a=0 is not in your domain

@Robert Israel 

Thank you - do you have a version as plain worksheet as well, containing meaning visible code instead of sliders?

@tomleslie Thank you, now I see

@tomleslie 

Neat!

I wanted to scale it to have N = 1 = 100%, but the solutions do not scale - how to do that in your setting?

@reza gugheri 

1. You should consult Maple's help about hypergeom( [n1, n2], [d1], z) - the input for the parameters n1,n2,d1 as list is a convention only.

2. You also could search a bit on the web, https://en.wikipedia.org/wiki/Hypergeometric_function (not all languages have the same info), the case is often denoted by 2F1. Roughly this stands for a powerseries in z = cos(x)^2

@Preben Alsholm : You are right, thank you.

evalf[2](1000 - 1); # preferred notation
                                1000.

evalf(1000 - 1, 2);
                                1000.

NB: For me the following is a bug:

evalf[2](1014 - 1007);
                                  7.

neat - frohe Feiertage!

@Mariusz Iwaniuk : or in Maple notation you just forgot the minus sign of your trick

with(inttrans):
'fourier(diff(F(x), x)*I/k, x, k)':
'%' = %;

                  /d      \
                  |-- F(x)| I
                  \dx     /
          fourier(-----------, x, k) = - fourier(F(x), x, k)
                       k

@Mariusz Iwaniuk 

I think that Maple's sign is correct, for k=1 the numerical answer is  - 1.5575878673834*I (one can use MMA online, which does not know the symbolic answer, but finds a numerical result)

What do you mean by that - can you give a specific example for that?

a=0.573179148515927e-1, v = 27.0772299601075 or
a=9.54497166461128*10^14, v = .811477602357159

seem to be the only solutions

I have Maple 2017.3 32 bit and get similar results (5 times faster, 10 times for accuracy)

Neat :-)

I once used William T. Shaw, Nick Brickman, http://arxiv.org/abs/0901.0638

@Rouben Rostamian  

Yes, simplifying diff(solution)/integrand gives 1.

Note that d=1 (see above) and a=1 (using cNew= c/a instead) can be assumed [and perhaps some of the assumptions can be weakend]