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The arithmetic mean of $y_i ... y_n$ is: $$\frac{1}{n}\sum_{i=1}^n~y_i $$

For a smooth function $f(x)$, we can find the arithmetic mean of $f(x)$ from $x_0$ to $x_1$ by taking $n$ samples and using the above formula. As $n$ tends to infinity, it becomes an integration: $$\int_{x_0}^{x_1} f(x)~dx \over x_1 - x_0$$

On the other hand, the geometric mean of $y_i ... y_n$ is: $$\left( \prod_{i=1}^n~{y_i}\right) ^{1/n}$$

Similarly, we can find the geometric mean of $f(x)$ by taking $n$ samples.

Here is my question: As $n$ tends to infinity, what do we call the resultant mathematical object? The geometric integration?

The geometric mean and the arithmetic mean, along with the quadratic mean (root mean square), the harmonic mean, etc, are special case of the generalized mean (with $p=0,1,2,-1$, respectively).

$$\left( \frac{1}{n}\sum_{i=1}^n~x_i^p\right) ^{1/p}$$

Do we have a generalized integration for different values of $p$?

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up vote 14 down vote accepted

You don't need to introduce a new concept for this. The geometric mean of $y_i$ is nothing but $\exp$ of the arithmetic mean of $\log y_i$, and this generalizes in the straightforward way to integration: $$\exp\left(\frac{\int_{x_0}^{x_1} \log f(x)dx}{\int_{x_0}^{x_1} dx}\right).$$ You can do the same thing with the generalized mean, replacing $\log$ and $\exp$ with raising to the power of $p$ and $1/p$ respectively. I'm not aware of whether this has a specific name, but it is quite similar to the concept of the $L^p$ norm of a function.

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The expression that Rahul wrote is the natural generalisation of $L^p$ norms when $p\to 0$. In particular, if you have a function $u$ such that $|u|^p$ is integrable on the interval for all $p\in[-\alpha,\alpha]$ (for example $u$ bounded above and below by positive numbers), then $\lim_{p\to 0} \left( \frac{1}{x_1-x_0}\int_{x_0}^{x_1}|u|^p dx\right)^{1/p}$ is precisely the expression Rahul wrote. –  Willie Wong May 4 '11 at 14:27
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There is a term for this (actually, more than one). It's called the product integral or multiplicative integral, and together with the corresponding derivative you get what's called product calculus or multiplicative calculus or non-Newtonian calculus. In addition, your idea of obtaining different versions of calculus by considering the generalized mean is discussed in this article by H. Vic Dannon.

The idea of product integration goes back at least to Vito Volterra in the late 1800s, and there are various applications of it. The standard reference is probably Dollard and Friedman's text in the Encyclopedia of Mathematics series, Product Integration with Application to Differential Equations. (The description says, "This book shows the beautiful simplifications that can be brought to the theory of differential equations by treating such equations from the product integral viewpoint.")

However, as Rahul Narain points out, it is easy to express the product integral in terms of usual integration. For that reason, some people don't believe it is really anything new.

And for more on the product integral, see one of the references in the Wikipedia page cited or this or this survey paper I wrote a few years ago.

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I do not know what you mean by generalized integration as integration is simply the concept of adding things up in a special way. It is not much different than normal summation. If I gave you some type of answer in terms of integrals it wouldn't be a generalization because it is in terms of integrals(it wouldn't be a new concept).

Now if your asking about some type of analog then you could come up with (1/L*int(f(x)^p))^(1/p) and this is actually used(almost at least)

http://en.wikipedia.org/wiki/Lp_space

As far as I know there is no thing that is analogous to products as integrals are to sums. But Lp spaces deal a lot with the sum you have given because of the way the metric is defined. i.e., your sum is simply the distance in these spaces(well, scaled at least).

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product integral would be the opposite of the form log base one of one: log base h of (f(xh)/ f(x), and the inverse. this super derivative can also be represented as x f('x)/f(x)

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Can you clarify what you mean? This makes no sense to me. –  robjohn Sep 30 '12 at 7:26
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