# Summation is to integration, as the large product operator is to…? [duplicate]

The integral is defined many ways but one that I am aware of is the Riemann Integral(?) which looks sorta like $\sum^n_{i=0} f(a +i\frac {b-a}n)*\frac {b-a} n$. An interesting thought is "is there a similar construct built from $\prod$". Then my second thought was "what would it be used for".

So I thought about and figured maybe the $f(x)$ term would look the same, but then have it raised to the power of $dx$ as that second term is more commonly known as.

Is this sort of thing done and used or is it just a silly construction I thought up randomly?

• This may be somewhat of a cop out answer, but I would regard it as just the exponential of an integral. – EuYu Jun 4 '16 at 5:33
• @EuYu It wouldnt be any answer. I never asked how the two proposed integrals related. I asked whether or not the prooduct one has been /studied/ or is /useful/. – user64742 Jun 4 '16 at 5:52
• A continuous product doesnt sound anything like what Im saying. – user64742 Jun 4 '16 at 5:57
• @TheGreatDuck Well, admittedly it's not a very good answer, but your construction is more or less the exponential of an integral, as Zev's link points out (although I didn't know there was a name for it). Variations of this construction show up in many places in mathematics, for example as integrating factors or as the Dyson series. So I wouldn't say it's a completely useless answer either. – EuYu Jun 4 '16 at 6:11
• Oh ok, @EuYu. I thought that was just an identity he was stating to ease integral calculations. – user64742 Jun 4 '16 at 16:25

$$\prod_{a}^{b}f(x)^{dx}=\lim_{\Delta x\to 0}\prod f(x_i)^{\Delta x}=\exp\left(\int_a^b\ln f(x)\,dx\right)$$
• I'm personally accustomed to the "Type I" definition myself, since it's consistent with the expression of the product derivative as $\exp\left(\frac{\mathrm d}{\mathrm dx}\log f(x)\right)$. This is an interesting preprint. (For the hardcore: there are fractional and $q$-generalizations of all of these entities, too.) – J. M. isn't a mathematician Jun 4 '16 at 5:40
• @TheGreatDuck, you can intuit it yourself if you recall that $\prod\limits_{k=1}^n a_k = \exp\left(\sum\limits_{k=1}^n \log a_k\right)$, with the necessary restrictions on the $a_k$. – J. M. isn't a mathematician Jun 4 '16 at 5:51
• @TheGreatDuck, it's just a restatement/generalization of the usual laws of exponents: $p^x p^y=p^{x+y}$. – J. M. isn't a mathematician Jun 4 '16 at 5:57