I could be wrong you know Pascal. Let's see then: I can draw the contour free-hand in Mathematica (see below) and "capture" the points by just doing a cut-and-paste then assigning the paste to some variable name which I called mycontourpts. I now have all the points (crudely) making up the contour. I can then extract those points and calculate interpolation functions for the x and y values and even calculate their derivatives. I then set $\displaystyle z(t)=x(t)+iy(t)$ and then calculate numerically the following integral:

$\displaystyle \frac{1}{2\pi i}\mathop\oint\limits_{\text{mycontourpts}}\frac{x '(t)+i y'(t)}{z(z^2-1)}\biggr|_{z=x(t)+iy(t)}$

This is the Mathematica code I used (minus the capture of mycontourpts):

Code:

lns = Cases[Normal[First[mycontourpts]],
Line[pts_] -> pts, {0, Infinity}];
myvals = First[lns];
myxval = (#1[[1]] & ) /@ myvals;
myyval = (#1[[2]] & ) /@ myvals;
myx = ListInterpolation[myxval]
myxd[t_] = D[myx[t], t]
myy = ListInterpolation[myyval]
myyd[t_] = D[myy[t], t]
i1 = NIntegrate[(myxd[t] + I*myyd[t])/
(z*(z^2 - 1)) /.
z -> myx[t] + I*myy[t], {t, 1, 989},
WorkingPrecision -> 25]
N[i1/(2*Pi*I)]

The result is $\displaystyle 1.49633 - 0.0043122 I$

It's kinda' close to 3/2 with a little bit of imaginary part that could be due to the crudeness of the method. Really, I would have liked to get it closer or else I made a mistake. I'm thinkin' another B at best here.