238 
MR. STOKES ON THE THEORY OF CERTAIN 
at M. It would not have been correct to integrate using the displa(;ement instead 
of the intensity, because the different colours cannot interfere. Suppose the angular 
extent, in the direction of x, of the system of diffraction bands which would be seen 
with homogeneous light, or at least the angular extent of the brighter part of the 
system, to be small compared with that of the spectrum. Then we may neglect the 
variations of B and of X in the integration, considering only those of | and f, and we 
may suppose the changes of § proportional to those of | ; and we may moreover sup- 
pose the limits of | to be — oo and -f-oo . Let §' be the value of §, and —nr that of 
when |=y, so that we may put §z=z^'-\-vr(p' — ^) ; and take/? instead of | for the in- 
dependent variable. Then putting for shortness 
^ nr-^^{4g+h+k)=g. 
( 13 .) 
we have for the intensity. 
I: 
dp 
f ^00 
J ^ {sin2 A^p-l-sin2 k,p-\-^ sin h,p . sin k,p . cos {§'—g,p) 
Now sin^ =/?,^y^^sin2 ^ Similarly,^y^^sin2 p 
Moreover, if we replace 
cos {^—g,p) by cos ^ . cos gp+ sin . sin gp, 
the integral containing sin will disappear, because the positive and negative elements 
will destroy each other, and we have only to find w, where 
sin h(p . sin kp . cos gp . 
Now we get by differentiating under the integral sign, 
dw /*“ . , ... dp 
j- =-y_^ sm hp . sin .sin^^p . 
P 
1 f . 
But it is well known that 
/ 
» sm sp 
00 p 
dp=‘7r, or = — !T, 
according as s is positive or negative. If then we use F(^) to denote a discontinuous 
function of s which is equal to -f- 1 or — 1 according as s is positive or negative, we 
get 
This equation gives 
=0, from g,= -oo to g,= -{h,+k,) 
