Integral of products

I have come up with a way to find continuous products the way an integral finds continuous sums. i.e., by multiplying very  small intervals. The way to find the continuous product of a function is: e^∫[ln f(x)] dx. Input limits a and b into the integral. The ln function is undefined at 0, and a continuous product is 0 at 0, so, if a function crosses 0 between a and b, the continuous product is irrelevant. Negative numbers are a curious case, because if there are an odd number of negative terms, the function is negative, whereas if there are an even number, it is positive. Since we're taking infinitely small intervals, we can't determine whether there are an odd or even number of negative terms. One could argue that this means that there the number must be even, since it's broken up into infinite parts, therefore, the region that it is negative for will also be divisible by 1/(infinity/2). If one assumes this stance, then the correct function is e^∫ln |fx| dx, if not, put a negative sign in front, It is an interesting question, though. 

Twin prime proof

Break up all numbers into sets of 6, because every 3rd odd number is composite. Therefore, we can conduct twin prime tests on pairs of numbers that are mod 6 ±1. The odds of a number being prime is approximately equal to (1-1/P1)*(1-1/P2)*(1-1/P3)...(1-1/Pn) Where Pn is the largest prime less than or equal to √x. Since these numbers are written in the form 6x ± 1, they're not divisible by 2 or 3. 

The probability that both these numbers (6x±1) are prime can be found using a variation of the earlier method. The odds that one of these numbers (picked before) is not divisible by 5 (it's not divisible by 2 or 3),  is 4/5.  The probability that a number not divisible by 5+2 is divisible by 5 is 3/4. Therefore, the probability that the numbers 6x±1 are both not divisible by 5 is 3/5. If one extends this for all primes, the probability of the pair 6x±1 not being divisible by a number P is (P-2)/P. Since the largest smallest prime factor of a number n can't be greater than √n, the probability of a pair 6x±1 being prime are approximately equal to (3/5)*(5/7)....*(Pn-2/Pn) Where Pn is the largest prime below √6x+1. 

This sequence yields a greater result than the sequence (3/5)*(5/7)....*(On-2/On) Where On is the largest odd number below √6x+1. That sequence's solution is 3/On. Therefore, the lowest possible probability that numbers of the form 6x±1 are both prime is equal to 3/√6x+1. Since we break the numbers up into sets of 6, the probability of this being true for any set of 2 numbers is 1/2√x+1. 

Therefore, the minimum number of paired primes between 1 and infinity is approximately equal to ∫1/(2[(x+1)] with limits 1 and infinity. Since the sum of probabilities ~ the average number of instances, as x tends to infinity. The integral of the above function=√(x+1), which becomes infinity as x reaches infinity. Therefore, there are an infinite number of twin primes.