Let $H$ be a given symmetric matrix, i.e., $H\in\mathbb{R}^{d\times d}_{\rm sym}$, $d\geq1$. The operator $T$ is given by $$TX=\int_0^1e^{(1-s)H}Xe^{sH}\mathrm{d}s,\qquad X\in\mathbb{R}^{d\times d}_{\rm sym}.$$ This is a fourth order tensor, that is also positive definite: $$TX\cdot X=\int_0^1|e^{\frac{1-s}2H}Xe^{\frac{s}2H}|^2\mathrm{d}s,$$ so $T^{-1}$ exists. Can we find an explicit formula for $T^{-1}$?
Based on the scalar case, where $TX=e^HX$ and $T^{-1}X=e^{-H}X$, I have tried to compute $$\int_0^1e^{-(1-t)H}\int_0^1e^{(1-s)H}Xe^{sH}\mathrm{d}s\,e^{-tH}\mathrm{d}t=\ldots=\int_0^1(1-u)(e^{uH}Xe^{-uH}+e^{-uH}Xe^{uH})\mathrm{d}u,$$ which however does not seem to evaluate to $X$.