Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question

1 Answer 1

Hint: If $f' > 0$, then $$y > x_0 \implies f(y) > f(x_0)$$ Do you see why, and how to use this fact?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.