what is going on in std::function operator() and std::forward?

I was looking at the std::function implementation and its call operator()

template<typename Ret, typename... ArgTypes>
Ret function< Ret (ArgTypes...)>::operator()(ArgTypes...args) const
{
  // some stuff
  return invoker(functor, std::forward<ArgTypes>(args)...);
}

I was particularly wondering, why it uses std::forward here? Does this anything have to do with perfect forwarding? Because perfect forwarding could only be done if operator() is a template with a variadic template declaration template<typename... Args> (which it is not). What is the intention of using std::forward here? I am confused :-)?