# 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 the summation operator $$\sum$$?

Thanks!

• Sure; take the exponential of the integral of the logarithms. Commented Apr 26, 2012 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 $f$ Commented Apr 26, 2012 at 20:35
• @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. Commented Nov 18, 2017 at 2:47