seanieg89
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Re: Extracurricular Integration Marathon
p(x)\,dx\right| \leq p(TM)\left|\int_{TM}^{T(M+1)}\phi(x)\, dx\right|)
if p monotonically decreases to zero and the integral of phi between mT and (m+1)T is const.(-1)^m.
(This comes from the alternating nature of the integrand and is a good exercise to prove.)
Equipped with this result we can truncate our integrals upper limit to m*pi with an explicit error term of O(1/m).
Now that we are working with finite integral, we Taylor expand e^(e^(ix))-e^(e^(-ix)) in powers of e^(ix) to m-th order. As we are evaluating the functions e^z only on the unit disk, we incur an O(1/(m+1)!) error here.
Of course, the interval of integration has length of order m, so overall this thing gives us an O(m/(m+1)!) error. The remaining main term is
}{x}\, dx.)
The integral occuring in the sum is pi/2, up to an order (1/(k*m*pi)) order error. (We have evaluated the integral earlier in this thread, and the error bound comes because it is of the oscillatory type discussed in this post.)
Summing 1/(k.k!) over ALL of the integers is convergent, so this error term sums to O(1/m) as well. The final error term comes from truncating the Taylor sum for e. This error obviously tends to zero as well, so we are left with the main term, which is simply (e*pi)/2.
Have been a bit busy the last couple of days, hence the delay. The method does indeed work and is based on the estimate:
if p monotonically decreases to zero and the integral of phi between mT and (m+1)T is const.(-1)^m.
(This comes from the alternating nature of the integrand and is a good exercise to prove.)
Equipped with this result we can truncate our integrals upper limit to m*pi with an explicit error term of O(1/m).
Now that we are working with finite integral, we Taylor expand e^(e^(ix))-e^(e^(-ix)) in powers of e^(ix) to m-th order. As we are evaluating the functions e^z only on the unit disk, we incur an O(1/(m+1)!) error here.
Of course, the interval of integration has length of order m, so overall this thing gives us an O(m/(m+1)!) error. The remaining main term is
The integral occuring in the sum is pi/2, up to an order (1/(k*m*pi)) order error. (We have evaluated the integral earlier in this thread, and the error bound comes because it is of the oscillatory type discussed in this post.)
Summing 1/(k.k!) over ALL of the integers is convergent, so this error term sums to O(1/m) as well. The final error term comes from truncating the Taylor sum for e. This error obviously tends to zero as well, so we are left with the main term, which is simply (e*pi)/2.
