Nth order order integration help

[Jhenrique]

If exist differentiation until the nth order, so, why "no exist" integration until the nth order too? I never saw a quadruple or quintuple integral, and if exist, it's always with respect to different variables. Why not difine an integral so?

$$\int\int\int f(x)dx^3$$

 

[mfb]

It is possible, I just don't know of any applications.

 

[arildno]

Why bother?
Even the first integral is non-unique, what point would it be to gain a long, polynomial tail from your subsequent antidifferentiations??

 

[Jhenrique]

There's a good reason for knowing solve this integral: http://en.wikipedia.org/wiki/Jerk_(physics)

 

[Mark44]

If exist differentiation until the nth order, so, why "no exist" integration until the nth order too? I never saw a quadruple or quintuple integral, and if exist, it's always with respect to different variables. Why not difine an integral so?

$$\int\int\int f(x)dx^3$$

I've never seen one written this way; i.e., with dx³. The usual way things are done is to have different variables of integration, something like this:
$$\int_a^b \int_c^d \int_e^f f(x, y, z) dz dy dx$$

or even like this:
$$\int_a^b \int_c^d ~...~\int_e^f f(x_1, x_2, ..., x_n) dx_1~ dx_2~...~ dx_n$$
Here we're integrating over a subset of Rⁿ.

 

[Curious3141]

It is possible to define a repeated integral, although the notation used is usually $dx...dx$ (n times) rather than $dx^n$. The nth such repeated integral can be denoted $f^{-n}(0)$.

And as long as the constants of integration at every step can be justifiably "ignored", e.g. $f^{-1}(0) = ... = f^{-n}(0) = 0$, then it's easy to derive and prove a simple formula for the general repeated integral. See: http://mathworld.wolfram.com/RepeatedIntegral.html

You can derive it with integration by parts and prove the form by induction. Wiki also has something on this: http://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration

The general formula has an application in defining fractional integration.