# 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