﻿110 
  Mr. 
  J. 
  H. 
  Michell 
  on 
  the 
  

  

  and 
  put 
  

  

  J{x,z)=Z 
  r 
  2< 
  n 
  l 
  A 
  m 
  cos 
  —j- 
  + 
  Br* 
  sin 
  -j- 
  j. 
  cos 
  wfc 
  --/*), 
  

  

  where 
  r 
  is 
  a 
  positive 
  integer. 
  

   By 
  Fourier's 
  method 
  

  

  I 
  / 
  (a?, 
  3) 
  cos 
  —j- 
  dx 
  = 
  Z 
  £ 
  M 
  A 
  rn 
  cos 
  ra(s 
  — 
  /*), 
  

  

  J 
  

  

  f(x, 
  z) 
  sin 
  -y- 
  dx 
  — 
  l 
  S„B 
  rM 
  cos 
  n[z 
  — 
  h) 
  

  

  -1 
  ' 
  ** 
  

  

  where 
  A 
  0M 
  is 
  to 
  be 
  halved 
  as 
  usual. 
  

  

  Since 
  the 
  functions 
  cos 
  n(z—h) 
  are 
  all 
  conjugate, 
  as 
  is 
  easily 
  

   proved, 
  from 
  these 
  we 
  get 
  

  

  f 
  ,h 
  C 
  l 
  irrx 
  C 
  h 
  

  

  1 
  f(x, 
  z) 
  cos— 
  =^-cosh(s 
  — 
  h)dxdz 
  — 
  lA 
  ril 
  S 
  cos 
  2 
  n(z— 
  h)dz, 
  

   „o 
  J-i 
  * 
  Jo 
  

  

  = 
  Z 
  A 
  ni 
  j— 
  (2n/i 
  + 
  sin 
  2w/t) 
  

  

  and 
  

  

  i 
  h 
  C 
  l 
  trrx 
  1 
  

  

  I 
  j 
  f(x, 
  z) 
  sin 
  - 
  j- 
  cos 
  n(z—7i)dxdz 
  = 
  I 
  B 
  rn 
  — 
  (2/iA 
  + 
  sin 
  2/i/t) 
  ; 
  

  

  where 
  A 
  00 
  is 
  to 
  be 
  halved; 
  and 
  the 
  coefficients 
  of 
  the 
  terms 
  

   given 
  by 
  the 
  imaginary 
  roots, 
  here, 
  as 
  always 
  below, 
  are 
  got 
  

   by 
  putting 
  n 
  = 
  in 
  / 
  ; 
  so 
  that 
  

  

  \ 
  \ 
  f( 
  x 
  , 
  z 
  ) 
  cos 
  —f 
  coshn(z 
  — 
  h)dxdz 
  = 
  l 
  A 
  rn 
  , 
  —,{2nli 
  + 
  smh27i'h) 
  

  

  and 
  so 
  for 
  B 
  r?l 
  '. 
  

  

  Hence 
  the 
  theorem 
  

  

  f{ 
  X> 
  z)=tX 
  ; 
  ( 
  tn 
  W 
  rinlSt) 
  £ 
  f 
  /(f 
  '° 
  C0S 
  T 
  *-•) 
  

  

  xcos??(£— 
  h)dZ>d% 
  

  

  ^ 
  4n' 
  coshnY^r 
  — 
  h) 
  C 
  l 
  f 
  h 
  ,,<. 
  ^ 
  irr 
  fJ 
  , 
  N 
  

   + 
  2 
  ' 
  ;(2n'A 
  + 
  sinh2n'A) 
  J_, 
  J 
  /(? 
  ' 
  ?) 
  C 
  ° 
  S 
  T 
  <*"*> 
  

  

  X 
  cosh 
  n(t£—h)d$d%. 
  

   Now 
  let 
  Z 
  become 
  infinite, 
  and 
  putting 
  

  

  irr 
  j 
  I 
  = 
  m 
  

   7r/Z 
  =dm, 
  

  

  