Bash prints syntax error when command in $(...) contains case-esac

[Juergen Gohlke]

Configuration Information [Automatically generated, do not change]:
Machine: i686
OS: cygwin
Compiler: gcc
Compilation CFLAGS:  -DPROGRAM='bash.exe' -DCONF_HOSTTYPE='i686' 
-DCONF_OSTYPE='cygwin' -DCONF_MACHTYPE='i686-pc-cygwin' 
-DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash' 
-DSHELL -DHAVE_CONFIG_H -DRECYCLES_PIDS   -I.  
-I/home/eblake/bash-3.2.9-11/src/bash-3.2 
-I/home/eblake/bash-3.2.9-11/src/bash-3.2/include 
-I/home/eblake/bash-3.2.9-11/src/bash-3.2/lib   -O2 -pipe
uname output: CYGWIN_NT-5.0 JONA 1.5.24(0.156/4/2) 2007-01-31 10:57 i686 
Cygwin
Machine Type: i686-pc-cygwin

Bash Version: 3.2
Patch Level: 9
Release Status: release

Description:
       If a command in $(...) contains a case-esac construction, the 
bash prints a syntax error instead
       of executing the code:
       bash: syntax error near unexpected token `;;'

Repeat-By:
       A simple example is: x=$(a=4; case $a in 3) echo a=3;; 4) echo 
a=4;; esac)
       (typed in an interactive shell or being part of a shell script).
       The same error has been seen on Windows Vista, Windows XP and 
Windows2000.
       The same error has been seen with an earlier version of bash, 
i.e. 2.05b.0(8)-release
       If $(...) is replaced by `...`, it works.