|
|
| Line 1: |
Line 1: |
| 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]]).
| | Hi there, I am Andrew Berryhill. Office supervising is my occupation. What me and my family love is bungee leaping but I've been taking on new issues lately. I've usually cherished residing in Alaska.<br><br>Here is my webpage psychic phone - [http://ustanford.com/index.php?do=/profile-38218/info/ ustanford.com] - |
| | |
| ==Scalar case==
| |
| Let ''ƒ'' be a continuous function on the real line. Then the ''n''th repeated integral of ''ƒ'' based at ''a'',
| |
| | |
| :<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>,
| |
| | |
| is given by single integration | |
| | |
| :<math>f^{(-n)}(x) = \frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t</math>.
| |
| | |
| 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]]:
| |
| | |
| :<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>;
| |
| | |
| where
| |
| | |
| :<math>f^{(-1)}(a) = \int_a^a f(t)\,\mathrm{d}t = 0</math>.
| |
| | |
| 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,
| |
| | |
| :<math>
| |
| \begin{align}
| |
| 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 \\
| |
| &= \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 \\
| |
| &= \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 \\
| |
| &= \frac{1}{n!} \int_a^x \left(x-t\right)^n f(t)\,\mathrm{d}t
| |
| \end{align}
| |
| </math> | |
| | |
| The proof follows.
| |
| | |
| ==Applications==
| |
| | |
| 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]]).
| |
| | |
| ==References==
| |
| | |
| *Gerald B. Folland, ''Advanced Calculus'', p. 193, Prentice Hall (2002). ISBN 0-13-065265-2
| |
| | |
| ==External links==
| |
| *{{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}}
| |
| | |
| [[Category:Integral calculus]]
| |
| [[Category:Theorems in analysis]]
| |
Hi there, I am Andrew Berryhill. Office supervising is my occupation. What me and my family love is bungee leaping but I've been taking on new issues lately. I've usually cherished residing in Alaska.
Here is my webpage psychic phone - ustanford.com -