显示来自php的标题包括页面？

im having some problems finding a way of doing this.

i have a page on my site called profile.php, this page currently has a page tittle 'user profile', my profile.php page looks like this:

<?php $page_title = "User Profile";
include('includes/headerframe.php'); ?>
<? // GET PROFILE ID FROM URL
if (isset ($_GET['id'])) {
    $profile_id = $_GET['id'];
}

$user_info_set = get_user_info();
if (!$user = mysql_fetch_array($user_info_set)) {
include ('includes/mod_profile/mod_noprofile.php');

}

else if (!isset($profile_id)) {
include("includes/mod_profile/mod_noprofile.php");

}

$profile_info_set = get_profile_info();
while ($profile = mysql_fetch_array($profile_info_set)) 

if (isset ($profile_id))
if ($user['account_status'] == "Active")
if ($user['account_type'] == "User") {
include("includes/mod_profile/mod_profile.php");

}

what im basically doing is using profile.php as an includes page to say if user exists include mod_profile else if the user doesnt exist include_modnoprofile.

so in mod_profile.php i echo out the user's display name though a query and im also wanting to try and put the users display name in the page title in profile.php instead of 'user',

so my question really is, is there a way to pull a page title from mod_profile and set it as the profile.php title or can i somehow include the page title in this part of the code where i get users information in profile.php:

$profile_info_set = get_profile_info();
while ($profile = mysql_fetch_array($profile_info_set)) 

if (isset ($profile_id))
if ($user['account_status'] == "Active")
if ($user['account_type'] == "User") {
include("includes/mod_profile/mod_profile.php");

would appreciate anyone's help with this, thanks in advance