201 
Ag A 
; 4nl | 
so=2fw (A) da + 2 fu (A) bin ah or 
rf je 
Jo=2f4 (m) dm + 2 fx (m) cos 42 lm dm a eee ee. , (9) 
+a st a 
Joe] yp (x) de + 2 g(x) cos 4nl (m Hede, 
= = 
Here 2, and 4, are the lengths of the smallest and of the largest of the 
waves present, hence m, and m, are the outmost frequencies on either 
side, m being the mean of them, so that we may put m=m +2. 
Further (xs) has been written for y (m + «). The region of 
frequency 2a of the beam is supposed very small by MicHeLson, 
so that he examines only a practically monochromatic beam. It 
should at once be pointed out that we shall not use this restriction 
in Our reasoning. 
m being a constant in the integration, the last equation of (3) 
can be put in the form: 
+a 
Jy = afs (w) dw + 2 cos 4x lm x C)—2sn4an lm « S(D). . (4) 
=e 
in this C(/) and S(/) are the following functions of 7: 
= @ 
+a 
C (U) =|» (x) cos An le dx, S (l) =| yp (a) sin4aledx, . (5) 
Our aim is to determine the function y(m), hence g(x). 
According to Fourtmr’s integral theorem we have generally: 
1 co Jo 1 oo + 
PD = = fe an de fy (8) cos Ea dS + af ux duly (S) sin Sa d5 . (6) 
7” n 
0 —o 0 —o 
fo 
+a 
In this fv (8) cos Eads may be replaced by fe (xv) cos 4m led, 
since g(x) for —a > > Ja is zero. 
+x ie 
Analogously | p(8) sin Sadi can be replaced by | op (a) sin An lada, 
— 09 0 
