Axel Vogt

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16 years, 210 days
Munich, Germany

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These are replies submitted by Axel Vogt

@digerdiga what are your ranges for t?

@vv for correcting me

The integral is (bounded and oscillating) * 1/k, so it does not converge

@vv 

I meant the limit over Gamma*exp (after partial integration w.r.t. Gamma(a,s)), 0<a,0<b.

That works:

limit(-1/b*exp(-b*s) * GAMMA(a,s),s = infinity);
eval(%, limit=MultiSeries:-limit);

                                 exp(-b s) GAMMA(a, s)
                     lim       - ---------------------
                s -> infinity              b

                                  0

While that does not work:

MultiSeries:-limit(exp(-b*s) * GAMMA(a,s),s = 0,right);

@vv 

it seems that "limit" does not always know or use https://fr.wikipedia.org/wiki/Limite_(math%C3%A9matiques)#Limite_et_op%C3%A9rations_alg%C3%A9briques

@maple2015 

Using t = 12^(1/4)*omega^(1/2) you can estimate your zeros by fsolve( cos(t)*(cos(t)-sin(t)), t=desiredGuess), which has an approximate symbolic presentation (but needs a cleanup like shown by Christian Wolinski, evalc shows your function is real valued).

In you case tGuess = 5/4*Pi ~ 3.9269... or omegaGuess = 25/192*Pi^2*12^(1/2) ~ 4.45...

A "proof" for the zero to be minimal is through RootFinding:-Analytic (needs a cleanup for the original task).

@vv 

At least a weakness  and that is (probably) the reason for acer's version (division by zero)

It is not the integrand per se, using the series for MeijerG it can be seen that the integrand is

(-3*polylog(4,x^Pi)+Pi*ln(x)*polylog(3,x^Pi))/Pi^4/(-1+x)^2

 

@acer , it was vv, not me ...

yours is better, indeed

@Markiyan Hirnyk 

It is (a kind of) averaging over the singularity s. More formally split in s and use a change of variable to bring the left side to the right, cancelling the singularity. While for CPV one can not use change of variables in general this one is linear. I do not have a precise Lemma at hand.

f:= t -> (abs(t-(1/6)*Pi)+1)*exp(-sec(t))*cos(t)/(-1/4+sin(t)^2)+1/sin(t-(1/6)*Pi)^5;

'Int(f(t),t = 0 .. Pi/2, CPV)';
``='Int(f(t),t = 0 .. 1/3*Pi, CPV)+Int(f(t),t = 1/3*Pi .. 1/2*Pi)';
``='Int(f(1/6*Pi+t)+f(1/6*Pi-t),t = 0 .. 1/6*Pi)+Int(f(t),t = 1/3*Pi .. 1/2*Pi)';
evalf[30](%); # increase because of numerical singularity, now in t=0
                                        Pi
                    Int(f(t), t = 0 .. ----, CPV)
                                        2

 

                   = 3.80259236620486251646441207770

 

@Markiyan Hirnyk so what the numerical "CPV" values for the integral over Pi/6 - 1/2 ... Pi/6 + 1/2?

(the 2nd example would not work in Pi/2, but that's no so important)

@vv 

It reminds me of needing 15+3 decimals to represent a double precision

Increase precision. For example begin with restart; Digits:=40; and finally use SK0_SUM; SKA_SUM; %-%%; which gives 0.

 

Which means: the first 40 decimals are the same.

 

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