CREATE FUNCTION dbo.is_different_integer (@old_value integer, @new_value integer)
RETURNS int
AS
BEGIN
    -- NULL has to be treated separately as comparison results in unknown!
    if @old_value IS NULL
        if @new_value IS NULL
            return 0
        else
            return 1

    if @new_value IS NULL
        return 1

    if @old_value <> @new_value
        return 1
    return 0
END
GO