# Passing an integral through an infinite sum

## Homework Statement

I want to show that
$$\tan^{-1}(x)=\sum\limits_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}x^{2n+1}.$$

## Homework Equations

$$\int\frac{1}{1+x^{2}}dx.$$

## The Attempt at a Solution

I want to be able to do the following:
$$\int\frac{1}{1+x^{2}}dx=\int\sum\limits_{n=0}^{\infty}(-1)^{n}x^{2n}dx=\sum\limits_{n=0}^{\infty}\int (-1)^{n}x^{2n}dx$$
but I am afraid that the infinite sum might create problems. Can anyone take a look? Thanks!

Gold Member
The order of the sum and integral can be switched, since they are both sums, so that's fine.

So it seems that $\frac{1}{1+x^2}$ can be expressed as an infinite geometric series so long as $|x^{2}|\leq 1$.

Where
$\frac{1}{1-r}=\sum_{i=0}^{∞} r^{i}$
when $|r|\leq 1$

we can say that

$\frac{1}{1+x^2}=\sum_{i=0}^{∞} (-x^2)^{i}=\sum_{i=0}^{∞} (-1)^{i}(x)^{2i}$
when $|x^{2}|\leq 1$

Hope this helps:)

Jufro
The order of the sum and integral can be switched, since they are both sums, so that's fine.

They cannot always be switched. Only under certain conditions is that statement valid. While this can be done so here it is bad habit to think the statement as always true.

To reference http://en.wikipedia.org/wiki/Fubini's_theorem

$$\int_{a}^{b} \sum_{n=0}^{∞} a_nx^n = \sum_{n=0}^{∞} \int_{a}^{b} a_nx^n$$
$$\frac{d}{dx} \sum_{n=0}^{∞} a_nx^n = \sum_{n=0}^{∞} \frac{d}{dx} a_nx^n$$
$$lim_{x→a} \sum_{n=0}^{∞} a_nx^n = \sum_{n=0}^{∞} lim_{x→a} a_nx^n$$