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I'm dealing with a problem for improper integrals reads the following:

Let $f(x)$ be a function defined on $[a,+\infty)$, monotonic down to $0$ as $x \to +\infty$. Prove that the following integrals $\int_{a}^{+\infty} f(x)dx$ and $\int_{a}^{+\infty} \cos^2 x.f(x)dx$ are convergent or disvergent simultaneously.

I'm quite confused with some words monotonic down to $0$. Once I thought that $f(x)$ is simply non-increasing (means if $a \le b$ then $f(a) \ge f(b)$) so that $f(x) \ge \lim_{x \to +\infty}f(x)=0, \quad \forall x \in [a,+\infty)$. Therefore, $$0\le \cos^2 x.f(x) \le f(x), \quad \forall x \in [a,+\infty).$$ This means that, the integrals $\int_{a}^{+\infty} f(x)dx$ and $\int_{a}^{+\infty} \cos^2 x.f(x)dx$ are convergent or disvergent simultaneously.

Is my solution correct? Otherwise, if is just equipped the assumption that $\lim_{x \to +\infty}f(x)=0$ without the non-increasing property. How can I come up with another proof?

user53541
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  • From the inequality $0\le f(x)\cos^2x\le f(x)$ you can deduce only that if $\int_a^\infty f(x),dx$ is convergent so is $\int_a^\infty f(x)\cos^2x,dx$, and that if $\int_a^\infty f(x)\cos^2x,dx$ is divergent, so is $\int_a^\infty f(x),dx$. – Julián Aguirre Mar 17 '14 at 17:23
  • @JuliánAguirre You're right. My proof had lots of mistakes. By reviewing some materials I testify that $f$ is monotonic down to $0$ as $x \to +\infty$ just means $f$ is monotonic and $\lim_{x \to +\infty}f(x)=0$ – user53541 Mar 17 '14 at 17:41

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Outline: For the harder direction, without loss of generality we may assume that we are integrating from $0$. Let $I_n=\int_{n\pi}^{(n+1)\pi} f(x)\cos^2 x \,dx$.

The sum $\sum I_n$ converges. Let $J_n=\int_{n\pi}^{n\pi+\pi/4} f(x)\cos^2 x\,dx$. Then $\sum J_n$ converges.

Let $K(n)=\int_{n\pi}^{(n+1)\pi} f(x)\,dx$. Conclude that $\sum K(n)$ converges.

André Nicolas
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  • Dear @AndréNicolas, what do you mean by Let $I_n=\int_{n\pi}^{(n+1)\pi} f(x)\cos^2 x f(x),dx$? – user53541 Mar 18 '14 at 02:52
  • I mean that I have a typo. Couldn't make up my mind whether to put $f(x)$ in front or behind, so did both. Thanks. – André Nicolas Mar 18 '14 at 02:55
  • I'm trying to follow your outline. First, suppose $\int_{0}^{+\infty} \cos^2 xf(x)dx$ converges then the sum $\sum I_n$ converges. Now I have two questions: 1. Why does $\sum J_n$ converge? Since we haven't proved $f(x)\ge 0$ for all $x \ge 0$, so we don't have $J_n \le I_n$. 2. If $\sum J_n$ does converge. Why $\sum K(n)$ converges? – user53541 Mar 18 '14 at 03:06
  • In your proof of the convergence of $\sum J_n$, you will have to use the fact that $f$ decreases monotonically to $0$. Well, at this stage you use very little of that, just the fact that it implies that $f(x)\ge 0$. Later you will need more. For your second question, here we will use decreasing to $0$. – André Nicolas Mar 18 '14 at 03:17