"The operator equation $$\sum_{i=1}^n A^{n-i}XB^{i-1} = Y$$ and a specific application"
Abstract: The solution of the linear operator equation $$\sum_{i=1}^n A^{n-i} XB^{i-1} = Y$$ is given by the integral formula $$X= \frac{\sin(\pi /n)}{\pi} \int_0^\infty t^{1/ n} (t+A^n)^{-1}Y(t+B^n)^{-1}dt$$ if $$A, B$$ are bounded normal, $$Y$$ is bounded on a complex Hilbert space with spectrum in $$\{z\neq 0, -\pi/(2n) < Arg(z) < \pi/(2n)\}$$.