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This article gives an expression for $\ln D$:

$$\ln D=\left( \begin{array}{cccccc} \psi (1) & 0 & 0 & 0 & 0 & . \\ 0 & \psi (2) & 0 & 0 & 0 & . \\ 0 & 0 & \psi (3) & 0 & 0 & . \\ 0 & 0 & 0 & \psi (4) & 0 & . \\ 0 & 0 & 0 & 0 & \psi (5) & . \\ . & . & . & . & . & . \\ \end{array} \right)-\ln x$$

I derived the same result using the following Mathematica code, starting from the first principles:

Table[DSolveValue[{f'[x] + s f[x] == x^n, f[0] == 0}, f[x], x] // FullSimplify, {n, 0, 5}]
Integrate[%, s] // FullSimplify 
Limit[%, s -> 0, Direction -> "FromAbove"] // FullSimplify 
Table[Coefficient[%, x, k], {k, 0, 5}] // Expand // MatrixForm

But when I try to exponentiate it, I do not obtain $D$. What's wrong?

Anixx
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    You have asked three or four questions on the same topic just today. You might consider merging them rather than asking a slight modification each time you get a response. – Cameron Williams Feb 02 '21 at 01:09
  • @CameronWilliams I received useful answers, accepted some, made progress and now I am at this point. – Anixx Feb 02 '21 at 01:10
  • What Mathematica code did you use to exponentiate it? It is not obvious. – Somos Feb 02 '21 at 03:33
  • @Somos Well, MatrixFunction – Anixx Feb 02 '21 at 03:36
  • Check out Higham's paper for a detailed investigation of this phenomenon. – greg Feb 03 '21 at 04:02
  • @greg thanks. I think, actually this definition of matrix logarithm is flawed, see here: https://math.stackexchange.com/questions/4009298/different-representations-of-ln-xd-operator-which-one-is-correct – Anixx Feb 03 '21 at 04:42

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