User:Vezér

The concept of the sequence of iterated integrals
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:

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

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

$$F^{(2)}(s):=\int^s_a F^{(1)}(u)du=\int^s_a \left( \int^t_a f(u)du \right ) dt   ,$$

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$$F^{(n)}(s):=\int^s_a F^{(n-1)}(u)du,$$

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Example 1
Let {a,b}=[0,1) and f(s)≡1. Then the sequence of iterated integrals of 1 is defined on [0,1), and

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

$$F^{(1)}(s)=\int^s_0 F^{(0)}(u)du=\int^s_0 1 du=s,$$

$$F^{(2)}(s)=\int^s_0 F^{(1)}(u)du=\int^s_0 u du={s^2 \over 2}   ,$$

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$$F^{(n)}(s):=\int^s_0 {u^{n-1}\over (n-1)!}du={s^n \over n!},$$

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Example 2
Let {a,b}=[-1,1] and f(s)≡1. Then the sequence of iterated integrals of 1 is defined on [-1,1], and

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

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

$$F^{(2)}(s)=\int^s_{-1} F^{(1)}(u)du=\int^s_{-1} (u+1) du={s^2 \over 2!}+{s \over 1!}+{1 \over 2!}={(s+1)^2 \over 2!}   ,$$

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$$F^{(n)}(s)={s^n \over n!}+{s^{(n-1)}\over {(n-1)!1!}}+{s^{(n-2)} \over (n-2)!2!}+ \dots +{1 \over n!} ={(s+1)^n \over n!}  ,$$

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Example 3
Let {a,b}=(-∞,0] and f(s)≡e2s. Then the sequence of iterated integrals of e2s is defined on (-∞,0], and

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

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

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

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

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