Since you've answered part of the question, let me elaborate on the Dunford integral. If $f:\Omega\to E$ is weakly measurable and satisfies $\langle x^* ,f\rangle\in L^1(\Omega)$ for all $x^*\in E^\*$ then it's possible to define $\int_X f\in E^{\*\*}$. This acts on $E^{\*}$ via the prescription
$$x^*\mapsto \int_X\langle x^{\*},f(\omega)\rangle d\omega.$$
I first learnt this in Talagrand's AMS memoir:

http://books.google.co.uk/books?id=YqJ3t2oT5WgC&lpg=PP1&dq=pettis%20integral%20and%20measure%20theory&pg=PP1#v=onepage&q=pettis%20integral%20and%20measure%20theory&f=false

but a good text on infinite dimensional analysis will probably mention it. $f$ is Pettis integrable precisely when the Dunford integral $\int_Xf$ lies in the canonical image of $E$ in $E^{\*\*}$ and then the two agree (modulo the spaces they live in).

Let's cook up an example. Take the function $f:[0,1]\to c_0$ defined by $f(x)=2^ne_n$ whenever $x\in [2^{-n},2^{-n+1})$. Define $f_n$ to be the obvious simple approximation: let it be $0$ on $[0,2^{-{n+1}})$ and agreeing with $f$ elsewhere.

The problem with these step functions: we compute
$$\int_0^1{\| f(s)-f_n(s)\|ds}=\sum_{k=n}^\infty{2^k(2^{-k+1}-2^{-k})}=\infty$$
and note that our dumb approximation doesn't work in the strong sense.

But pick any $g\in\ell^1$ and note that
$$\int_0^1{\langle f(s)-f_n(s) , g\rangle ds} =\sum_{k=n}^\infty g(k) \to 0$$
as $n\to\infty$. So our function can be weakly approximated by step functions - in fact you can easily see $s\mapsto \langle f(s),g\rangle$ is $L^1$ with $\int\langle f(s),g\rangle=\lim_{n\to\infty}\int\langle f_n(s),g\rangle$. So $f$ is Dunford integrable.

Note that the integrals
$F_n=\int_0^1{f_n(s)ds}=\sum_{k=1}^{n-1}{e_k}$
do not converge to anything in $c_0$. We can think of them as members of $\ell^\infty$ and of course here they converge (in weak* topology) to the identity. In fact, $\int_0^1{\langle f(s), g\rangle ds}=\sum_{k=1}^\infty g(k)$ - so by definition $\int_0^1{f(s)ds}=1 \in \ell^\infty$. This is an example of the Dunford integral, when the Petis/Bochner integrals won't do.

Another example: modify the function $f$ in our example of Dunford integrability, so that $f(x)=2^ne_n/n$ instead of $2^ne_n$. The obvious step function approximation still can't be strong, as $\int_0^1{\|f(s)-f_n(s)\|}=\sum_{k=n}^\infty{1/k}$. Now the sequence of approximations $F_n$ converges in norm to $(1/k)_{k=1}^\infty$. Thought of as an element of $\ell^\infty$ this is the Dunford integral, but as an element of $c_0$ it's just the Pettis integral. (Note, I'm not even saying this function is *not Bochner integrable*, but the weak approximations make the Pettis integral very natural in this context).