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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a few questions from my h.w, I hope someone can help me.

the question is:

$f:[a,\infty) \rightarrow \mathbb{R}$ is a continuous and periodic function, with period of $T>0$ .

$g:[a,\infty) \rightarrow \mathbb{R}$ is a monotonic function and $\lim \limits_{x\rightarrow\infty} g(x) = 0$

Assume that $\displaystyle \int_{a}^{a+T} f(x) dx = 0$, and that $g$ is differentiable and its derivative is continuous. Prove that $\displaystyle\int_{a}^{\infty} f(x)g(x) dx$ converges.

I hope the question is clear. Thank you

share|cite|improve this question
There're directions in the FAQ section how to properly write mathematics in this site with LaTeX... – DonAntonio Apr 20 '13 at 15:39
up vote 0 down vote accepted

Hint: (many details need to be added) $$ F(x)=\int_a^xf(x)\,\mathrm{d}x\tag{1} $$ Show that $F$ is periodic with period $T$. Furthermore, since $g$ is monotonic, $$ \begin{align} \int_a^\infty|g'(x)|\,\mathrm{d}x &=\left|\int_a^\infty g'(x)\,\mathrm{d}x\right|\\[6pt] &=|g(a)|\tag{2} \end{align} $$ Finally, $$ \begin{align} \int_a^bf(x)g(x)\,\mathrm{d}x &=\int_a^bg(x)\,\mathrm{d}F(x)\\ &=g(b)F(b)-g(a)F(a)-\int_a^bF(x)g'(x)\,\mathrm{d}x\\ &=g(b)F(b)-\int_a^b F(x)g'(x)\,\mathrm{d}x\tag{3} \end{align} $$ Show that $(3)$ has a limit as $b\to\infty$.

share|cite|improve this answer
I think there is a problem. Did you use $\lim_{x\to\infty}g'(x)=0$ in your proof? This condition is not given. – xpaul Apr 22 '13 at 14:48
@xpaul: nope. But we do know that $\int_a^\infty|g'(x)|\,\mathrm{d}x=|g(a)|$ and $F(x)$ is bounded. – robjohn Apr 22 '13 at 15:06

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.