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S c a l a r c a s e
T o g g l e S c a l a r c a s e s u b s e c t i o n
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The Cauchy formula for repeated integration , named after Augustin-Louis Cauchy , allows one to compress n antiderivatives of a function into a single integral (cf. Cauchy's formula ).
Scalar case
[ edit ]
Let f be a continuous function on the real line. Then the n th repeated integral of f with base-point a ,
f
(
−
n
)
(
x
)
=
∫
a
x
∫
a
σ
1
⋯
∫
a
σ
n
−
1
f
(
σ
n
)
d
σ
n
⋯
d
σ
2
d
σ
1
,
{\displaystyle f^{(-n)}(x )=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n-1}}f(\sigma _{n})\,\mathrm {d} \sigma _{n}\cdots \,\mathrm {d} \sigma _{2}\,\mathrm {d} \sigma _{1},}
is given by single integration
f
(
−
n
)
(
x
)
=
1
(
n
−
1
)
!
∫
a
x
(
x
−
t
)
n
−
1
f
(
t
)
d
t
.
{\displaystyle f^{(-n)}(x )={\frac {1}{(n-1)!}}\int _{a}^{x}\left(x-t\right)^{n-1}f(t )\,\mathrm {d} t.}
Proof
[ edit ]
A proof is given by induction . The base case with n=1 is trivial, since it is equivalent to:
f
(
−
1
)
(
x
)
=
1
0
!
∫
a
x
(
x
−
t
)
0
f
(
t
)
d
t
=
∫
a
x
f
(
t
)
d
t
{\displaystyle f^{(-1)}(x )={\frac {1}{0!}}\int _{a}^{x}{(x-t)^{0}}f(t )\,\mathrm {d} t=\int _{a}^{x}f(t )\,\mathrm {d} t}
Now, suppose this is true for n , and let us prove it for n +1. Firstly, using the Leibniz integral rule , note that
d
d
x
[
1
n
!
∫
a
x
(
x
−
t
)
n
f
(
t
)
d
t
]
=
1
(
n
−
1
)
!
∫
a
x
(
x
−
t
)
n
−
1
f
(
t
)
d
t
.
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left[{\frac {1}{n!}}\int _{a}^{x}\left(x-t\right)^{n}f(t )\,\mathrm {d} t\right]={\frac {1}{(n-1)!}}\int _{a}^{x}\left(x-t\right)^{n-1}f(t )\,\mathrm {d} t.}
Then, applying the induction hypothesis,
f
−
(
n
+
1
)
(
x
)
=
∫
a
x
∫
a
σ
1
⋯
∫
a
σ
n
f
(
σ
n
+
1
)
d
σ
n
+
1
⋯
d
σ
2
d
σ
1
=
∫
a
x
[
∫
a
σ
1
⋯
∫
a
σ
n
f
(
σ
n
+
1
)
d
σ
n
+
1
⋯
d
σ
2
]
d
σ
1
{\displaystyle {\begin{aligned}f^{-(n+1)}(x )&=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}\,\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}\left[\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}\right]\,\mathrm {d} \sigma _{1}\\\end{aligned}}}
Note, the term within square bracket has n-times succesive integration, and upper limit of outermost integral inside the square bracket is
σ
1
{\displaystyle \sigma _{1}}
. Thus, comparing with the case for n=n , and replacing
x
,
σ
1
,
⋯
,
σ
n
{\displaystyle x,\sigma _{1},\cdots ,\sigma _{n}}
of the formula at induction step n=n with
σ
1
,
σ
2
,
⋯
,
σ
n
+
1
{\displaystyle \sigma _{1},\sigma _{2},\cdots ,\sigma _{n+1}}
respectively to obtain
∫
a
σ
1
⋯
∫
a
σ
n
f
(
σ
n
+
1
)
d
σ
n
+
1
⋯
d
σ
2
=
1
(
n
−
1
)
!
∫
a
σ
1
(
σ
1
−
t
)
n
−
1
f
(
t
)
d
t
{\displaystyle {\begin{aligned}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}&={\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}\left(\sigma _{1}-t\right)^{n-1}f(t )\,\mathrm {d} t\\\end{aligned}}}
Putting this expression inside the square bracket results in
=
∫
a
x
1
(
n
−
1
)
!
∫
a
σ
1
(
σ
1
−
t
)
n
−
1
f
(
t
)
d
t
d
σ
1
=
∫
a
x
d
d
σ
1
[
1
n
!
∫
a
σ
1
(
σ
1
−
t
)
n
f
(
t
)
d
t
]
d
σ
1
=
1
n
!
∫
a
x
(
x
−
t
)
n
f
(
t
)
d
t
.
{\displaystyle {\begin{aligned}&=\int _{a}^{x}{\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}\left(\sigma _{1}-t\right)^{n-1}f(t )\,\mathrm {d} t\,\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}{\frac {\mathrm {d} }{\mathrm {d} \sigma _{1}}}\left[{\frac {1}{n!}}\int _{a}^{\sigma _{1}}\left(\sigma _{1}-t\right)^{n}f(t )\,\mathrm {d} t\right]\,\mathrm {d} \sigma _{1}\\&={\frac {1}{n!}}\int _{a}^{x}\left(x-t\right)^{n}f(t )\,\mathrm {d} t.\end{aligned}}}
It has been shown that this statement holds true for the base case
n
=
1
{\displaystyle n=1}
.
If the statement is true for
n
=
k
{\displaystyle n=k}
, then it has been shown that the statement holds true for
n
=
k
+
1
{\displaystyle n=k+1}
.
Thus this statement has been proven true for all positive integers.
This completes the proof.
Generalizations and applications
[ edit ]
The Cauchy formula is generalized to non-integer parameters by the Riemann-Liouville integral , where
n
∈
Z
≥
0
{\displaystyle n\in \mathbb {Z} _{\geq 0}}
is replaced by
α
∈
C
,
ℜ
(
α
)
>
0
{\displaystyle \alpha \in \mathbb {C} ,\ \Re (\alpha )>0}
, and the factorial is replaced by the gamma function . The two formulas agree when
α
∈
Z
≥
0
{\displaystyle \alpha \in \mathbb {Z} _{\geq 0}}
.
Both the Cauchy formula and the Riemann-Liouville integral are generalized to arbitrary dimensions by the Riesz potential .
In fractional calculus , these formulae can be used to construct a differintegral , allowing one to differentiate or integrate a fractional number of times. Differentiating a fractional number of times can be accomplished by fractional integration, then differentiating the result.
References
[ edit ]
Augustin-Louis Cauchy : Trente-Cinquième Leçon . In: Résumé des leçons données à l’Ecole royale polytechnique sur le calcul infinitésimal . Imprimerie Royale, Paris 1823. Reprint: Œuvres complètes II(4 ), Gauthier-Villars, Paris, pp. 5–261.
Gerald B. Folland, Advanced Calculus , p. 193, Prentice Hall (2002). ISBN 0-13-065265-2
External links
[ edit ]
R e t r i e v e d f r o m " https://en.wikipedia.org/w/index.php?title=Cauchy_formula_for_repeated_integration&oldid=1236083276 "
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