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

*To*: mathgroup at smc.vnet.net*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)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

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.