If you integrate 1/x from 0 to infinity, you will get infinity as your answer, even though 1/x is asymptotic to the x-axis. The normal distribution function however, which is also asymptotic to the x-axis, has a definite value, 0.5, for the integral from 0 to infinity. Is there a way of telling which asymptotic functions will have a definite sum to infinity from their equations? Is it do with the rate at which they converge with the x-axis?

So, let's take the function f(x)=x⁴ e^-x as an example. For large enough R, x⁴ < e^x/2 (do you know how to prove this?) and so for x > R f(x) < e^x/2e^-x=e^-x/2, and we know ò₁^¥ e^-x/2 dx is finite. If x < R then x⁴ < R⁴ and e^-x < 1 so f(x) < R⁴. So ò₁^¥ x⁴ e^-x dx is finite (using the above).

Good functions h(x) whose integrals from 1 to infinity converge are x^t for t < -1, e^-x and e^-x2. The best function g(x) whose integral from 1 to infinity is infinite is 1/x. This collection of functions is usually enough to prove that any integral of a positive function that you come across converges or diverges. It's more difficult to show things for functions which are sometimes positive and sometimes negative, but if you can show that ò₁^¥ |f(x)| dx is finite then also ò₁^¥ f(x) dx is finite.

Could you explain how to prove e^x/2 > x⁴ for some number R?

Hmm, it depends on what you know. Do you know that e^x =1+x+x² /2+...+xⁿ /n!+... ? If not, write back and I'll see if I can prove it another way. If so, then for x> 0 we have e^x > xⁿ⁺¹ /(n+1)! (since this is just one term of the series). If x> (n+1)! then e^x > (x/(n+1)!)xⁿ > xⁿ since x/(n+1)!> 1. So for any integer n, if x> (n+1)! then e^x > xⁿ . Obviously if you are thinking about x/2 instead of x you have to make a slight change to that argument.