Let denoted by $G = \mathbb R^n\ltimes SO(n)$ the motion group of $\mathbb R^n$, a typical element $g\in G$ is written as $g = (x, k), x \in \mathbb R^n, k\in SO(n)$. Let $f\in L^{\infty}(G)$, I would like to know why this we have the following equality
$$\int_{SO(N)}f(x,k k')\, \overline{\pi_{i,j}}(k') \, dk' \, \underbrace{=}_{??} \, \sum_{s=1}^r (f_{s, j}\otimes \pi_{s,i}) (x,k),$$ where $f_{i, j}(x)=\int_{SO(N)}f(x,k')\, \overline{\pi_{i,j}}(k') \, dk'$ and where $\pi_{i,j}$ is a matrix coefficient of an irreducible unitary representation $\pi$ of of rank $r$ of $SO(n)$.
Thank you in adavance
The semidirect product is irrelevant since $x$ is fixed throughout, and for fixed $x$, $k \mapsto f(x, k)$ is a bounded function on $\rm{SO}(n)$. So the general setting is : $G$ is a compact group, $\pi: k \mapsto (\pi_{i,j}(k))$ is a not necessarily irreducble finite dimensional unitary matrix representation of $G$, and $f \in L^\infty(G)$. Let $$ f_{i,j} = \int_G f(k) \overline\pi_{i,j}(k)\, dk. $$ Calculate $$ \begin{aligned} &\int_G f(k k') \overline\pi_{i,j}(k')\, dk' = \int_G f(k') \overline\pi_{i,j}(k^{-1} k')\, dk' \\ &= \int_G f(k') \sum_s \overline \pi_{i, s}(k^{-1}) \overline \pi_{s, j}(k') \, dk' = \sum_s f_{s, j}\ \overline \pi_{i, s}(k^{-1}) \\ &= \sum_s f_{s, j}\ \pi_{s, i}(k). \end{aligned} $$
The short summary answer: to calculate this quantity, or variants, e.g. with $k$ replaced by $k^{-1}$ or $k k'$ replaced by $k' k$, etc., you will use translation invariance on the left or right of Haar measure, and the properties of the unitary matrix representation.