I didn't check all the details, but you probably know that ?
Hi. I am trying to evaluate the following integral:
I =
where -1 < p < 1 and 0 < < 2*Pi. One can easily show the following:
When we transfer this over to the complex plane this means we have a branch point at x = 0 and a couple of simple poles at x = and x =
I am attempting to apply the pacman or keyhole contour here where the branch cut is made along the positive real axis.
I end up with the following:
A = = = I = I
But by the residue theorem we have:
A =
which implies:
I = =
but this cannot be right since we are integrating a real valued function over the real line... it shouldnt have any imaginary components. What have I done wrong?
I believe you're using the wrong residue for the pole at based on the branch you're using: You define:
Then for
Also, try it first for and or something easy like this. Do it numerically first in Mathematica, then do it numerically via the Residue Theorem:
Code:In[56]:= p = 1/4; t = Pi/4; NIntegrate[1/(x^p*(1 - 2*x*Cos[t] + x^2)), {x, 0, Infinity}] r1 = Exp[-7*Pi*(I/16)]/ (Exp[(-Pi)*(I/4)] - Exp[Pi*(I/4)]); r2 = Exp[(-Pi)*(I/16)]/(Exp[Pi*(I/4)] - Exp[(-Pi)*(I/4)]); N[((2*Pi*I)/(1 - Exp[-2*Pi*I*p]))* (r1 + r2)] Out[58]= 3.4907507252185646 Out[61]= 3.4907507252151033 + 3.487868498008632*^-16*I