[Eratosthenes]

I think that anything with an absolute value in it is suspect, because the function will have at least one point where it is not differentiable.

Taking ideas from everyone else's solutions, here's mine:

[HIDE="answer"]C(1-cos(2.pi.x))/e^(x^2), where C = 1/(sqrt(pi).(1-e^-(pi^2))), where C normalizes the integral. [/HIDE]

[Latro]

Another solution, slightly simpler than my last one; its origins are really just my last one to be honest:

( Show Spoiler )

(-e^(2*pi)*sin(2*pi*x+3*pi/2))/(pi*(1+x^2))

[SirJac]

( Show Spoiler )

For the first requirement, use a trig function like sin(pi*x) will give your zeros at all integers and it and all derivatives are defined for all real numbers.

For the second requirement, we need to introduce exponential decay so the integral will converge. For this I chose the core of the probability density function: e^(-x^2) which will cause exponential decay on both sides of zero and it and it's derivatives are defined everywhere.

The third requirement can be fulfilled by instroducing some scalar constant that ensures that the definate integral over the entire real real line converges to 1, call it C.

Put it together and f(x)=C*sin(pi*x)*e^(-x^2)

Problem, the integral of sin over those limits is zero. So try a variation:

f(x)=C*(1-cos(2pi*x))e^(-x^2)

value of the integral when c = 1 from Wolfram Alpha is (1-e^(-pi^2))*sqrt(pi) and the inverse of this is C so the final function is then:

f(x) = (1-cos(2pi*x))e^(-x^2)/((1-e^(-pi^2))*sqrt(pi))

ps thnx for the link to Wolfram Alpha Monte, looks like a very useful site!