The shift operator $T^{t}$ (where $t \in \mathbf{R}$ ) takes a function $f$ on $\mathbf{R}$ to its translation $f_{t}$, $T^{t} f(x)=f_{t}(x)=f(x+t)$. A practical operational calculus representation of the linear operator $T^{t}$ in terms of the plain derivative $\frac{d}{d x}$ was introduced by Lagrange, $$T^{t}=e^{t\frac{d}{d x}} $$
But what about $$T^{it}=e^{it\frac{d}{d x}} $$ where $f$ is a real valued function of a real variable, $i$ the imaginary unit and $t\in \mathbb{R}$. It's meaningful to consider its action on a real valued function? $$T^{it}f(x)=e^{it\frac{d}{d x}}f(x)=???$$ What would be the expected result?
It is to be noted that the action of the shift operator on a given function, namely $$ e^{t\partial_x}f(x) = \sum_{n=0}^\infty \frac{t^n}{n!} \frac{\partial^n}{\partial x^n} f(x) = f(x+t), $$ necessitates an analytic $f$ in order to be correctly defined. Let's recall that an analytic function is infinitely differentiable and thus can be represented by its Taylor series. If this representation possesses an analytical continuation on (some region of) the complex plane, then the shift operator acts the same way as in the real case, such that $e^{\tau\partial_x}f(x) = f(x+\tau)$, with $\tau \in \mathbb{C}$.