++sn1987a (when the Vote Fairy next visits) for an ingenious observation! Unfortunately, I don’t think it answers the OP’s question, as masking with bitwise-AND does remove the sign bit, even with use integer in effect:

#! perl
use strict;
use warnings;

print "\nno integer\n";

my $bit = -1 >> 31;
printf "unmasked: 0x%x = %d\n", $bit, $bit;

$bit = (-1 & 0x80000000) >> 31;
printf "masked:   0x%x = %d\n", $bit, $bit;

use integer;

print "\nuse integer\n";

$bit = -1 >> 31;
printf "unmasked: 0x%x = %d\n", $bit, $bit;

$bit = (-1 & 0x80000000) >> 31;
printf "masked:   0x%x = %d\n", $bit, $bit;
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Output:

 1:29 >perl 868_SoPW.pl

no integer
unmasked: 0x1ffffffff = 8589934591
masked:   0x1 = 1

use integer
unmasked: 0xffffffffffffffff = -1
masked:   0x1 = 1

 1:29 >perl -v

This is perl 5, version 18, subversion 2 (v5.18.2) built for MSWin32-x
+86-multi-thread-64int
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:-(

Update: As BrowserUk observes below, masking removes the sign bit (in this case) only in perls built with 64-bit ints. For perls built with 32-bit ints, sn1987a’s observation does answer the OP’s question. (Moral for self: test, test, test!)