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I'm approximating an integral with only exponentials. i.e., it is equal to

$\displaystyle \int_{-\pi}^\pi{\frac{\displaystyle\sum_{j=a}^b{c_j e^{i\cdot d_j \cdot t}}}{\displaystyle\sum_{k=a}^b{r_k e^{i \cdot s_k \cdot t}}} dt}$ where $c_j, d_j, r_k, s_k \in \mathbb{Z}$

In other words, it's a fraction of exponentials over exponentials.

I realized that we can integrate the individual terms of the power series expansion to integrate the expression. I also found that for the particular integrals that I'm working with, only the first $N$ terms (for some $N$) significantly contribute to the value of the integral.

My idea

We can use feedback and recurrences to approximate this integral. For example, the first two terms of the power series expansion of the function will be something like $c_0 + c_1x$. If we attempt to integrate these terms, we should get $c_0x + \frac{c_1}{2}x$ evaluated at the two endpoints of the interval of integration. But we can approximate this second function. Suppose we take the original power series expansion ($c_0 + c_1x$), rewrite $x$ as $x/2$, and then multiply by $x$. Bingo! We have the integrated function.

Now we can make "guesses" at the integral using the rewriting technique above. Without going into detail, I then have a way to determine the error of the guess. So we can introduce a feedback system to get closer and closer to the actual integral.

My Question

Does a method like this already exist? I've seen techniques for approximating integrals using intervals, but this technique attempts to do it using feedback, instead. Has a similar technique been used? I would like to see the method, if it exists.

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I'm skeptical of the performance of series methods on integrals with complex exponentials; most certainly if the $d_j$ and $s_k$ parameters are at least medium-sized. Could you at least give a numerical example? –  J. M. Nov 16 '10 at 3:39
    
I'm still researching this, but I've tried with one example:$\frac{1-e^{6t\cdot i}}{1-e^{2t\cdot i}}$. I know that the series should converge to one. It does so after around 30 terms, that is, to a value that rounds to one. –  Matt Groff Nov 16 '10 at 4:12
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2 Answers

up vote 2 down vote accepted

Are you aware that integral of this type can be computed exactly by means of the residue theorem? The price to pay is that you have to find the roots of the equation $\sum_a^b r_k z^{s_k}=0$ in the unit disk. (You may assume all $s_k\ge0$.)

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Thanks for the alternative. My concern is that I'm working with huge powers. I really only want to determine if the integral evaluates to a nonzero value, but I need a method that does this in time logarithmic to the size of the powers of the exponentials. –  Matt Groff Nov 16 '10 at 14:36
    
Be damn careful: if you get a (near) zero in the denominator somewhere, all your talk about "only the first N terms matter" and "time logarithmic in the size of the powers" may become total nonsense... –  fedja Nov 26 '12 at 4:45
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You can read about Robmerg Integration in section 4.3 of Numerical Recipes or in Wikipedia and the references. It is a more formal approach to what you are describing. You talk about using feedback, but it is not clear where you get the feedback from. If your rewriting is really term-by-term integration (see your previous question and my answer) you do have feedback from that. The Romberg routine follows the changes in the integral as the order rises and makes a guess based on that. As explained earlier in the book regarding fitting functions, there are risks in this-maybe the function goes wild in a small area.

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You're integrating a ratio of sums of complex exponentials (i.e., sines and cosines), over a (probable) period. Romberg can actually retard the convergence of the trapezoidal rule, which converges remarkably fast for integrating sines and cosines. An adaptive method might be useful, but certainly not Romberg. (In any event, the exponential growth of abscissas ($2^n$) can be overwhelming; the Bulirsch sequence might be a better choice if one insists on Richardsonian methods.) –  J. M. Nov 16 '10 at 6:14
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