James Henderson 
To find the value of the integral on the right-hand side, consider the term 
Cs (— Dy in the function cf) (— D). Its contribution to the integral is 
where 6' = 6 ^f]). 
On integrating by parts the term between limits vanishes owing to the factor 
e-iz^_ Hence the integral 
and ultimately 
J -QO V Ztt 
J — 00 
dz 
IT 
dz 
V27r 
Therefore the whole integral on the right is 
i.e. . e-J'« ( 1 - = '■Jf cf) ( ^/pd) eii'^\ 
or (/) (\/p9) = e-i'fl-^^'f- ( 1 - 0)-i\ 
and (f){'^p9) = Co + Ci{\^pO)-i- c.i'Jpdy + ... +c,{'JpdY + ... 
^co + c,'0 + c.:6'+ +c;e'+..., 
where = (Vp)* or c, = c/ (Vjj)-". 
Now e-pe-^pe'^(l _ 
= g-?)9-^/)9--/>loga-e) 
= e -/ifl - ipfl- + iiJ^^ + + ipeH ■ . . 
= h, + b,e + b,e' + h,0' + h,e' + ..., 
where 
b, = l, b, = h, = 0, b,^ ip, b, = Ip, br, = Ip, h = ii/> + k W -=i^PiP + 3), etc. 
But v'jJC,,' = bn, therefore 
Co = ^ , c/ = c.; = 0, c; = ^Vp, c/ = c,' = iVj), etc. 
so that Co = ~ , Ci = c, = 0, c-j = X - , c^ = |- —'7= , c, = ^ -„ , etc. 
vp ' P 2) Vp P' 
For numerical purposes these coefficients are much more usefully obtained in 
the following way: 
Let e-"'-ip'' {l-0)-p = bo + he + b£- + ... etc. 
Take the differential of the logarithms of both sides ; then 
{-p-pe + pjl-O) (b, + b,d + M-' + ■ . ■ + bfi^ +...) = 6i + 2/; J + . . . + shO'-' + . . . , 
i.e. pd-' {b,+ b,e+ ...+b,e'+ ...) = {l-e) (b, + 2b.fi + . .. + sb.e^-' +...). 
