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

*To*: mathgroup at smc.vnet.net*Subject*: [mg128920] Re: Integral that should converge, but Mathematica says it does not*From*: Daniel <dosadchy at its.jnj.com>*Date*: Tue, 4 Dec 2012 04:11:10 -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

Hi, you got me looking there. The proof in the link seems correct. Also I can see why Mathematica (or any other generic numerical method) will fail here. Lets see how the function looks like for a large x, particularly x = pi*n+eps for large integer n and small eps<<1. (pi*n+eps)/(1 + (pi*n+eps)^6 Sin[pi*n+eps]^2) For very large n and eps<<1 we can approximately write: (pi*n)/(1+(pi*n)^6 eps^2) ~ x/(1+x^6 eps^2) Now you can see that this function describes a narrow peak centered on pi*n, having amplitude x and a very small width of approximately 1/x^3. It is very logical that no general sampling method can catch those narrow peaks for numerical integration. We can continue to estimate the integral by proposing that the area under each peak is: x*(2/x^3) ~ 2*pi*n/(pi*n)^3 = 2/(pi*n)^2 Although this is just estimation, we also get a series of peaks with areas that behave like ~1/n^2. The rigorous proof in the link arrives to an upper bound which also behaves like 1/n^2.