# All Questions

Let $D=\dfrac{d}{dx}.$ Consider the operator $$D_{h,x}=\frac{e^{hD}-1}{h}.$$ Question. What is explicit form of the operator $D^{-1}_{h,x}?$ I think that $$D^{-1}_{h,x}=\frac{h}{e^{hD}-1},$$ ...