Hi simran,
Yes, the 2F1 expression does consider the pole, as you can see by the fact that the expression gets large as z appoaches 1 from below. I am going to consider 1/(z^n-1) rather than 1/(1-z^n) since it's easier to keep track of the nth roots of unity on the unit circle. It's easy enough to multiply the solution by -1 after the calculations are done. Also, n is odd here..
The solution for |z| < 1 is found by expanding the denominator in powers of z and integrating term by term, as mentioned by Torsten. I'm kind of surprised that syms got that far, but it's helpful. For |z| >1 you can expand z^(-n)/(1-z^(-n)) and do the same thing. This results in
y = -z*2F1(1,1/n;1+1/n,z^n) |z| < 1
y = z*(2F1(1,-1/n;1-1/n,z^(-n)) -1) |z| > 1
Here y(0) = 0, y(z) -> 0 as z-> +-inf and since the integral from -inf to inf is nonzero, constants of integration have to be added to these solutions to make them match up. syms does not seem to be capable of doing definite integrals in this case, but calculations using the log expression (see below) give
n = 5;
% some definite integrals
% Int(-inf,inf) 1/(z^n-1) = -C
% Int(-inf,0) 1/(z^n-1) = -D
% Int(0,inf) 1/(z^n-1) = -E
a = exp(2*pi*i/n); % poles at z=a^m, for 0<m<n-1 pole at z=1, no pole at z=-1
sqa = sqrt(a);
% use real to eliminate very small nuisance imag values
C = real( (-i*pi/n)*(1+sqa)/(1-sqa) ) % >0
D = real( (-2*i*pi/n)*sqa/(1-a) ) % >0
E = real( (-i*pi/n)*(1+a)/(1-a) ) % >0
% calculate and plot
z1 = -10:.01:-1;
z2 = -1:.01:.99;
z3 = 1.01:.01:10;
H1 = z1.*(hypergeom([1 -1/n],1-1/n,z1.^(-n))-1);
H2 = -z2.*hypergeom([1 1/n],1+1/n,z2.^n);
H3 = z3.*(hypergeom([1 -1/n],1-1/n,z3.^(-n))-1);
H1 = H1 - H1(end) + H2(1); % match at z = -1
H3 = H3 - E; % effectively match at z = inf
figure(1)
plot(z1,H1,z2,H2,z3,H3)
grid on
You can see the effect of the pole at z = 1. The pole can be integrated across by using the so-called principal value (which is how E was determined).
The 2F1 function that syms came up with was pretty good, but a more useful expression is
1/(z^n-1) = (1/n) Sum{0,n-1} a^m/(z-a^m)
and integrating this gives the indefinite integral
y = Int 1/(z^n-1) = (1/n) Sum{0,n-1} a^m log(z-a^m)
since log(p*q) = log(p) + log(q), the argument of each log term can be divided by -a^m and the only consequence is introduction of some constants of integration. This results in
y = Int 1/(z^n-1) = (1/n) Sum{0,n-1} a^m log(1-z/a^m)
which equals 0 at z=0, just as the 2F1 expression does. Now this has to be modifed for the m=0 (pole at z = a^0 = 1) term to use the principal value, log(|1-z|) instead of log(1-z):
y = Int 1/(z^n-1) = (1/n) Sum{1,n-1} a^m log(1-z/a^m) + (1/n) log(|1-z|)
and the function can be calculated in one go, rather than piecewise.
mm = (1:n-1)';
z4 = [z1 z2 z3];
[z m] = meshgrid(z4,mm);
% again remove small nuisance imag
A4 = real((1/n)*(sum(a.^m.*log(1-z./a.^m)))) + (1/n)*log(abs(1-z4));
A plot of z4,A4 is identical to the one above.
