Skip to main content
180 ON THE PASSAGE OF WAVES THROUGH [375
in which cos 6 cos a passes through zero within the range of integration. It will be shown that (84) vanishes ultimately when x' = 0. To, this end the range of integration is divided into three parts: from 0 to «1} where «! < a, from ax to a2, where «2 > a> and lastly from a2 to TT. In evaluating the first and third parts we may put x 0 at once. And if z tan \Q
dO I t( dz dz
f cos 6 cos a sin a ] (tan ^a + z tan £« Sin a being omitted, the first and third parts together are thus
. z + t , t + tl, tz~t
log----- 4- log -- + log r-7,
where t = tan -^a, ia = tan £ala £2 = tan ^<x2, and z is to be made infinite.
It appears that the two parts taken together vanish, provided tl} L are so chosen that i? tJ2.
It remains to consider the second part, viz.,
fa* d6 (cos Q ~ cos a)
I (cos ^^cos~aFT«'2' "*.....................^ '
* &i \ y
in which we may suppose the range of integration «2 «i to be very small. Thus
sin2 a («2 a)2 + x'n-
~ 2 sin a sin2 a (a - a^f + x'2 '
and this also vanishes if az~- a a a.l, a condition consistent with the former to the required approximation. We infer that in (83)
so that, with the aid of a suitable multiplier, (73) can be satisfied. Thus if XF = J.V(6S - 2/2)> (73) gives A = ikjir, and the introduction of this into (74) gives (78). We have now to find what departure from (86) is entailed when kb is no longer very small.
Since, in general,
d*D/da? + d*D/dyz + /clD = 0, we find, as in (81),
and for the present NP" has the value defined in (82). The first term on the right of (87) may be treated in the same way as (28) of the former problem, the difference being that V(&2 - 2/2) occurs now in the numerator instead ofe integral on the right in (83) the first yields TT when x = 0. For the second we have to consider