by a Semi-infinite Screen with a Straight Edge. 671 

 Now 



I £ COS- + 77 Sill a ) = ^(7' + t f COS a+j/ Sill «), 



i. e. mi = Jr cos 2 J(0 — a) where a-\-iy = re i ; 

 and similarly 



&> 2 2 = ^7'COS 2 i(^-}- a). 



Using these results and also writing 2k£ 2 = u 2 , we get, 

 finally 



• • • (M) 

 giving the well-known solution for the diffraction of an 

 harmonic wave train. 



§4. To deal with the diffraction of a disturbance of 

 arbitrary wave form, it will be convenient to transform 

 the integrals so that the limits are constants. Using (12), 

 we have 



x/^'^^df^ra^^f^ 1 ^ . (15) 

 V wt^o wrJo Jo 



Now write w = £tan ijr, and we get 



— 1 tig . \ e * Y £ sec- \jr d\jr . (o^ +ve).. (16) 

 117 Jo Jo 



Changing the order of integration, this becomes 



i*f%fV* (Ssee ^i<*(r S ec*)? . . (17) 



771 Jo Jo 



= _1 p^savsec^^i)^ ...... (18) 



7T 



= \^- ( 2 e m ^^ 2 ^df (19) 



* ^ Jo 



If gjj is negative, the upper limit of integration becomes 

 — -, which is equivalent to changing the sign throughout 

 (16)-(19). After ihis is applied to (14) the solution 



