digerdiga

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10 years, 126 days

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These are replies submitted by digerdiga

@acer Thank you. Can you also try the second example "for general a"?

So is it working now with 2019?

@vv Nonono...The answer was not useless since you reminded me about algcurves:-puiseux

I just said that I also "solved" it "manually", and apart from asympt/series/MultiSeries I'm not very familiar with other methods of which there are algcurves:-puiseux , but also else?

@vv That's how I did also solve it, but I thought there must have been some build in function as well...

 

Btw: What do you need the [] in alias(alpha=indets(a,RootOf)[]):  for ?

 

If I have

restart;
f:=RootOf(x^2+1,x)+RootOf(x^2+2,x)*RootOf(x^2+x+3,x);
alias(alpha=indets(f,RootOf)[1]);
evalf(alpha)

 

I can use it to select an operand, but in your case there is only one, so it should be obsolete or?

@vv 

Actually thanks...I completely forgot about algcurves:-puiseux

But do you know why series/asympt has issues with the zeros in rootof?

@Christian Wolinski Not sure what G[1]('%', indets('%', t));  is supposed to do. It's not doing anything for me.

@Kitonum Thanks first, but what if

f:=ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln(2*(1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))

 

?

 

I see that I can first do

op(f)

and then proceed termwise:
 

@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?

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