@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.

@acer Thanks. Why can I then use multiple arguments in ?collect() without getting an error?

Try:

restart;

f := simplify(GAMMA(2*q-2)*GAMMA(-2*q+L)*hypergeom([-L+1, 2*q-2], [-L+2], 1)/((2*(L-1))*(-1+2*q)*(4*q-3)*GAMMA(L+2*q-3))+GAMMA(-2*q+L)*GAMMA(2*q-2+L)*hypergeom([1, 1, 2*q-2+L], [2, L+1], 1)/((2*(-1+2*q))*(4*q-3)*GAMMA(L+2*q-3)*GAMMA(L+1)));

collect(f, hypergeom, simplify);

collect(f, hypergeom, simplify, factor)

@Kitonum If I have a list with function expressions how do I use frontend ?

o:=[f1,f2,f3]

frontend~(expand,o) or frontend~(expand,[o])

does not work.

@acer 

I can write

collect(f,[k,GAMMA],simplify,combine)

and each operand is simplified and combined separately. I always assumed that this is done in the order specified so in this case first the simplification and then the combination. In essence this is not precisely equivalent to what you are showing me:

collect(numer(f),[k,GAMMA],u->u/denom(f),simplify)

I expected to first collect the numerator, then do u/denom(f) for each operand and for each operand the result is simplified. In fact doing this above the denominator is completely lacking.

@acer What goes wrong with this map u->u/denom(f) when I add additional arguments such as ?simplify() ?

I would have expected that each term is simplified seprately as is usually the case.

@Mariusz Iwaniuk Ah yeah right. Could have thought about it. I think this is probably even intended, while still confusing. I think it should be either 0 or 4 defaultwise with the option for both.

I still have a similar problem with this where the "convert" trick doesn't work

restart;

((1-t)^2)^q;

`assuming`([combine(simplify(%))], [1-t > 0, q > 0, q < 1])

@Carl Love Since q € (0,1) why is it only true for t € (0,Pi) ? The domain (-Pi,0) is then also mapped to (-Pi*q,0) which is not in the upper half of the domain and thus distinct.

@acer What do you mean by "fortuitous Change and Split might allow resolution of the `abs`" ?

Are you talking about abs(cos(theta1)+cos(theta2)) in the denominator of the arctan of your reintegrated Q ?

Do you think this integral can be actually done analytically?

@vv Yes the precise value is expected to be 1/8 (see putnams 1992 A6)

It's a straight-forward but not very elegant way of solving it...