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

Thanks!

user64742
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chaohuang
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    Sure; take the exponential of the integral of the logarithms. – Qiaochu Yuan Apr 26 '12 at 20:33
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    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
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    @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. – silvascientist Nov 18 '17 at 02:47

1 Answers1

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

AD - Stop Putin -
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    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
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    @GEdgar I have never seen any real application of this. Perhaps you have a reference for further study. – AD - Stop Putin - Apr 26 '12 at 21:45