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Cauchy's repeated integration formula is as follows: Let $f^{(-n)}$ be a continuous function on the real line. Then the $n^{th}$ repeated integral of $f$ based at $a$,

$$f^{(-n)}(x) = \int_a^x \int_a^{\sigma_1} \cdots \int_a^{\sigma_{n-1}} f(\sigma_{n}) \, \mathrm{d}\sigma_{n} \cdots \, \mathrm{d}\sigma_2 \, \mathrm{d}\sigma_1$$

is given by single integration

$$f^{(-n)}(x) = \frac{1}{(n-1)!} \int_a^x\left(x-t\right)^{n-1} f(t)\,\mathrm{d}t$$

source: http://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration

Now, consider something like: $$I(c)= \int_0^c \int_b^2 \int_0^a ({2-x})\, \mathrm{d}x\, \mathrm{d}a\, \mathrm{d}b$$

The lower limits are different ($0,b,0$) and I want to apply Cauchy's formula, what do I do? Is there any generalized variant for this?

I tried a change of variables, but was unsuccessful:

Set $a=2-t.$ So when $a=b$,we have $t=2-b$, and when $a=2$, we have $t=0.$

Also, $\, \mathrm{d}a = -\, \mathrm{d}t$.

Substituting $t$ into the integral and reversing the limits of integration by utilizing the negative sign , I got: $$I(c)= \int_0^c \int_0^{2-b} \int_0^{2-a} ({2-x})\, \mathrm{d}x\, \mathrm{d}a\, \mathrm{d}b$$

Now, I applied Cauchy's formula since all the lower bounds on the integrals are equal. According to the formula, my answer should be: $$I(c)=\frac{1}{2} \int_0^c {(c-t)}^2 ({2-t})\, \mathrm{d}t $$

However, when calculated step-by-step, integrating thrice, $I(c)$ turns out to be different and does not match. Why so?

Thank you for your time.

EDIT: Besides, is it even possible to find a closed-form for this one? If not, can you suggest why?

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