digerdiga

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11 years, 112 days

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These are replies submitted by digerdiga

@Mariusz Iwaniuk I see... ;)

 

Must have misread.

@Mariusz Iwaniuk What do you mean by "trivial" corrections? Do you have an example and do you expect a large jump with 2019?

Can you provide a list of a few "weaknesses" of int, that are going to be fixed?

@Mariusz Iwaniuk why so "late"?

When I do

simplify(polylog(3,1+I) + polylog(3,1-I))

is it possible to look up how maple simplifies ? Apparently it can be transformed into a polylog(3,I).

@Christian Wolinski  How do I replace it?

@vv I actually get "Kernel lost" in the B1 part...

@vv Hey, I attached a worksheet so you can try yourself and maybe find some tweaks to improve the numerical integration, better than I did so far.

 


 

# comparison of integral (f1) and calculation by residues (f2) assuming b lies in lower C-plane, a>0, and epsilon>0 (here e, because of the epsilon in the errorbound of the integral). I'm not showing here that the integral over a big circle in the upper half plane vanishes for epsilon>0 and t>0, but it's not too difficult to show. So since Im(b)<0 then (k-b)^epsilon is holomorphic in the upper half plane where 1/(1+a^2*sin^2(k)) has poles at k_n=n*Pi + i*arcsinh(1/a). Summing over all these residues from n=-infinity..infinity.

restart; Digits := 15; a := sqrt(2); b := -I; e := 3/2; t0 := 1.312; f1 := Int(exp(I*k*t)/((1+a^2*sin(k)^2)*(k-b)^e), k = -infinity .. infinity); f11 := unapply(op(1, combine(IntegrationTools:-Change(f1, u = arctan(k), u))), t); f1 := proc (t) options operator, arrow; evalf(Int(f11(t), u = -(1/2)*Pi .. (1/2)*Pi, method = _d01ajc, epsilon = 0.1e-2, maxintervals = 10000000)) end proc; f2 := proc (t) options operator, arrow; evalf(Pi*(a/(1+sqrt(a^2+1)))^t*(sum(exp(I*Pi*t*n)/(n*Pi+I*arcsinh(1/a)-b)^e, n = -infinity .. infinity))/sqrt(a^2+1)) end proc; r := [f1(t0), f2(t0)]

ratio := r[1]/r[2]

15

 

2^(1/2)

 

-I

 

3/2

 

1.312

 

Int(exp(I*k*t)/((1+2*sin(k)^2)*(k+I)^(3/2)), k = -infinity .. infinity)

 

[-.348164061581795-.348164061581795*I, -.347637832603616-.347637832603616*I]

 

1.00151372758896+0.*I

(1)

``


 

Download Oscillating_integral_by_residue_theorem.mw

 

 

 

edit: BTW using this sum formula over the residues, it is easy to see (by summation by parts) that it converges for epsilon>0 and t not an even integer, because then the Dirichlet Kernel sum(exp(I*Pi*t*n),n=-N..N) is bounded as N -> infinity and sum( 1/(n*Pi+c)^epsilon - 1/((n+1)*Pi+c)^epsilon , n=-infinity..infinity) converges absolutely.

@vv I'm not sure what you try to tell me, but the integrand is not singular as in 1/x in your example, so I'm not sure why you compare these integrands?

 

Which conditions are not fulfilled concerning the residue theorem? The cut is not within the integration contour and when the integral converges and the result of the residue calculation is finite, I presume both results should be the same. In particular this is indeed what I find numerically for certain values, so can you clarify?

@vv If it helps you can transform the integral into a sum of the form

sum(exp(I*Pi*t*k)/(k+c)^(epsilon),k=-infinity..infinity)

where Im(c)>0 if Im(b)<0 using the residue theorem. (The arc in the upper plane should vanish when epsilon>0 and t>0)

Do you post the solution here if you uploaded it to the application center (where do I find it?)

I'm still suprised about the discontinuities at t=0 and t=2 for example. If Im(b)<0 there is no singularity on the axis, so why do you need to calculate the integral by CPV anyhow?

 

Apparently when t=2 in this sum you get 1^k=1 and a badly blowing up series if epsilon <= 1. So in that sense I'm surprised why it is at all finite (as you say as a CPV at t=2\IZ) just with discontinuities.

@vv Do you mind to add the proof or send me the notes? I'd like to see.

How do you see it diverges for t=2?

It doesn't even converge in the sense of CPV as in the case t=0?

@Axel Vogt I guess -Pi..Pi is enough, but you can as well assume t€R.

@vv I pressume you integrated the expression exp(I*k*t)/(1+a^2*sin(k)^2) by parts. Don't you need to know that the anti-derivative is for this is bounded?

@vv Well thanks, but as it happens by coincidence that for t=1 you have an analytic anti-derivative ;)

Or is there actually one for arbitrary t?

@Axel Vogt But int(exp(I*k*t)/(k+I),k=-infinity..infinity) should be defined too?

I should have never labeled 18, but version 17 I think which is the first I had. At work I have access to 2015 and at home to 2018.

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