The translation operator in one dimension is defined as
$$
\hat T \psi(x) = \psi(x-\alpha) .
$$
This can be written as an integral transformation,
$$
\begin{align\*}
 \hat T\psi(x) = \langle x|\hat T|\psi\rangle
 = \int\mathrm{d} x' \langle x|\hat T|x'\rangle\langle x'|\psi\rangle
 = \int\mathrm{d} x' T(x,x')\psi(x')
 = \int\mathrm{d} x' \delta(x-\alpha-x')\psi(x') = \psi(x-\alpha) 
\end{align\*}
$$
$$
\implies T(x,x') = \langle x | \hat T | x' \rangle = \delta(x' - (x-\alpha)).
$$

On the other hand, the same operation can be represented by a differential operator using the series expansion of $\psi(x-\alpha)$,
$$
\psi(x-\alpha) \approx \psi(x) - \alpha\frac{\partial\psi}{\partial x} + \frac{(-\alpha)^2}{2!}\frac{\partial^2\psi}{\partial^2x} + ...
 = \sum\limits\_{k=0}^{\infty}\frac{(-\alpha)^k}{k!}\frac{\partial^k\psi}{\partial^k x}\\
= \left[ \sum\limits\_{k=0}^{\infty} \frac{1}{k!} \left(-\alpha\frac{\partial}{\partial x}\right)^k\right]\psi(x)
 = e^{-\alpha\frac{\partial}{\partial x}}\psi(x).
$$

So, we have two different but equivalent representations of $\hat T$,
$$
\hat T[\psi] = \int\mathrm{d}x'T(x,x')\psi(x') = \int\mathrm{d}x'\delta(x' - (x-\alpha))\psi(x')
\\
\hat T[\psi] = e^{-\alpha\frac{\partial}{\partial x}}\psi(x).
$$
**Questions:**

* How can I determine if an integral transformation with kernel $K(x,x')$ has a corresponding differential operator?
* If there exists a differential operator for $K(x,x')$, how can I calculate it?
* How to calculate $K(x,x')$ if the differential operator is known 
(e.g. $\hat K[\psi] = a\partial\_x \psi(x) + b e^{-\partial^2\_x}\psi(x) + U(x)\psi(x)$) ?