... (z)=i√π2e−z2[1 −erfc(−iz)] . (6 .10. 2)A remarkable approximation for F (x), due to Rybicki[1],isF(z) = limh→01√πn odde−(z−nh)2n(6 .10. 3)What makes equation (6 .10. 3) unusual is that its ... byDover Publications, New York), Chapters 5 and 7.6 .10 Dawson’s IntegralDawson’s Integral F (x) is defined byF (x)=e−x2x0et2dt (6 .10. 1)The function can also be related to the complex ... -n*(n+1);b=Cadd(b,Complex(4.0,0.0));d=Cdiv(ONE,Cadd(RCmul(a,d),b)); Denominators cannot be zero.6 .10 Dawson’s Integral259Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-4 3108 -5)Copyright (C) 1988-1992 by Cambridge...