contour integration example 1

Contour integration example 1 PhL 6.18.10 Consider this integral I = !Syntax Error, Idθ / α ≥ 1 I know we can look this up, but I want to understand this integral as a contour integral, that is the subject of this doc. The usual steps are these z = Reiθ dz = izdθ dθ = (1/i) dz/z z/R + R/z = eiθ + e-iθ = 2 cosθ cosθ = (1/2) (z/R + R/z) = (z2+R2)/(2Rz) I = (1/i) dz/z (1/) = -i dz/z (1/) = -i dz(1/z) (2Rz)1/2 (1/) = -i (2R)1/2dz (1/) (1/) Examine the argument of the radical f(z) = 2αRz - z2 - R2 = (-1)(z2-2αRz + R2) = (-1)(z-z1)(z-z2) = (z1-z)(z-z2) B2-4AC = 4α2R2 - 4R2 = 4R2(α2-1) z1,2 = αR ±R = R( α ± ) z1 = the + I claim that [ α + ] is ≥ 1 = 1 when α = 1 [ α - ] is ≤ 1 = 1 when α = 1 So we now have I = -i (2R)1/2dz (1/) (1/) Now, the whole exercise here is to try to understand what this looks like in the z plane and how to interpret how it looks. Question: what "sheets" does this circular contour lie on? I guess we realize that we can redraw the cuts like this and then there is no "mystery" as to which sheet the contour is on. That is to say, at least the contour is all on one sheet, it does not cross any cuts in some strange way. We can then shrink the contour around the little segment shown. Then write for points above the cut z-z2 = |z-z2| eiπ => (z-z2)-1/2 = (z2-z)-1/2 e-iπ/2 = -i (z2-z)-1/2 So the disc is this above - below = -i (z2-z)-1/2 - [+i (z2-z)-1/2] = -2i (z2-z)-1/2 Then we have I = -i (2R)1/2dz z-1/2 (z1-z)-1/2 (z-z2)-1/2 = (-i ) (2R)1/2!Syntax Error, Idx x-1/2 (z1-x)-1/2[-2i (z2-x)-1/2] = 2i2 (2R)1/2 !Syntax Error, Idx x-1/2 (z1-x)-1/2(z2-x)-1/2 The integral is now of this form ∫ dx (1/R) where R is a quadratic and we can then use page 97 GR7 and we could then finish off the integral this way. When I started this doc, I thought I was going to have the circular contour stuck crossing some cut, so that the integral did not even make sense. But that did not happen.