# Proving an improper integral diverges

Let $f:\mathbb R \to \mathbb R$, with $f \in C^1$ and $f'>0$. Suppose there exists $x_0 \in \mathbb R$ such that $f(x_0)>0$.

Prove that $\space \space$ $\int_0^{+\infty} f(t)dt$ $\space$ is divergent.

I don't have a any idea how could I prove this.

I thought about series, and one necessary condition for a series to converge, which is $lim_{n \to +\infty} a_n=0$, maybe I could think of an analogue condition for this integral, i.e, $lim_{x \to +\infty} f(x)=0$. I am not so sure if what I am saying is correct, if I follow this path, I would not be using the hypothesis $f \in C^1$. I would appreciate any suggestions on how could I solve the problem.

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Hint: If $f' > 0$, then $$y > x_0 \implies f(y) > f(x_0)$$ Do you see why, and how to use this fact?
Hmm, if $x_0<0$, then $0<f(x_0)<f(0)<f(y)$ $\forall y>0$. If I consider the function $g(x)=f(0)$ for all $x\geq 0$, then $\lim_{m \to +\infty} \int_0^m g(t)dt=+\infty$, so the integral diverges. By the comparison test, I would conclude that the original improper integral diverges. If $x_0>0$, I could express the original integral as the sum of two integrals and conclude the same thing for $\int_{x_0}^{+\infty} f(t)dt$. Is this correct? I didn't use that $f$ is $C^1$. –  user100106 Dec 18 '13 at 4:49