﻿of 
  Equal 
  Order 
  and 
  Argument. 
  367 
  

  

  Since 
  r 
  is 
  an 
  even 
  function 
  of 
  0, 
  we 
  easily 
  deduce 
  that 
  

  

  Jo 
  wrj 
  

  

  (r 
  + 
  1/r) 
  si 
  n 
  6{dr/rdd) 
  - 
  (r— 
  1/r) 
  cos 
  ™ 
  > 
  .. 
  

   X 
  r 
  2 
  + 
  1/r 
  2 
  - 
  2 
  cos 
  20 
  l 
  J 
  

  

  Now 
  it 
  is 
  easy 
  to 
  show 
  that 
  

   (r 
  4- 
  1/r) 
  sin 
  6(dr/rd6) 
  — 
  (r- 
  1/r) 
  cos 
  

   r 
  2 
  + 
  l/r 
  2 
  -2cos20 
  

  

  1 
  d 
  ^rsinfl 
  

  

  ~ 
  2 
  d6^ 
  n 
  1-r 
  2 
  

  

  1 
  d 
  f\ 
  . 
  rsm0 
  . 
  , 
  rsm^ 
  ] 
  

  

  ^ 
  -rx 
  ■{ 
  tan 
  z 
  -x 
  4- 
  tan 
  J 
  ^ 
  V 
  

  

  2d0\ 
  l 
  + 
  rcos0 
  l-rcos0J 
  

  

  But 
  the 
  contour 
  starts 
  from 
  (1, 
  0) 
  in 
  a 
  direction 
  making 
  

   an 
  angle 
  \ir 
  with 
  the 
  line 
  joining 
  this 
  point 
  to 
  the 
  origin 
  ; 
  

   and 
  so 
  tan 
  -1 
  \r 
  sin 
  0/(1 
  — 
  r 
  cos 
  0)} 
  decreases 
  from 
  §7r 
  to 
  

   as 
  increases 
  f 
  rom 
  to 
  7r, 
  while 
  tan 
  -1 
  {r 
  sin 
  0/(1 
  + 
  r 
  cos 
  0) 
  

   increases 
  from 
  to 
  it 
  in 
  the 
  same 
  circumstances. 
  

  

  We 
  therefore 
  write 
  equation 
  (4) 
  in 
  the 
  form 
  

  

  on 
  integrating 
  by 
  parts 
  ; 
  that 
  is 
  to 
  say 
  

  

  f 
  J 
  n 
  {nx)dx= 
  M" 
  V*™ 
  {Tr-tan- 
  1 
  ^^! 
  F'(0)<*0, 
  (5) 
  

  

  it 
  being 
  observed 
  that 
  F(0)=0, 
  F(7r) 
  = 
  oo, 
  so 
  that 
  the 
  

   integrated 
  part 
  vanishes 
  at 
  each 
  limit. 
  

  

  Now 
  F'(0) 
  is 
  positive* 
  when 
  0^= 
  0=?tt, 
  so 
  that 
  F(0) 
  is 
  a 
  

   steadily 
  increasing 
  function 
  of 
  0. 
  Moreover, 
  we 
  can 
  show 
  

   that 
  7T 
  — 
  tan" 
  1 
  {2r 
  sin 
  0/(1 
  — 
  r 
  2 
  )} 
  is 
  a 
  steadily 
  decreasing 
  

   function 
  of 
  0. 
  For 
  we 
  have 
  

  

  1 
  d 
  f 
  ^rsinfll 
  

  

  isi'-*" 
  1 
  1=5 
  =rj 
  

  

  _ 
  (r 
  + 
  1/r) 
  sin 
  0(dr/rd0) 
  - 
  (r- 
  1/r) 
  cos 
  

   - 
  r 
  2 
  + 
  1/r 
  2 
  -2 
  cos 
  20 
  

  

  0+ 
  sin 
  cos 
  0— 
  20 
  2 
  cot0 
  

  

  2(0 
  2 
  - 
  sin 
  2 
  0cos 
  2 
  0) 
  v 
  /(0 
  2 
  - 
  sin 
  2 
  0) 
  

   * 
  Proc. 
  London 
  Math. 
  Soc. 
  (2) 
  xvi. 
  p. 
  153. 
  

  

  sin 
  2 
  0. 
  . 
  (6) 
  

  

  