144 Mr Sharpe, On the Reflection of Sound at a Paraboloid. 

 Putting these values in (111) we get 



/•OO It A 



I i =\ cos (2x 2 + A log x) dx = 2-1 e~ ir (p 2 + q 2 )*cos(E+ B) 

 Jo 



p OO IT A 



1 6 = I sin (2a; 2 + A log x)dx = - 2~* e _T (p 2 + q 2 )* sin (E + B)\ 



Jo ) 



(117). 



By means of (112) we thus get a 3rd corroboration of Stokes's 

 formula (106). It will be observed that the formulae (117) for 

 I 5 and I 6 are of value only when A does not exceed 1, for if A be 

 > 1 we see from (116) that the series for E is divergent. 



We will next examine what results we get when in the above 

 investigations — A is put for A, for the Physical Problem 

 requires both of these cases to be considered. Suppose I s , I 6 

 then become I 7 and I s respectively. It will be found that instead 

 of (111) we shall get 



,-ao tfA 



1 7 = I cos (2a; 2 — A log x) dx = 2~^ e 4 (— q sin B 1 + p cos Bj, 



Jo 



«ao irA 



1 8 = I sin (2x 2 — A log x) dx = 2 -i e 4 (— q cos Bj — p sin Bj), 



Jo 



where B x = — A log 2* — j , 



and p and ^ have the same values as before. 

 Instead of (117) we shall get 



/. oo irA 



I 7 = cos (2x 2 - A log x)dx = - 2~* e" 4 " (/>* + q 2 f sin (# + 5) 



Jo 



poo ■jrA 



I e = I sin (2x 2 -A log a) da; = 2"* e 4 (p 2 + q 2 )* cos (# + B)\ 



(H8X 



where B, E and {p 2 + q^ are known from (110), (116), and (112). 

 It can be easily shown that the limit of (i\ — | log r) when r 



is oo is !y + loge 2 (7 being Euler's constant and = "5772, &c), so 

 that instead of (116) we may put 



E = A(±y+\og2)-^r s + ^r 5 -Yr 7 + &c (119). 



It must be remembered in using the formulae (117) and (118) 

 that in them A ;J> 1. 



44. We are now in a position to complete the case of A = 0. 

 We see from (103) and (105) that in order to find what X x and X 2 



