The concept of the sequence of iterated integrals
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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),\ }](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/34c96171fb430644a2ba5a2dace63937f5cdb63d)
![{\displaystyle F^{(1)}(s):=\int _{a}^{s}F^{(0)}(u)du=\int _{a}^{s}f(u)du,}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/eec878d62b819eaaf3a20f57735f8afa57fed5ec)
![{\displaystyle F^{(2)}(s):=\int _{a}^{s}F^{(1)}(u)du=\int _{a}^{s}\left(\int _{a}^{t}f(u)du\right)dt,}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/7e2b0efcc7a49d8b00dc9f549a0c858c14f41e2a)
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![{\displaystyle F^{(n)}(s):=\int _{a}^{s}F^{(n-1)}(u)du,}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/dbf0ddfdf73828fe78688f05ce5395e75dd426c9)
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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,\ }](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/96276e1a8bc28921c5c0c6fb11030ae9690cbc6f)
![{\displaystyle F^{(1)}(s)=\int _{0}^{s}F^{(0)}(u)du=\int _{0}^{s}1du=s,}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/5ab29a3055eef4a8fc1dcf898325b61abb4c9334)
![{\displaystyle F^{(2)}(s)=\int _{0}^{s}F^{(1)}(u)du=\int _{0}^{s}udu={s^{2} \over 2},}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/6b68d4d83763ccb84a50eca20a03e39fde487ac9)
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![{\displaystyle F^{(n)}(s):=\int _{0}^{s}{u^{n-1} \over (n-1)!}du={s^{n} \over n!},}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/a29ff5dcf8693caa5c4a5bd0c85f9e27e78b2bb0)
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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,\ }](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/96276e1a8bc28921c5c0c6fb11030ae9690cbc6f)
![{\displaystyle F^{(1)}(s)=\int _{-1}^{s}F^{(0)}(u)du=\int _{-1}^{s}1du=s+1,}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/64c05236eccf8f5a2f971c4eea733e86b421ea67)
![{\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!},}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/40c5f70fd4ebaa48c99ca14303f45f7f5a5efeff)
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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!},}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/7b1f47b28c758fe0dc49297ae1377a44cd6a9c62)
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Let {a,b}=(-∞,0] and f(s)≡e2s. Then the sequence of iterated integrals of e2s is defined on (-∞,0], and
![{\displaystyle F^{(0)}(s)=e^{2s},\ }](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/198bdbba93481872c4bf297986996c5f574ac90c)
![{\displaystyle F^{(1)}(s):=\int _{-\infty }^{s}F^{(0)}(u)du=\int _{-\infty }^{s}e^{2u}du={e^{2s} \over 2},}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/75192e71616317a7001aa29545c5cea240da138a)
![{\displaystyle F^{(2)}(s):=\int _{-\infty }^{s}F^{(1)}(u)du=\int _{-\infty }^{s}{e^{2u} \over 2}du={e^{2s} \over 2^{2}},}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/9fe7b0c78b29b6b034a5e6684c89e39ae486fcce)
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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}},}](https://wikimedia.riteme.site/api/rest_v1/media/math/render/svg/230a45437a4987cf85fabeac2780f9dd483a6478)
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