks i gots it nows.