702
xand equating real and imaginary parts, we find that
1
(^00) cos ax 7r •
(PV) 2 2 dx = - - smap
-oo x - p p
(PV) loo sin ax dx = 0
-oo x2 - p2
Chapter 9
(9.11-24)
(9.11-25)
If p = 1r /2a, then 7r /2a and -rr /2a are no longer poles but removable
singularities of cos ax/ ( x^2 - p^2 ) since
cos ax -a sm ax a^2
lim = lim =--
x->±Tr/2a x^2 - (rr^2 /4a^2 ) x-+±Tr/2a 2x 7rThus (9.11-24) is only an improper integral of the first kind which converges
in the ordinary sense since 1/(x^2 - p^2 ) = 0(1/x^2 ). Therefore, the PV
notation can be dropped, and we may write
or
-----= -2a
1
(^00) cosaxdx
-oo x2 - ( rr2 /4a2)
[
00
cos ax dx =
Jo 4a2x2 - rr2
1
4a
the integrand being even.
- To evaluate fo
00
(sinx/x)dx. Consider Ic+(eiZ/z)dz, where c+ is
the contour shown in Fig. 9.19, namely, c+ = [-R, -r] + ,-+ [r, R] + r+,
0 < r < R. The indentation about the origin is made in order to avoid
the simple pole z = 0 of the integrand. Since no singularities are enclosed
by c+' we have
1
-r eix dx + J eiz dz+ 1R eix dx + J eiz dz = 0 (9.11-26)
-R X z r X z
-y- r+