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I am very new to Mathematica, and I searched the documentation and google now for quite some time.

Let's say, I want to implement a mathematica function, that does the following mapping: $f(t)\mapsto \int_0^t f(s)\ \mathrm{d}s+f'(t)$. ($f$ is assumed to be differentiable and integrable.)

Edit: It should also work for $\mathbb{R}\longrightarrow\mathbb{R}^n$ functions.

How would you do that? It should be a function like

IntAndDiff[f_]=Integrate[f,{s,0,t}]+D[f,t]

which does of course not work. How am I able to "access" the variable of $f$ in this case, so that I can define the integration and differentiation accordingly?

Edit: I also want this function to behave like an operator, so the output should be a $\mathbb{R}\longrightarrow\mathbb{R}^n$ function again. I.e. I want to be able to apply this (and other such operators) on the result again...

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  • $\begingroup$ ClearAll[IntAndDiff];IntAndDiff[f_] = Integrate[f@s, {s, 0, t}] + D[f@t, t]? $\endgroup$ – kglr Dec 31 '20 at 0:07
  • $\begingroup$ As @kglr suggested, except use SetDelayed, i.e., IntAndDiff[f_] := Integrate[f[s], {s, 0, t}] + D[f[t], t] Examples: IntAndDiff /@ {Sin, Sin[#] &, Cos, Cos[#] &, Sqrt, Sqrt[#] &, #^2 + 2 # - 3 &} $\endgroup$ – Bob Hanlon Dec 31 '20 at 0:35
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$f(t)\mapsto \int_0^t f(s)\ \mathrm{d}s+f'(t)$

A little analysis:

Input is actually a function $f(t)$, the output is also a function $F(t)=\int_0^t f(s)\ \mathrm{d}s+f'(t)$.

So an easy way is to introduce $t$

F[f_, t_]:= Integrate[f[s], {s, 0, t}] + f'[t]
F[Sin,t] (*1*)

Then use Function to make it a pure function

IntAddDiff[f_] := Function[t, Evaluate[Integrate[f[s], {s, 0, t}] + f'[t]]]

Then use # and & to remove t

IntAddDiff[f_] := Evaluate[Integrate[f[s], {s, 0, #}] + f'[#]] &
IntAddDiff[Sin] (*1 &*)
IntAddDiff[#^2 &] (*2 #1 + #1^3/3 &*)
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  • $\begingroup$ For me this looks like the solution that works best for my application. I will do some testing, and after this accept and mark the question as answered. $\endgroup$ – NG98 Jan 3 at 10:29
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Reply the new question.

By using Through we can deal with multiple function {F,G,H} map at t

intAndDiff[f___][t_] := 
  Integrate[Through[{f}@s], {s, 0, t}] + 
   Through[(Derivative[1] /@ {f})@t];
intAndDiff[Sin, Cos, #^2 &][x]

{1, 0, 2 x + x^3/3}

Original

IntAndDiff[f_][t_] := Integrate[f[s], {s, 0, t}] + D[f[t], t];
IntAndDiff[#^2 &][x]
IntAndDiff[Sin][x]

2 x + x^3/3

1

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  • $\begingroup$ I think this suggestion is very close to the solution I search for. It does however not work as intended in the case, where f is a function, that takes a scalar as input and returns a vector, as I cannot "concatenate" it. I.e. I search for an operator, so I take a R->R^n function as input and get a R->R^n function as output... $\endgroup$ – NG98 Jan 2 at 10:34
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Same as @cvgmt's solution but defined as an operator, so that it also works on $\mathbb{R}\to\mathbb{R}^n$ functions. Both the input and the output are now pure functions:

IntAndDiff[f_] := Function[t, Evaluate[Integrate[f[s], {s, 0, t}] + D[f[t], t]]]

IntAndDiff[#^2 &]
(*    Function[t$, 2 t$ + t$^3/3]    *)

IntAndDiff[Sin]
(*    Function[t$, 1]    *)

IntAndDiff[{#, #^2, #^3} &]
(*    Function[t$, {1 + t$^2/2, 2 t$ + t$^3/3, 3 t$^2 + t$^4/4}]    *)
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