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Differentiation can be expressed $k$ times in a row as either $f^{(k)}(x)$ or as $\frac{d^n}{dx^n}f(x)$. How do I express indefinite integration $k$ times in a row?

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    $\begingroup$ There is not a standard notation for this. I would suggest using something like $I(f) = I^1(f)= \int f dx$ and $I^k(f) = \int I^{k-1} dx$ for $k \ge 2$. Then $I^n(f)$ is what you’d want. You could also use $I^0(f) = f(x)$ for consistency too. $\endgroup$ – JavaMan Aug 15 at 2:45
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    $\begingroup$ One thing to keep in mind is that this $I^k$ maps functions into the space of functions modulo degree $k-1$ polynomials, which is a little awkward and may explain why its not been given an established notation $\endgroup$ – Calvin Khor Aug 15 at 3:08
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    $\begingroup$ Say "I will in the course of this writing use the notation $... something you made up .....$ to indicate indefinite integration iterated $k$ times". Then use the thing you made up. $\endgroup$ – fleablood Aug 15 at 6:57
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$$\underbrace{ \int\cdots\int }_k f(x) \; \underbrace{ \mathrm dx\cdots\mathrm dx }_k$$

but the lazy sometimes just use one $\int$.

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A prior question: what do you mean by "integration $k$ times in a row"?

If you mean the sequence of functions $f_n$ given by $$\begin{align}f_0(x)&=f(x)\\ f_{n+1}(x)=\int_a^xf_n(t)\mathrm{d}t \end{align}$$ for some fixed value $a$, then in fact

$$f_n(x)=\frac{1}{(n-1)!}\int_a^x(x-t)^{n-1}f(t)\mathrm{d}t\text{,}$$ a result that motivates the definition of the Riemann–Liouville fractional integral $$I^{\nu}_af(x)\stackrel{\text{def}}{=}\frac{1}{(\nu-1)!}\int_a^x(x-t)^{\nu-1}f(t)\mathrm{d}t\text{.}$$ Unfortunately, there appears to be no consensus notation for the latter either (I've seen both $I$ and $J$ used, as well as $1$, $2$, or $3$ superscripts/subscripts).

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  • $\begingroup$ I meant indefinite integration i.e. the sequence $f_{n+1}(x)=\int f_n(x)dx$ with $f_0(x)=f(x)$. My question has been edited to reflect this; apologies for the confusion. $\endgroup$ – Tesseract Aug 15 at 2:41
  • $\begingroup$ Using some variant of $I^{\nu}$ or $J^{\nu}$ is probably still the way to go $\endgroup$ – K B Dave Aug 15 at 2:47

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