A few days ago I asked a question: If the integral from one to plus infinity of a continues function exists, does it mean that the limit when x approaches infinite of this function is zero?
The answer to this is: no. In this post I want to show that even stronger claim is true. I will built a function that is continues in the interval [1,plus infinty), and unbounded both from above and below, and has no limit when x approaches plus infinity, but the integral of this function from 1 to plus infinity exists.
Firstly, lets define the following function:
n is a natural number. This function is continues, and has no limit when x approaches infinity - it is easy enough to see. Also for any natural number k, f(k)=k. Therefore f is not bounded from above. The integral of this function is the area below the graph. For this particular function it is very easy to find. The graph is just triangles separated by intervals on which f is zero. Therefore all we need is to find the area of all these triangles. The area is different from triangle to triangle. For the n-th triangle the area is:
So the integral is equal to:
This new function is continues, because it is a multiplication of two functions. It doesn't has a limit in infinity, and it is not bounded from above or below, because sin(k) is not 0 for all natural k. Therefore, g(k)=sin(k)*k and it is perfectly possible to find a k large enough for this to be bigger than any positive M or smaller than any negative P. But the integral still exists because of two simple theorems.
The first theorem says that if is finite so is (The limits of the integration is 1 and plus infinity.).
The second theorem says that if h(x)>f(x)>0 for all x, than if the integral of g(x) exists so is the integral of f(x), if f and g are both continues.
Since |sin(x)|<=1 for all x, we get that |g(x)|=|sin(x)*f(x)|=|sin(x)|*|f(x)|<=|f(x)|=f(x). And by using the above mentioned theorems we get that the integral of g(x) exists.