﻿£ 
  

  

  366 
  Prof. 
  G. 
  N. 
  Watson 
  on 
  Bessel 
  Functions 
  

  

  In 
  the 
  last 
  integral 
  of* 
  all, 
  we 
  deform 
  the 
  contour 
  into 
  a 
  

   circular 
  arc 
  of 
  indefinitely 
  great 
  radius, 
  starting 
  and 
  ending 
  

   at 
  — 
  oo. 
  Since 
  n 
  is 
  supposed 
  to 
  be 
  positive 
  the 
  integrand 
  

   is 
  0(£~ 
  2 
  ) 
  on 
  the 
  deformed 
  contour, 
  and 
  so 
  the 
  integral 
  is 
  zero. 
  

   We 
  thus 
  obtain 
  the 
  formula 
  

  

  J^^X^U^^-l}^. 
  (2) 
  

  

  Now 
  that 
  the 
  large 
  variable 
  n 
  only 
  occurs 
  in 
  a 
  single 
  

   term 
  of 
  the 
  integrand, 
  we 
  proceed 
  to 
  apply 
  the 
  methods 
  of 
  

   Debye 
  by 
  choosing 
  a 
  contour 
  on 
  which 
  

  

  i(*-l/*)-log* 
  

   is 
  purely 
  real. 
  

  

  Writing 
  t 
  = 
  re 
  ie 
  , 
  we 
  see 
  that 
  the 
  contour 
  has 
  to 
  satisfy 
  the 
  

   condition 
  (r+1/r) 
  gin 
  0_ 
  2 
  = 
  o 
  . 
  

  

  this 
  equation 
  is 
  satisfied 
  if 
  = 
  or 
  if 
  

  

  r=0cosec0{l 
  + 
  v/(l-0- 
  2 
  sin 
  2 
  0}. 
  

  

  Taking 
  the 
  upper 
  sign, 
  so 
  that 
  

  

  r^cosecflU+v^l-^sin 
  2 
  ^)}, 
  . 
  . 
  (3) 
  

  

  we 
  obtain 
  a 
  contour 
  of 
  the 
  required 
  type 
  if 
  we 
  take 
  6 
  to 
  

   vary 
  from 
  — 
  it 
  to 
  it. 
  

  

  The 
  contour, 
  which 
  is 
  symmetrical 
  with 
  respect 
  to 
  the 
  

   real 
  axis, 
  passes 
  through 
  (1, 
  0) 
  and 
  has 
  an 
  abrupt 
  change 
  of 
  

   direction 
  at 
  that 
  point 
  ; 
  its 
  direction 
  immediately 
  above 
  the 
  

   real 
  axis 
  is 
  inclined 
  ^ir 
  to 
  the 
  positive 
  direction 
  of 
  the 
  real 
  

   axis 
  *. 
  

  

  We 
  thus 
  get 
  

  

  where 
  r 
  is 
  given 
  as 
  a 
  function 
  of 
  6 
  by 
  equation 
  (3), 
  t 
  = 
  re 
  ie 
  

   and 
  f 
  

  

  F(0) 
  = 
  i(r 
  - 
  1/r) 
  cos 
  6 
  - 
  log 
  r. 
  

  

  Now 
  . 
  i 
  

  

  t 
  2 
  - 
  1 
  (r 
  — 
  1/r) 
  cos 
  + 
  i(r 
  + 
  1/r) 
  sin 
  

   _ 
  (r 
  — 
  1/r) 
  cos 
  6 
  — 
  i(r 
  + 
  1/r) 
  sin 
  6 
  

   "' 
  r 
  2 
  + 
  1/r 
  2 
  -2 
  cos 
  26 
  ' 
  

  

  * 
  This 
  is 
  easily 
  proved 
  ; 
  it 
  is 
  suggested 
  by 
  the 
  fact 
  that 
  ±(t 
  — 
  I/t) 
  — 
  logt 
  

   has 
  a 
  triple 
  zero 
  at 
  t=l. 
  Various 
  properties 
  of 
  the 
  contour 
  are 
  given 
  

   in 
  the 
  first 
  of 
  my 
  papers 
  to 
  which 
  reference 
  is 
  made 
  in 
  § 
  1. 
  The 
  curve 
  

   obtained 
  by 
  giving 
  the 
  lower 
  sign 
  to 
  the 
  radical 
  is 
  the 
  inverse 
  of 
  the 
  

   curve 
  obtained 
  by 
  taking 
  the 
  upper 
  sign. 
  

  

  t 
  In 
  my 
  previous 
  paper, 
  it 
  was 
  convenient 
  to 
  call 
  this 
  function 
  F(0, 
  1). 
  

  

  