Originally Posted by

**shawsend** After checking it, I think this should be:

Here's a numeric check of this relationship with n=6 (because this sort of thing is new for me too). Remember, we're going around twice in the second integral so I just multiplied it by two and went around once:

Code:

In[65]:= NIntegrate[(u^(2*n)/(u + 1)^(n + 1))*I*
Exp[I*t] /. {n -> 6,
u -> -1 + Exp[I*t]}, {t, 0, 2*Pi}]
N[2*Pi*I*((-1)^n(2*n)!/n!^2) /. n -> 6]
NIntegrate[((z^2 - 1)^(2*n)/z^(2*n + 1))*
I*Exp[I*t] /. {n -> 6,
z -> Exp[I*t]}, {t, 0, 2*Pi}]
Out[65]= -9.379164112033322*^-13 +
5805.66322740536*I
Out[66]= 0. + 5805.663223833938*I
Out[67]= -1.3515644212702682*^-6 +
5805.663226625173*I