With the Cauchy formula for multiple integration , named after the French mathematician Augustin Louis Cauchy , certain -th iterated integrals of a function can be expressed in a single integral.
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Cauchy's formula
Let be a continuous function on the real axis.
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Then the -th iterated integral of is at the point
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{\ displaystyle I ^ {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}}
given by the following integral:
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{\ displaystyle I ^ {n} (x) = {\ frac {1} {(n-1)!}} \ int _ {a} ^ {x} \ left (xt \ right) ^ {n-1} f (t) \, \ mathrm {d} t}
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proof
The proof is obtained by complete induction . Since is continuous, the beginning of induction can be derived from the fundamental theorem of analysis .
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{\ displaystyle {\ frac {\ mathrm {d}} {\ mathrm {d} x}} I ^ {1} (x) = {\ frac {\ mathrm {d}} {\ mathrm {d} x}} \ int _ {a} ^ {x} f (t) \, \ mathrm {d} t = f (x)}
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in which
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{\ displaystyle I ^ {1} (a) = \ int _ {a} ^ {a} f (t) \, \ mathrm {d} t = 0}
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Let's say the formula is correct for . Now it is time to prove that the formula is correct.
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{\ displaystyle {\ frac {\ mathrm {d}} {\ mathrm {d} x}} \ left [{\ frac {1} {n!}} \ int _ {a} ^ {x} \ left (xt \ right) ^ {n} f (t) \, \ mathrm {d} t \ right] = {\ frac {1} {(n-1)!}} \ int _ {a} ^ {x} \ left (xt \ right) ^ {n-1} f (t) \, \ mathrm {d} t}
The derivation of the integral can be derived from the Leibniz rule.
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{\ displaystyle {\ begin {aligned} I ^ {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} {\ 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 (xt \ right ) ^ {n} f (t) \, \ mathrm {d} t \ end {aligned}}}
That concludes the proof.
Riemann-Lioville integral
Cauchy's formula has a limitation, the factorial, which is only defined for positive natural numbers. The Riemann-Liouville integral allows multiple integration not only for real but also for complex numbers by exchanging with , where the gamma function denotes:
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{\ displaystyle I ^ {\ alpha} f (x) = {\ frac {1} {\ Gamma (\ alpha)}} \ int _ {a} ^ {x} f (t) (xt) ^ {\ alpha -1} \, dt}
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Applications
With a few transformation steps it is also possible to find a formula for the -th derivative.
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Here you can also find applications such as:
Web links
Individual evidence
Gerald B. Folland, Advanced Calculus , p. 193, Prentice Hall (2002). ISBN 0-13-065265-2
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