Spatial Translations in QM as described in Sakurai In Sakurai, page 43, he defines the operator $\hat{J}$ which is the infinitesimal translation by $d \vec{x}'$ as $$\hat{J}(d \vec{x})| \vec{x}' \rangle = | \vec{x}' + d \vec{x}' \rangle.$$
On an arbitrary state ket $| \alpha \rangle$ he then writes that $$\hat{J}(d \vec{x}')| \alpha \rangle = \hat{J}(d \vec{x}') \int d^3 x' | \vec{x}' \rangle \langle \vec{x}' | \alpha \rangle = \int d^3 x' | \vec{x}+ d \vec{x}' \rangle \langle \vec{x}'| \alpha \rangle.$$ 
How do we know that we can take the function $\hat{J}(d \vec{x})$ into the integral in the second equation?
Also, this seems like a strange way to define an operator, first on a specific vector (position eigenket) and then using that to extend the definition to an arbitrary vector.
Thanks for any assistance.
 A: Sakurai is being rigorous on the level of physicists. We will give a rigorous proof of Eq. (1.6.36) in Sakurai,
$$\tag{1}J(\mathbf a)=\exp\left(-\frac{\mathrm i}{\hbar}\mathbf a\cdot\mathbf p\right).$$
Let $\mathscr H$ denote the Hilbert space $L^2(\Bbb R^3)$, with its usual norm $||\cdot||$. We note first that since the momentum operator is unbounded on $\mathscr H$, (1) cannot actually be interpreted by the power series (1.6.37) in Sakurai. That series only makes sense for $X\in\mathscr B(\mathscr H)$, the Banach space of bounded operators on $\mathscr H$. Instead, we must interpret (1) in terms of the functional calculus for self-adjoint operators. Note that $\mathbf p$ is essentially self-adjoint on $\mathscr H$, so we may apply the spectral theorem to obtain a spectral measure $\mu^j$ for each $p_j$. Letting $\sigma_j$ denote the spectrum of each $p_j$, we define $$\exp\left(-\frac{\mathrm i}{\hbar}a_jp_j\right)=\int_{\sigma_j} \mathrm e^{-\mathrm i a_j\lambda/\hbar }\,\mathrm d\mu^j(\lambda).$$
We then define 
$$\exp\left(-\frac{\mathrm i}{\hbar}\mathbf a\cdot\mathbf p\right)=\exp\left(-\frac{\mathrm i}{\hbar} a_xp_x\right)\exp\left(-\frac{\mathrm i}{\hbar}a_yp_y\right)\exp\left(-\frac{\mathrm i}{\hbar}a_zp_z\right).$$
Even though $\mathbf p$ is not defined on all of $\mathscr H$, these exponentials are bounded operators, so by the BLT theorem admit extensions to all of $\mathscr H$.
The actual proof of (1) relies on Stone's theorem. Before we give Stone's theorem, we state a definition. Here $\mathscr H$ is an arbitrary separable Hilbert space.

A one parameter unitary group on $\mathscr H$ is a homomorphism $U:\Bbb R\to \mathrm U(\mathscr H)$. This group is strongly continuous if $$\lim_{s\to t}U(s)\psi=U(t)\psi$$
  for all $\psi\in\mathscr H$. The operator $A$ given by $$A\psi=\lim_{t\to 0}\frac{1}{\mathrm i}\frac{U(t)\psi-\psi}{t}$$
  is called the infinitesimal generator of $U$. The domain of $A$ is the subspace of $\mathscr H$ for which this limit exists in the topology of $\mathscr H$. 
Stone's Theorem. Suppose $U$ is a strongly continuous one-paramter unitary group on $\mathscr H$. Then the infinitesimal generator $A$ of $U$ is densely defined and self-adjoint, and $$U(t)=\exp(\mathrm itA)$$ for each $t\in\Bbb R$. The exponential here is defined via the functional calculus from above.

Letting $\mathscr H=L^2(\Bbb R^3)$ again, we let $U_\mathbf a$ denote the translation operator $$U_\mathbf a(t)\psi(\mathbf x)=\psi(\mathbf x-t\mathbf a).$$
Clearly this is a one-parameter unitary group (Lebesgue measure is translation invariant, hence the unitarity). Strong continuity is a little harder, but using the density of $C_0^\infty$ in $L^2$ can be easily established. We claim that the infinitesimal generator of $U_\mathbf a$ is $\mathbf a\cdot\mathbf p/\hbar$. It suffices to verify this on a dense subspace of $\mathscr H$. So take $\psi\in C^\infty_0(\Bbb R^3)$, then we have $$A\psi(\mathbf x)=\lim_{t\to 0}\frac{1}{\mathrm i}\frac{\psi(\mathbf x-t\mathbf a)-\psi(\mathbf x) }{t}=\frac{1}{\mathrm i}\mathbf a\cdot\nabla \psi(\mathbf x)=\frac{\mathbf a\cdot\mathbf p}{\hbar}\psi(\mathbf x).^\dagger$$
Then, noting that $J(\mathbf a)=U_\mathbf a(1)$, we have 
$$J(\mathbf a)=\exp\left(-\frac{\mathrm i}{\hbar}\mathbf a\cdot\mathbf p\right).$$
The power series for $J(\mathbf a)$ cannot be understood rigorously, but we can give the following interpretation: for $\psi\in C^\omega(\Bbb R^3)\cap\mathscr H$, we have $$J(\mathbf a)\psi(\mathbf x)=\sum_{k=0}^\infty\frac{(-\mathrm i)^k}{\hbar^kk!} (\mathbf a\cdot\mathbf p)^k\psi(\mathbf x),$$
since the series converges by analyticity of $\psi$. Thus, formally, 
$$J(\mathbf a)``="\sum_{k=0}^\infty\frac{(-\mathrm i)^k}{\hbar^kk!} (\mathbf a\cdot\mathbf p)^k\quad\text{on $C^\omega(\Bbb R^3)$.}$$
But not all functions in $L^2$ are analytic, so this has its limitations. 

$^\dagger$ Don't be too hasty here. This might seem like just the definition of the directional derivative, but the limit is taken here in the $L^2$ norm. The fact that difference quotients converge to derivatives uniformly in $C_0^\infty$ is essential here. 
A: 
Also, this seems like a strange way to define an operator, first on a specific vector (position eigenket) and then using that to extend the definition to an arbitrary vector.

For a linear operator, it's perfectly appropriate and very common to write down what it does to each element of a particular basis. This forms a definition of the linear operator, because there is always exactly one unique linear operator that does that. (For a finite-dimensional space, this procedure is equivalent to defining a linear operator by writing down its matrix in a particular basis.)
Taking the J(dx) into the integral is legitimate precisely because J(dx) is a linear operator. (It takes some practice before you can immediately see that kind of thing ... I suggest that you set aside a few minutes to stare at the definition of "linear operator", and at that equation. Remember that an integral is a sum. :-D )
