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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.. – AD. Dec 18 '11 at 21:29
The canoical example to look into is $ \sum 1/n^2 $ – Adam Dec 19 '11 at 2:21
@JonasMeyer Removed my stupid comment. – AD. Dec 19 '11 at 8:03
up vote 16 down vote accepted

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... – ivt Dec 18 '11 at 20:58
I added a sentence on how to deal with it "practically" as well. – Dan Petersen Dec 18 '11 at 21:24
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? – ivt Dec 18 '11 at 21:40
Note that if the sum converges absolutely it doesn't matter how you sum the terms. – Qiaochu Yuan Dec 18 '11 at 21:42
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... – ivt Dec 18 '11 at 21:51

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 are expressed as sums over $\mathbf Z$, too. – Dylan Moreland Dec 19 '11 at 0:44

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! – ivt Dec 22 '11 at 23:00

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! – ivt Dec 18 '11 at 21:00

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 are a particularly prominent example. – J. M. Dec 22 '11 at 1:06

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