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MathGroup Archive 2012

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Re: Integral that should converge, but Mathematica says it does not

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  • Subject: [mg128955] Re: Integral that should converge, but Mathematica says it does not
  • From: Daniel <dosadchy at its.jnj.com>
  • Date: Wed, 5 Dec 2012 03:15:38 -0500 (EST)
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I was thinking about this some more, and in my opinion, this integral is almost impossible to compute numerically (in a "standard" way).

As you recall from my previous post in this thread, the function looks like a series of peaks, centered at x=pi*n, with height x and width ~2/x^3 with area proportional to 1/n^2.

The error between the infinite sum and the sum up to n behaves like 1/n, so if you want to compute this integral to 6 digits accuracy,  you need to sum up to n =~ 10^6. The corresponding peak is centered at pi*10^6, but for the sake of simplicity lets say x=10^6. the width of this peak is ~1/10^18. I'd expect to be able to put at least 10 points inside this peak, so I'll need resolution of 1/10^19, and remember, this is at x=10^6. This means you have to compute at precision of at least 25 digits to have 6 digit precision! (and this is still a gross underestimation of the required precision).

So, I don't think a bug report is necessary here.



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