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So, I decided to dig a little deeper into numerical integration because we hardly had any of that in my analysis class. I've come across this method for improper integrals: Метод Самокиша (not in English, unfortunately).

What scares me though, is that the series in the last formula starts with $- \infty$. Is that even possible? We haven't studied series yet, but from what I understand the series usually starts with $1$ (or $0$) and goes to infinity.

I really hope you can help me figure this out. Thanks!

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    I acted before thinking - of course you are right! I think my example is called symmetric limit or something similar, it is very bad in many senses..2011-12-18
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    The canoical example to look into is $ \sum 1/n^2 $2011-12-19
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    @JonasMeyer Removed my stupid comment.2011-12-19

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We have the following definition: $$ \sum_{n=0}^\infty a_n = \lim_{N \to \infty} \sum_{n=0}^N a_n,$$ if this limit exists. It is now clear how to make sense of a sum which is infinite on both sides: $$ \sum_{n=-\infty}^\infty a_n = \lim_{N\to \infty} \lim_{M \to -\infty} \sum_M^N a_n, $$ if this limit exists. Moreover, if this latter limit exists, then it is also equal to $$ \lim_{N\to\infty}\sum_{n=-N}^N a_n,$$ which you can evaluate numerically in the usual way -- sum up more and more terms and keep track of how small the summands are getting...

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    Wow, never seen a double limit, but it makes perfect sense from the theoretical point of view. However, I still don't get it how to calculate the actual terms of the series...2011-12-18
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    I added a sentence on how to deal with it "practically" as well.2011-12-18
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    Thanks for still being on this thread! I understand this much better now. Just one last thing, what do I start with? For example, in harmonic series you start with 1 and go to infinity, getting 1, 1/2, 1/3, 1/4...Maybe when evaluation the improper integral from -inf to inf, I should break it down into two series, from 0 to infinity and from 0 to -infinity, right?2011-12-18
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    Note that if the sum converges absolutely it doesn't matter how you sum the terms.2011-12-18
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    For example, how do I calculate that: Σ from -∞ to ∞ (1/n). I know it doesn't converge, but I still should be able to get the terms. Just don't know where to start...2011-12-18
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    Well, of course I am making it the best answer, but just in case you read this again, it would be fantastic if you explained that last part. Thank you!2011-12-18
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    Your example doesn't converge, and the $n=0$ term doesn't make sense. But just consider any sequence $\{a_n\}$ such that both the positively and negatively indexed terms converge. Then you can write the sum as $a_0 + a_1 + a_{-1} + a_2 + a_{-2} + \ldots$ and evaluate this like you would an ordinary sum. Alternatively, split the sum into positive and negative parts and sum separately. (As Qiaochu notes, if you moreover assume absolute convergence then just sum the terms in any order whatsoever.)2011-12-18
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    @Dan This is a nice answer. An alternate equivalent definition is to break up the doubly infinite series into two singly infinite series: $$\sum_{n=-\infty}^{\infty} a_n = \sum_{j=0}^{\infty} a_j + \sum_{k=1}^{\infty} a_{-k} .$$2011-12-19
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There are in fact nice examples of doubly infinite series occurring in practice. The Jacobi theta functions can be defined as doubly infinite Fourier series, e.g.

$$\vartheta_2(z,q)=\sum_{n\in\mathbb Z} q^{\left(n+\frac12\right)^2}\exp((2n+1)iz)$$

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    Functions with [essential singularities](http://en.wikipedia.org/wiki/Essential_singularity) are expressed as sums over $\mathbf Z$, too.2011-12-19
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What scares me though, is that the series in the last formula starts with $−\infty$. Is that even possible? We haven't studied series yet, but from what I understand the series usually starts with $1$ (or $0$) and goes to infinity.

You should not be scared. Series of the type: $$\sum_{n=-\infty}^\infty a_n \qquad \text{(also denoted by } \textstyle\sum_{n\in \mathbb{Z}} a_n\text{)}$$ are usually called bilateral series. A bilateral series converges iff the limit: $$\lim_{N,M\to \infty} \sum_{n=-M}^N a_n$$ exists; otherwise, it is said to diverge. If you want, you can think a convergent bilateral series as a sum of two "standard" series, i.e.: $$\sum_{n=-\infty}^\infty a_n = \sum_{n=0}^\infty a_n +\sum_{n=1}^\infty a_{-n}$$

For example, the bilateral series: $$\sum_{n=-\infty}^\infty \frac{1}{(2n+1)^2}$$ converges: in fact, for fixed $N,M\in \mathbb{N}$ you get: $$\begin{split} \lim_{N,M\to \infty} \sum_{n=-M}^N \frac{1}{(2n+1)^2} &= \lim_{N\to \infty} \sum_{n=0}^N \frac{1}{(2n+1)^2} +\lim_{M\to \infty} \sum_{n=1}^M \frac{1}{(1-2n)^2} \\ &= \sum_{n=0}^\infty \frac{1}{(2n+1)^2} +\sum_{n=1}^\infty \frac{1}{(2n-1)^2}\end{split}$$ for both series $\sum 1/(2n+1)^2$ and $\sum 1/(1-2n)^2$ converge; in particular: $$\sum_{n=-\infty}^\infty \frac{1}{(2n+1)^2} =\frac{\pi^2}{4}\; .$$

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    Thank you! That was a very good explanation!2011-12-22
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It is possible. A way to conceptually verify this is to consider a function whose integral from -infinity to +infinity converges such as probability density functions (I.e. normal curves) whose aforementioned integral converges to one.

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    Thanks for the reference about the probability density functions. I imagined something like that to visualize an improper integral from -infinity to +infinity but now I know how it's called. Thank you!2011-12-18
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For conditional convergence, you're against the wall and fighting with however you decide to define a doubly infinte series I'm afraid.

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    The [Madelung constants](http://mathworld.wolfram.com/MadelungConstants.html) are a particularly prominent example.2011-12-22