Suppose that −∞≤a<b≤∞, and let f:{a,b}→ℝ be an integrable real function, where {a,b} denote any kind of the finite type intervals or {a,b}=(−∞,b) or (−∞,∞).If a=−∞, then the function f supposed to be integrable in the improper sense. Under the above conditions the following sequence of functions is called the sequence of iterated integrals of f:

${\displaystyle F^{(0)}(s):=f(s),\ }$

${\displaystyle F^{(1)}(s):=\int _{a}^{s}F^{(0)}(u)du=\int _{a}^{s}f(u)du,}$

${\displaystyle F^{(2)}(s):=\int _{a}^{s}F^{(1)}(u)du=\int _{a}^{s}\left(\int _{a}^{t}f(u)du\right)dt,}$
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${\displaystyle F^{(n)}(s):=\int _{a}^{s}F^{(n-1)}(u)du,}$
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Let {a,b}=[0,1) and f(s)≡1. Then the sequence of iterated integrals of 1 is defined on [0,1), and

${\displaystyle F^{(0)}(s)=1,\ }$

${\displaystyle F^{(1)}(s)=\int _{0}^{s}F^{(0)}(u)du=\int _{0}^{s}1du=s,}$

${\displaystyle F^{(2)}(s)=\int _{0}^{s}F^{(1)}(u)du=\int _{0}^{s}udu={s^{2} \over 2},}$
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${\displaystyle F^{(n)}(s):=\int _{0}^{s}{u^{n-1} \over (n-1)!}du={s^{n} \over n!},}$
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Let {a,b}=[-1,1] and f(s)≡1. Then the sequence of iterated integrals of 1 is defined on [-1,1], and

${\displaystyle F^{(0)}(s)=1,\ }$

${\displaystyle F^{(1)}(s)=\int _{-1}^{s}F^{(0)}(u)du=\int _{-1}^{s}1du=s+1,}$

${\displaystyle F^{(2)}(s)=\int _{-1}^{s}F^{(1)}(u)du=\int _{-1}^{s}(u+1)du={s^{2} \over 2!}+{s \over 1!}+{1 \over 2!}={(s+1)^{2} \over 2!},}$
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${\displaystyle F^{(n)}(s)={s^{n} \over n!}+{s^{(n-1)} \over {(n-1)!1!}}+{s^{(n-2)} \over (n-2)!2!}+\dots +{1 \over n!}={(s+1)^{n} \over n!},}$
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Let {a,b}=(-∞,0] and f(s)≡e^2s. Then the sequence of iterated integrals of e^2s is defined on (-∞,0], and

${\displaystyle F^{(0)}(s)=e^{2s},\ }$

${\displaystyle F^{(1)}(s):=\int _{-\infty }^{s}F^{(0)}(u)du=\int _{-\infty }^{s}e^{2u}du={e^{2s} \over 2},}$

${\displaystyle F^{(2)}(s):=\int _{-\infty }^{s}F^{(1)}(u)du=\int _{-\infty }^{s}{e^{2u} \over 2}du={e^{2s} \over 2^{2}},}$

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${\displaystyle F^{(n)}(s):=\int _{-\infty }^{s}F^{(n-1)}(u)du=\int _{-\infty }^{s}{e^{2u} \over 2^{(n-1)}}du={e^{2s} \over 2^{n}},}$

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