1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

11 .9 • THE SCHWARZ-CHRlSTOFFEL TRANSFORMATION 489

J dz 1 = iarcsinz.

(z^2 - l)'

J dz i = log ( z + ( z2 - 1) t) -i;.

(z^2 - l)"

dz

J - 2 --

1

= arctan z.

z +

I

...!!:=_= ~lo i+z.

z^2 +1 2 g i-z

J

dz. 1

---~ = -arcsm -.

z(z2-l)t z

J dz 1 =ilog. [1 ( -+^1 ) ~]

z(z2 - IP z^2 z - 1.

dz 1

J 1 = -2arctanh (z + l)".

z(z+l)~

J dz =logl-(z+l)!.

z(z+l)! l+(z+l)^2

J (l-z2)~ dz=! [z(l -z^2 )~ +arcsinz).

J ( 1 -z^2 )! dz = ~ [ z ( z^2 - 1) t + log ( z + ( z^2 - 1) ~)).

Table 11.2 Indefinite integrals.

Equation (11-40) gives a representation for f in terms of an indefinite inte-
gral. Note that these integrals do not represent elementary functions unless the
image is an infinite region. Also, the integral will involve a multivalued function,
and we must select a specific branch to fit the boundary values specified in the
problem. Table 11.2 is useful for our purposes.

• EXAMPLE 11.26 Use the Schwarz- Christoffel formula to verify that the

function w = f (z) = Arcsinz maps the upper half-plane Im (z) > 0 onto the

semi-infinite strip -
2
7r < u < ~· v > 0 shown in Figure 11.72.
-'Ir 7r 7r
Solution If we choose x 1 = - 1, X2 = 1, W1 = 2, and w2 = 2, then °'1 = 2
7r.
and a2 = '2, and Equation ( 11-39) for f ' ( z) becomes

!' (z) = A(z + 1 )-("/^2 )/" (z-1)- ("/^2 )/,,. = A.
(z2 - l)t

Then, using Table 11.2, the indefinite integral becomes

f (z) = AiArcsinz + B.