Can I cause BASH to echo all source when it is invoked?

Suppose I have a complex bash script:

x.sh
#! /bin/bash
. include/this/script.sh
. include/this/other/script.sh 
...
foo="foo$bar"
echo $foo

Is there some setting or way to cause BASH to echo all the source code it sources, including the main script when I invoke it, so I know exactly what is going on?

>Solution :

The proper way, without debugging garbage:

set -v
. include/this/script.sh
. include/this/other/script.sh