Passing an integral through an infinite sum

  • #1
DeadOriginal
274
1

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


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

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!
 

Answers and Replies

  • #2
jfizzix
Science Advisor
Insights Author
Gold Member
757
355
The order of the sum and integral can be switched, since they are both sums, so that's fine.

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

Where
[itex]\frac{1}{1-r}=\sum_{i=0}^{∞} r^{i}[/itex]
when [itex]|r|\leq 1[/itex]

we can say that

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

Hope this helps:)
 
  • #3
Jufro
92
8
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
 
  • #4
DeadOriginal
274
1
Thanks guys! That helps a lot.
 
  • #5
STEMucator
Homework Helper
2,075
140
Have you heard about uniform convergence? Long story short, it allows us to verify if the following relationships actually hold :

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

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