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| The '''Cauchy formula for repeated integration''', named after [[Augustin Louis Cauchy]], allows one to compress ''n'' [[antidifferentiation]]s of a function into a single integral (cf. [[Antiderivative#Techniques of integration|Cauchy's formula]]). | | The name of the writer is Jayson. Office supervising is exactly where my primary income arrives from but I've usually wanted my own company. To play lacross is some thing he would never give up. Her family members lives in Ohio.<br><br>Here is my web page online psychic reading ([http://chorokdeul.co.kr/index.php?document_srl=324263&mid=customer21 chorokdeul.co.kr]) |
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| ==Scalar case==
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| Let ''ƒ'' be a continuous function on the real line. Then the ''n''th repeated integral of ''ƒ'' based at ''a'',
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| :<math>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</math>,
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| is given by single integration | |
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| :<math>f^{(-n)}(x) = \frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t</math>.
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| A proof is given by [[mathematical induction|induction]]. Since ''ƒ'' is continuous, the base case follows from the [[Fundamental Theorem of Calculus|Fundamental theorem of calculus]]:
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| :<math>\frac{\mathrm{d}}{\mathrm{d}x} f^{(-1)}(x) = \frac{\mathrm{d}}{\mathrm{d}x}\int_a^x f(t)\,\mathrm{d}t = f(x)</math>;
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| where
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| :<math>f^{(-1)}(a) = \int_a^a f(t)\,\mathrm{d}t = 0</math>.
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| Now, suppose this is true for ''n'', and let us prove it for ''n+1''. Apply the induction hypothesis and switching the order of integration,
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| :<math>
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| \begin{align}
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| 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 \\
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| &= \frac{1}{(n-1)!} \int_a^x \int_a^{\sigma_1}\left(\sigma_1-t\right)^{n-1} f(t)\,\mathrm{d}t\,\mathrm{d}\sigma_1 \\
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| &= \frac{1}{(n-1)!} \int_a^x \int_t^x\left(\sigma_1-t\right)^{n-1} f(t)\,\mathrm{d}\sigma_1\,\mathrm{d}t \\
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| &= \frac{1}{n!} \int_a^x \left(x-t\right)^n f(t)\,\mathrm{d}t
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| \end{align}
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| </math> | |
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| The proof follows.
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| ==Applications==
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| In [[fractional calculus]], this formula can be used to construct a notion of [[differintegral]], allowing one to differentiate or integrate a fractional number of times. Integrating a fractional number of times with this formula is straightforward; one can use fractional ''n'' by interpreting (''n''-1)! as Γ(''n'') (see [[Gamma function]]).
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| ==References==
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| *Gerald B. Folland, ''Advanced Calculus'', p. 193, Prentice Hall (2002). ISBN 0-13-065265-2
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| ==External links==
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| *{{cite web|author=Alan Beardon|url=http://nrich.maths.org/public/viewer.php?obj_id=1369|title=Fractional calculus II|publisher=University of Cambridge|year=2000}}
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| [[Category:Integral calculus]]
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| [[Category:Theorems in analysis]]
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