1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/Applications 723

y y

z=iy

-1 x

(a) (b)

Fig. 9.26

For instance, g 0 (i) = -.J2 while g 0 (-i) = +.J2. For z = x with x > 1,

we have 01 + 02 = 0, so

go(x) = i~ = iVx2="1

However, if z = x < -1, then 01 + 02 = 21l', and

go(x) = i~eirr = -iVx2="1

Now, if z = x with -1 < x < 1 we must distinguish two cases:

(a) The point is taken on the upper boundary of the slit. In this case

81 + 02 = 1l', and

go(x) = i~eirr/2 = -~

(b) The point is taken on the lower boundary. Then 81 + 02 = 31l', and

go(x) = i~ei3rr/2 = +~

Returning to the problem of evaluating the given integral, consider a

circle r+: z = Rei^8 , 0 :::; 0 :::; 21l', with R > 1, two small circles 'Yi:

z-i = rei^8 , 0:::; 8:::; 21l', 1!: z+i = rei^8 , 0:::; 8:::; 27r, with 0 < r < R-1,
around the poles of the integrand, and the contour>. depicted in Fig. 9.27,
consisting of the lower boundary or the cut from -1 + E to 1-E (0 < E < 1 ),
the circle /t: Z - 1 = EeiO, -?l' :::; 0 :::; 'ff, the upper boundary from 1 - E to
-1 + E, and finally, the circle 'Yt: z + 1 = Eei^8 , 0:::; 0:::; 27r. By the strong
form of the Cauchy-Goursat theorem, as applied to several contours with
fo(z) = 1/(1 + z^2 )go(z), we have

J fo(z)dz = J fo(z)dz+ J fo(z)dz+ [

1

1

~<< fo(x)dx

r+ 'Yt 'Yi