X-Rays from Imperfect Crystals. 815 



enough cothme will tend to unity. This will bring the 

 denominator to the form it had in D.ii., viz. : 



\ika cos 6 . u + i\/ (ka cos 6 . u) 2 —q 2 \ 2 . 



But if we put 7i = before allowing m to become large, the 

 corresponding expression is 



\ika cos 6 . u + \/(ka cos 6 .u) 2 — q 2 cot m\/(ka cos 6 . u) 2 — q' 2 \ 2 . 



The integral still converges, but to a different value. To 

 avoid this difficulty we must suppose that the crystal is so 

 thin that mh is small — it must be less than about 10 5 layers 

 thick. For such a crystal the real part of e will not matter 

 and we may put h = and write the relations (6*4) straight 

 into ($'&)- In spite of its unpromising appearance the inte- 

 gral can b9 evaluated and leads to the remarkably simple 

 result * 



JW^HToP^^Trtanh^. . . (6-7) 



— cc 



If in this we allow m to become infinite the result is 



I T 1 2 7r -^ — - — - n . whereas the true value from D.ii. should be 

 1 ' ka cos 6 



l^o| 2 o7T~^ — d' so even ^ n tms ex treme case the error is 



O rCCl COS C7 



only 18 per cent, for taking h zero before putting m infinite. 

 This shows that the approximation may be expected to hold 

 for quite deep crystals with considerable accuracy. 



Now take the same problem and work it out on the 

 principles of § 4. It is easily found that 



. m 1 — exp — 2mika cos 6 . u 



b = — iq 1 ^rr- ^ 



1 2ika cos . u 



which leads to the result 



r 



\^\ 2 du = \T \ 2 T -^—, 7 r.mq. . . (6'8) 

 11 ' ' ka cos 6 * v ' 



* This result was first discovered by obtaining an expansion in terms 

 of viq. The complete proof may be constructed as follows. "Write the 

 denominator as the product of two conjugate imaginaries and split it into 

 partial fractions. Next express it as a complex integral with argument 

 e. . . . necessitating a cut "between +q in the e-plane. It may then be 

 proved that the poles of either fraction in the integrand lie entirely on 

 one or other side of the path of integration. Hence the path may be 

 replaced by a circle at infinity which contributes nothing, together with 

 a small circle round e = q which introduces the hyperbolic tangent. 



