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Value of $\sum x^n$

How do you rewrite this: $$ \int_{0}^{\infty} \cfrac{1}{1+x^4} $$

to: this$$\int_{0}^1 \sum_{n=0}^\infty (-1)^n x^{4n} + \int_{1}^\infty x^{-4} \sum_{n=0}^{\infty} (-1)^n x^{-4n}$$

and how do you solve it?

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marked as duplicate by t.b., Did, Srivatsan, Sasha, Asaf Karagila Jan 1 '12 at 19:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Do you know the closed formula for $\sum_k^n x^k$? What happens if $n$ tends to inifinity? And what would be the formula for $\sum x^{-k}$? –  user20266 Jan 1 '12 at 17:46
    
@Thomas - What is close formula? –  Victor Jan 1 '12 at 17:47
    
it's the formula for the geometric series. –  user20266 Jan 1 '12 at 17:49
2  
Next time you're lurking in chat, just ask there... –  t.b. Jan 1 '12 at 17:57
1  
@Victor It is not totally different. Hint: substitute $x=-t^4$ to the formula behind t.b.'s link. Your textbook probably explains, when it is permissible to integrate a series term-by-term (that is not automatic!). –  Jyrki Lahtonen Jan 1 '12 at 18:30

1 Answer 1

up vote 1 down vote accepted

To rewrite it, you need to know the geometric series expansion, $$ \frac{1}{1+z} = 1-z+z^2-z^3+...=\sum_{n=0}^{\infty} (-1)^{n}z^{n}, $$ and know that it converges absolutely for all $|z|<1$. There are two different geometric series expansions for $(1+x^4)^{-1}$; one is the Taylor series expansion, good for small $x$, and the other is the Laurent series expansion, good for large $x$. They are $$ \frac{1}{1+x^4} = 1-x^4+x^8-...=\sum_{n=0}^{\infty}(-1)^{n}x^{4n}$$ and $$ \frac{1}{1+x^4}=\frac{1}{x^4}\frac{1}{1+x^{-4}}=\frac{1}{x^4}\left(1-x^{-4}+x^{-8}-...\right)=\sum_{n=0}^{\infty}(-1)^n x^{-4-4n}.$$ These series converge absolutely when $|x^4|<1$ and when $|x^{-4}|<1$ respectively (i.e., for $|x|<1$ and for $|x|>1$). So we have $$ \begin{eqnarray} I=\int_{0}^{\infty}\frac{dx}{1+x^4}&=&\int_{0}^{1}\frac{dx}{1+x^4}+\int_{1}^{\infty}\frac{dx}{1+x^4} \\ &=&\int_{0}^{1}dx\sum_{n=0}^{\infty}(-1)^n x^{4n}+\int_{1}^{\infty}dx\sum_{n=0}^{\infty}(-1)^{n}x^{-4-4n}\\ &=&\sum_{n=0}^{\infty}(-1)^{n}\left(\int_{0}^{1}x^{4n}dx + \int_{1}^{\infty}x^{-4-4n}dx\right), \end{eqnarray} $$ where we need the absolute convergence of the sums to justify exchanging the sums and the integrals. The integrals are elementary: $$ \int_{0}^{1}x^{4n}dx = \frac{x^{4n+1}}{4n+1}\Bigg\vert_{0}^{1} = \frac{1}{4n+1} $$ and $$ \int_{1}^{\infty}x^{-4-4n}dx = \frac{x^{-3-4n}}{-3-4n}\Bigg\vert_{1}^{\infty} = \frac{1}{4n+3}, $$ giving the result $$ \begin{eqnarray} I &=& \sum_{n=0}^{\infty}(-1)^{n}\left(\frac{1}{4n+1} + \frac{1}{4n+3}\right) \\ &=& \sum_{n=0}^{\infty} \frac{\sigma_{n}}{2n+1} \\ &=& 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{13} - \text{...}, \end{eqnarray} $$ where $\sigma=\{1,1,-1,-1,1,1,-1,-1,...\}$.

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