Inspired by this question I was wondering whether "recursive" integrals have been studied or if they appear anywhere in applications. What I mean is the following: Let $I(x, y) = \int_x^y g(t) dt$ and define

$$\begin{aligned} I_0 &= \int_x^y g(t) dt &&= I(x, y)\\ I_1 &= \int_{\int_y^xg(t)dt}^{\int_x^yg(t)dt}g(t) dt &&= I(-I(x,y), +I(x, y)) \\ I_2 &= \int_{\int^{\int_y^xg(t)dt}_{\int_x^yg(t)dt}g(t) dt}^{\int_{\int_y^xg(t)dt}^{\int_x^yg(t)dt}g(t) dt} g(t) dt &&=I(-I(-I, +I), +I(-I, +I)) \\ &&\vdots \\ I_{n+1} &= I(-I_n, +I_n) \end{aligned}$$

What can be said about convergence criteria for this process? It should be straight-forward to show that for non-negative functions a blow up happens if $\int^x g(t) dt$ grows asymptotically faster than $x$. Otherwise it should converge I think. But who knows what could happen when fast oscillating functions taking both positive and negative values are involved.

  • $\begingroup$ Let $F(u) = \int_{-u}^u g(t)\,dt$. Then you ask about the iterates of $F$; a subject with vast literature. So, how does the assertion that $F(u) = \int_{-u}^u g(t)\,dt$ for some $g$ restrict the choice of the function $F$? $\endgroup$
    – GEdgar
    Feb 6, 2022 at 15:46

1 Answer 1


Let $$z_0 = \int\limits_x^y g(t)\,\text dt,\quad F(z)=\int\limits_{-z}^z g(t)\,\text dt.$$

Then we have a simple iteration $$z_{i+1}=F(z_i),$$ with the sufficient condition of convergency in the form of $$|F(z)|<\gamma|z|,\qquad(|z|\le|z_0|)$$ where $\;\gamma<1.\;$

  • $\begingroup$ very nice answer. trivial but nice $\endgroup$
    – mick
    Jan 21, 2023 at 23:39

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