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
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0
)
(
s
)
:=
f
(
s
)
,
{\displaystyle F^{(0)}(s):=f(s),\ }
F
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1
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s
)
:=
∫
a
s
F
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0
)
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u
)
d
u
=
∫
a
s
f
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u
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d
u
,
{\displaystyle F^{(1)}(s):=\int _{a}^{s}F^{(0)}(u)du=\int _{a}^{s}f(u)du,}
F
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2
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s
)
:=
∫
a
s
F
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1
)
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u
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d
u
=
∫
a
s
(
∫
a
t
f
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u
)
d
u
)
d
t
,
{\displaystyle F^{(2)}(s):=\int _{a}^{s}F^{(1)}(u)du=\int _{a}^{s}\left(\int _{a}^{t}f(u)du\right)dt,}
F
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n
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(
s
)
:=
∫
a
s
F
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n
−
1
)
(
u
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d
u
,
{\displaystyle F^{(n)}(s):=\int _{a}^{s}F^{(n-1)}(u)du,}
Example 1
edit
Let {a,b}=[0,1) and f(s)≡1 . Then the sequence of iterated integrals of 1 is defined on [0,1) , and
F
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0
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(
s
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=
1
,
{\displaystyle F^{(0)}(s)=1,\ }
F
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1
)
(
s
)
=
∫
0
s
F
(
0
)
(
u
)
d
u
=
∫
0
s
1
d
u
=
s
,
{\displaystyle F^{(1)}(s)=\int _{0}^{s}F^{(0)}(u)du=\int _{0}^{s}1du=s,}
F
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2
)
(
s
)
=
∫
0
s
F
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1
)
(
u
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d
u
=
∫
0
s
u
d
u
=
s
2
2
,
{\displaystyle F^{(2)}(s)=\int _{0}^{s}F^{(1)}(u)du=\int _{0}^{s}udu={s^{2} \over 2},}
F
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n
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(
s
)
:=
∫
0
s
u
n
−
1
(
n
−
1
)
!
d
u
=
s
n
n
!
,
{\displaystyle F^{(n)}(s):=\int _{0}^{s}{u^{n-1} \over (n-1)!}du={s^{n} \over n!},}
Example 2
edit
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
,
{\displaystyle F^{(0)}(s)=1,\ }
F
(
1
)
(
s
)
=
∫
−
1
s
F
(
0
)
(
u
)
d
u
=
∫
−
1
s
1
d
u
=
s
+
1
,
{\displaystyle F^{(1)}(s)=\int _{-1}^{s}F^{(0)}(u)du=\int _{-1}^{s}1du=s+1,}
F
(
2
)
(
s
)
=
∫
−
1
s
F
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1
)
(
u
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d
u
=
∫
−
1
s
(
u
+
1
)
d
u
=
s
2
2
!
+
s
1
!
+
1
2
!
=
(
s
+
1
)
2
2
!
,
{\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!},}
F
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n
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s
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=
s
n
n
!
+
s
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n
−
1
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(
n
−
1
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!
1
!
+
s
(
n
−
2
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(
n
−
2
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!
2
!
+
⋯
+
1
n
!
=
(
s
+
1
)
n
n
!
,
{\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!},}
Example 3
edit
Let {a,b}=(-∞,0] and f(s)≡e2s . Then the sequence of iterated integrals of e2s is defined on (-∞,0] , and
F
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0
)
(
s
)
=
e
2
s
,
{\displaystyle F^{(0)}(s)=e^{2s},\ }
F
(
1
)
(
s
)
:=
∫
−
∞
s
F
(
0
)
(
u
)
d
u
=
∫
−
∞
s
e
2
u
d
u
=
e
2
s
2
,
{\displaystyle F^{(1)}(s):=\int _{-\infty }^{s}F^{(0)}(u)du=\int _{-\infty }^{s}e^{2u}du={e^{2s} \over 2},}
F
(
2
)
(
s
)
:=
∫
−
∞
s
F
(
1
)
(
u
)
d
u
=
∫
−
∞
s
e
2
u
2
d
u
=
e
2
s
2
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}},}
F
(
n
)
(
s
)
:=
∫
−
∞
s
F
(
n
−
1
)
(
u
)
d
u
=
∫
−
∞
s
e
2
u
2
(
n
−
1
)
d
u
=
e
2
s
2
n
,
{\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}},}
