Rotation with axis $\hat{k}$ and angle $\theta$ in $\mathbb{R}^3$ is represented by $$ R = I + (\sin \theta) K + (1-\cos \theta) K^2 $$ where $K$ is the matrix for left cross product by $\hat{k}$. This is called the Rodrigues' rotation formula. (http://en.wikipedia.org/wiki/Rodrigues%27_rotation_formula)
In quantum mechanics course, finite rotation operator with axis $\hat{k}$ and angle $\theta$ is calculated as $$ D(\theta, \hat{k}) = \exp ( - \frac{i}{\hbar} \theta \hat{k} \cdot \vec{L} ) $$ where $\vec{L}$ is the angular momentum operator $\vec{L} = \vec{R} \times \vec{P} $.
It is usually calculated through somewhat cumbersome way such that at first calculate infinitesimal rotation operator and take exponentiation through limiting process. But in my opinion this kind of calculation somewhat boring and I think there would be more illuminating and direct deduction of $ D(\theta, \hat{k}) $ from the very definition of rotatation operator $$ D(\theta, \hat{k}) \psi (x,y,z) = \psi ( (x,y,z)R ) $$ Where $R$ is the rotation matrix calculated above via Rodrigues' formula.
Is there any more succinct and enlightening method to find the explicit form of rotation operator $D(\theta, \hat{k})$ from above equation?
Thanks in advance.