Dual space of Bochner space

Let $B$ be a refexive Banach space. I want to show that $$(L^2(0,T;B))^* = L^2(0,T;B^*)$$ and that the dual pairing is $$\langle F,f \rangle_{L^2(0,T;B^*), L^2(0,T;B)} = \int_0^T \langle F(t), f(t) \rangle_{B^*,B}.$$ Can anyone help me with either part? Thanks.

No proof in the generality that is desired is going to be very brief. But all the essential ideas come from the usual, scalar case. First we need the following:

Definition. A Banach space $V$ is said to have the Radon-Nikodym property with respect to a measure space $(X, \Sigma, \mu)$ if, for every $V$-valued vector measure of bounded variation $\nu$ which is absolutely continuous with respect to $\mu$, there exists a Bochner integrable $f \in L^1(X,V)$ such that $\nu(E) = \int_E f \, d\mu$ for every $E \in \Sigma$.

In other words, spaces with the Radon-Nikodym property are precisely those in which the generalization of the Radon-Nikodym theorem for scalar functions holds. Then the key result that one needs to prove is due to Phillips:

Reflexive Banach spaces have the Radon-Nikodym property.

A proof of this requires a decent amount of machinery and a proof of this fact and other related facts is the subject of Chapter 2 of Vector Measures by Diestel and Uhl. Now, with this fact under our belt, we have the following result:

Theorem. Let $(X, \Sigma, \mu)$ be a finite measure space, and let $V$ be a Banach space such that $V'$ has the Radon-Nikodym property. Then, for $p \in [1, \infty)$, $L^p(X,V)' \cong L^q(X,V')$ isometrically, where $1/p + 1/q = 1$.

Proof. Suppose $g \in L^q(X,V')$. Then, for $f \in L^p(X,V)$, the map $\Lambda_g(f) = \int_X \langle f, g \rangle \, d\mu$ is clearly a continuous linear functional, so $L^q(X,V') \subseteq L^p(X,V)'$. To prove the reverse inclusion, let $\Lambda \in L^p(X,V)'$. For $h \in L^{\infty}(X,V)$, $|\Lambda(h)| \le C \|h\|_p \le C \|h\|_{\infty}$, and hence $\Lambda$ is also a continuous functional on $L^{\infty}(X,V)$. Now, $\Lambda$ induces a $V'$-valued vector measure absolutely continuous with respect with $\mu$ given by $$ \langle \nu(E), x \rangle = \Lambda(x 1_E)$$ for each $x \in V$. Then, as $V'$ has the Radon-Nikodym property, then there exists $g \in L^1(X,V')$ such that $\nu(E) = \int_E g \, d\mu$. Then, by approximating each $f \in L^p(X,V)$ by simple functions, we have that $\Lambda(f) = \int_X \langle f, g\rangle \, d\mu$. Finally, it is straightforward to show that $g$ is in fact also in $L^q$.