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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

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


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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Start again with $I(c)=\int_0^c\int_b^2\int_0^a(2-x)dxdadb$.


That is probably the most general method to applying Cauchy's formula. Notably the left integral is easy, since there is no $b$ to integrate with, and we apply Cauchy's formula on the right.

The step you messed up at was forgetting that the upper bound of the integral must be the same as the variable you are integrating with respect to. Obviously, $2-a\ne a$, which is why you can't apply the formula.

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