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



present case a will be 7r/2 and the most important part of the 

 integral will be near x = ir/2. To get W x for v = A and both large, 

 we may integrate from lir — fa to \it, fa being a small finite angle. 

 Put x = 7r/2 — <f> where <f> is small. 



or cb cd 



Then cos x — log cot = = — ^- — ^r + &c, 



r* 1 Ad> s ( 3 \» r°° 



and W 1 =\ cos — ^- c?^> = ( -j J x| cos -^r 3 dty (74). 



And cZ Tfj/cfo = — I cos <r sin (wcosa? — J. log cot ^ J dx 



= 1 (jjsm-^- dc})= sm-^- x ^- = ^i-jj x J smx^dx 



(75). 



We have next to find 



/• 00 /.00 



I cos ^dyjr and I sin a$ ^ 

 Jo Jo 



By Art. 260, Todhunter's Integral Calculus (1857), it can be 

 shewn that 



/, 



8 , -i . is -2«-2 



1 + * 



We may pat A; = — ^- whence A; 2 = i, and 



f 



Jo 



3 -, -. . -, -2m-2 







whence 



3 3 





-i y n+\ „ j.. _ » T J- n / n T ± \ w / J- T t \ 3 



/;( 



cos y n+1 — i sin 2/ w+1 ) dy 



■ T • 1 -1 \ / \ -2m -2 



r — — . cos-7+ism- 6 (7b). 



3 V 3 J'\ 4 4 



First put n = 0. It will be found that the function of it on the 

 right hand will take three possible values, which are 



(1) cos g -i sing, (2) cos y -ism — , (3) -cos-- 1 sin ^ 



(.00 



As I sin y 3 dy is finite and positive, (2) must be rejected, and we 



J 

 have to discriminate between (1) and (3). 



