I am aware of the the following Stackexchange question: When Bochner and Pettis integrals coincide?, but I feel like it never really got the traction it deserves. What you will see below is my attempt at making sense of what is happening with the Bochner and Pettis integral. This is mainly a proof-verification post.
The following link gives a good review of the Bochner integral: https://encyclopediaofmath.org/wiki/Bochner_integral. The crux of the matter is that it seems to be a straightforward generalization of the usual constructions in measure theory to Banach space valued functions. If we take the Banach space $\mathbb{C}$, then we should regain the usual notion of an integral for complex valued functions. To make things really simple let us start with a measure space $(G,\mathfrak{G},\mu)$ with $\mu$ a $\sigma$-additive measure on a $\sigma$-algebra $\mathfrak{G}$ on $G$. In addition, let $\mathcal{B}$ be a separable Banach space, so that for a Bochner integrable function $f:G\to \mathcal{B}$, we denote its Bochner integral via $$(B)\int_G f(s)d\mu(s)\in \mathcal{B}.$$ A most curious property of Bochner integrals is that for any bounded linear operator $T\in \mathcal{L}(\mathcal{B},\mathcal{B}')$ between Banach spaces $\mathcal{B}, \mathcal{B}'$, we have: If $f: G\to \mathcal{B}$ is Bochner integrable, then so is $T\circ f: G\to \mathcal{B}'$ and $$T\left((B)\int_G f(s)d\mu(s) \right) = (B)\int_{G} (T\circ f)(s) d\mu(s). \tag{1}$$
Now let us take the time to review what Pettis measureability and integrability means: https://encyclopediaofmath.org/wiki/Pettis_integral. A good summary is that it uses one of the corollaries of the Hahn-Banach theorem to define what the Pettis integral is. The Banach space valued function $f: G\to \mathcal{B}$ is Pettis-measurable if for all $\varphi \in \mathcal{B}^{\ast}$, $\varphi\circ f: X\to \mathbb{C}$ is measurable in the usual sense. Pettis-integrability is defined the same way, the punch-line is that if $f: G\to \mathcal{B}$ is Pettis integrable, then we can define the Pettis-integral in the following way: $\displaystyle (P)\int_G f(s)d\mu(s)$ is the unique element in $B$ completely determined (via Hahn-Banach) by the following formula: for each $\varphi\in \mathcal{B}^*$ $$\varphi\left((P)\int_G f(s)d\mu(s)\right) = \int_G (\varphi\circ f)(s) d\mu(s). \tag{2}$$
Now, it seems to be a basic result that when we have the apt assumptions, i.e. separability of $\mathcal{B}$, then it is true (IIRC: they also call this ``Pettis-Measurability Theorem'') that Bochner measurability is equivalent to Pettis measurability. Unfortunately, the result does not exactly extend to integrability, there exists functions that are (equivalently) Bochner and Pettis measureable, which are Pettis integrable, but not Bochner integrable. What can be seen from the sentence preceding equation (1) is that Bochner-integrability implies Pettis-integrability. We just change the bounded linear operators $T$ in into linear functionals $\varphi\in \mathcal{L}(\mathcal{B}, \mathbb{C})= \mathcal{B}^*$ , and we find that if $f: G\to \mathcal{B}$ is Bochner integrable, then $\varphi \circ f : G\to \mathbb{C}$ is also integrable (in the usual sense). Therefore $f$ is Pettis-integrable as well.
More is true it seems, let $\varphi \in \mathcal{B}^*$, then it follows from both equation (1) and (2) that $$\varphi\left((B)\int_G f(s)d\mu(s) - (P)\int_Gf(s)d\mu(s) \right)=0.$$ Since $\varphi$ is arbitrary, does it not follow from Hahn-Banach that $$(B)\int_Gf(s)d\mu(s) = (P)\int_Gf(s)d\mu(s)?$$ Therefore whenever $f:G\to \mathcal{B}$ is Bochner integrable, then the Bochner and Pettis integral coincide no? Can someone check my work for me?