# Is there a “continuous product”?

Is there a "continuous product" which is the limit of the discrete product $\Pi$, just like the integral $\int$ is the limit of summation $\sum$.

Thanks!

• Sure; take the exponential of the integral of the logarithms. – Qiaochu Yuan Apr 26 '12 at 20:33
• You could take the logarithm of $f$ and take the integral of that, then take the exponent. You'll have a hard time defining this operator if $f$ is allowed to be negative, since it is unclear when multiplying a continuum of $-1$ whether the product should be $1$ or $-1$. But the logarithm works for positive $f$ – Thomas Andrews Apr 26 '12 at 20:35
• Products are inherently continuous. This does not even make sense. – The Great Duck Jun 4 '16 at 6:05
• @Typhon That's like saying summation is inherently continuous, so integration does not make any sense. The apostrophes around "continuous product" indicate OP is using that terminology in a looser sense to convey the idea of multiplying over a "continuous family" of factors, much like integration is intuitively representative of a summation over a "continuous" domain of terms. – Sir Jective Nov 18 '17 at 2:47
• @MonstrousMoonshiner that was in response to one of my questions being marked as an exact duplicate and I was likely responding purely to the title and not the body which is perfectly acceptable. Furthermore, why are you here responding to a post and comment from over a year ago? – The Great Duck Nov 18 '17 at 3:32

## 1 Answer

Yes, this is known as the Product integral which you can read about on this Wikipedia link.

• And when you have commutative multiplication it is computed fairly easily as shown in the link. But when you have noncommutative multiplication (such as matrix multiplication) the product integral becomes more interesting and useful. – GEdgar Apr 26 '12 at 21:26
• @GEdgar I have never seen any real application of this. Perhaps you have a reference for further study. – AD. Apr 26 '12 at 21:45