Prove there exists $\alpha \ge 0$ s.t $\int_0^\alpha f(x)dx =\int_0^\infty g(x)dx$ given that $f,g\ge 0$, $F(x)$ diverges and $G(x)$ converges This is one of the problems we got as an assignment:
if $f(x),g(x)$ are two integrable functions on $[0,t]$ for any $0<t\in \Bbb{R}$.
and suppose that:

*

*$f(x)\ge 0,\ g(x)\ge 0$, for all $x\ge 0$

*$\int_0^\infty f(x)dx$ diverges and $\int_0^\infty g(x)dx$ converges.

Prove that there exists some $\alpha \ge 0$ such that $\int_0^\alpha f(x)dx =\int_0^\infty g(x)dx$
So I obviously see that if $\int_0^\infty g(x)dx=0$ then $\alpha =0$
I also know that both functions are non-negative, so they are increasing.
therefor if $\int_0^\infty g(x)dx = S$ then $S>0$
But now I don't understand how this gets me to the a value that im trying to get to...
also with integrals I try to somehow visualize and I don't understand how this is true at all, I mean if $\int_0^\infty f(x)dx$ diverges how can I find such  specific value? I mean if it diverges it can "start diverging" at any point..

If $\int_0^{\infty}g(x)dx=0$,then clearly $\alpha=0$
If not , then let $\int_0^{\infty}g(x)dx=L\gt 0$  (since $g\ge 0$)
Now consider $F(x)=\int_0^{x}f(t)dt$ .
Then $F(0)=0$ and $F(x)$ is continous and $\displaystyle\lim_{x\to \infty}F(x)=\infty$  (Why? Because $F(x)$ is increasing and thus the above limit diverges to infinity)
Can you do from here?
