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Let $D$ denote the differentiation operator on the real functions in $C^1$. For $\alpha\in \Bbb{R}$ and $f\in C^1$ define ${1\over D-\alpha}$ as $${1\over D-\alpha} \,f(t)=e^{\alpha t}\Bigl( \int_0^tf(u)\, e^{-\alpha u} \, du + C\Bigr)$$ where $C\in \Bbb{R}$ is the constant of integration.
Clearly, ${1\over D-\alpha}$ is also an operator.

Question: Do the operators $(D-\alpha)$ and ${1\over D-\alpha}$ commute?
I think the answer is no.

Atom
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  • Have you tried calculating $(D-\alpha)\frac1{D-\alpha}f$ and $\frac1{D-\alpha}(D-\alpha)f$ and comparing the results? – Arthur May 18 '19 at 14:32
  • Yes. They come out to be different. – Atom May 18 '19 at 14:32
  • But how then our operator analysis of differential operators (Heaviside) work for constant coefficients linear differential equations? – Atom May 18 '19 at 14:34
  • Why is there a constant of integration? This is a definite integral, right? – Theo Bendit May 18 '19 at 16:00
  • I wanted to make it a function of $t$. And hence I had to introduce the definite integral but retain the arbitrary constant term. – Atom May 18 '19 at 16:30

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