1. ## intergration

Show that by expanding 1/(1+x^4) as a geometric series and integrating termwise.

2. Originally Posted by bigb
Show that by expanding 1/(1+x^4) as a geometric series and integrating termwise.
We know that $\displaystyle \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in fty}(-1)^nx^{4n}$ so \displaystyle \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\ &=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}

So \displaystyle \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0 ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\ &=\frac{\sqrt{2}\pi}{4}\end{aligned}

$\displaystyle \forall{x}\in(0,1)~\frac{1}{1+x^2}=\sum_{n=0}^{\in fty}(-1)^nx^{2n}$

So \displaystyle \begin{aligned}\int_0^1\frac{dx}{1+x^2}&=\int_0^1\ sum_{n=0}^{\infty}(-1)^nx^{2n}dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} \end{aligned}

But note that $\displaystyle \int_0^1\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}$

So $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$

Also note that the interchanging of all the summations and integrals was due to the integrability of a power seris for all points on the interior of its interval of convergence.

3. Originally Posted by Mathstud28
We know that $\displaystyle \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in fty}(-1)^nx^{4n}$ so \displaystyle \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\ &=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}

So \displaystyle \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0 ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\ &=\frac{\sqrt{2}\pi}{4}\end{aligned}

$\displaystyle \forall{x}\in(0,1)~\frac{1}{1+x^2}=\sum_{n=0}^{\in fty}(-1)^nx^{2n}$

So \displaystyle \begin{aligned}\int_0^1\frac{dx}{1+x^2}&=\int_0^1\ sum_{n=0}^{\infty}(-1)^nx^{2n}dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} \end{aligned}

But note that $\displaystyle \int_0^1\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}$

So $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$

Also note that the interchanging of all the summations and integrals was due to the integrability of a power seris for all points on the interior of its interval of convergence.
I am suppose to show in a three step process that the

and this was the first step part a)

4. Originally Posted by bigb
I am suppose to show in a three step process that the

and this was the first step part a)
Hmm, but this just simply is not true?

But for the sum look here http://www.mathhelpforum.com/math-he...415-post2.html and note then that your sum should be equal to $\displaystyle \frac{1}{16}\bigg[\frac{\pi}{2\frac{1}{4}\sin\left(\frac{\pi}{4}\rig ht)}-\frac{1}{2\left(\frac{1}{4}\right)^2}\bigg]=\frac{\sqrt{2}\pi-4}{8}$ which is what your first equation indicates.

5. Originally Posted by Mathstud28
Hmm, but this just simply is not true?

But for the sum look here http://www.mathhelpforum.com/math-he...415-post2.html and note then that your sum should be equal to $\displaystyle \frac{1}{16}\bigg[\frac{\pi}{2\frac{1}{4}\sin\left(\frac{\pi}{4}\rig ht)}-\frac{1}{2\left(\frac{1}{4}\right)^2}\bigg]=\frac{\sqrt{2}\pi-4}{8}$ which is what your first equation indicates.
What do you think is simply not true??

This portion of your first post is my second part of this problem, where i gotta show that the integral from my first post is equal to sqrt(2)*pi/4 like you have shown.

We know that $\displaystyle \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in fty}(-1)^nx^{4n}$ so \displaystyle \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\ &=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}

So \displaystyle \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0 ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\ &=\frac{\sqrt{2}\pi}{4}\end{aligned}

is there a way to conclude from here to prove

6. Originally Posted by bigb
What do you think is simply not true??

This portion of your first post is my second part of this problem, where i gotta show that the integral from my first post is equal to sqrt(2)*pi/4 like you have shown.

We know that $\displaystyle \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in fty}(-1)^nx^{4n}$ so \displaystyle \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\ &=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}

So \displaystyle \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0 ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\ &=\frac{\sqrt{2}\pi}{4}\end{aligned}

is there a way to conclude from here to prove
What is untrue is that $\displaystyle \int_0^1\frac{1+x^2}{1+x^4}dx\ne\sum_{n=0}^{\infty }\frac{(-1)^n}{2n+1}=\int_0^1\frac{dx}{1+x^2}$

And try not to quote it like that, it is hard to discern what I wrote and what you wrote.