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I'm struggling with the following calculus question.

Let there be two functions $f,g : [a, \infty) \to \mathbb R$ such that:

  1. $g$ is monotonic, differentiable and has a limit at zero

  2. $f$ is continuous such that $$\int_a^\infty f(x)dx < M \in \mathbb R$$

Prove that integral $$\int_a^{\infty} f(x)g(x)dx$$ converges.

While I do know how to prove the theorem using the Second Mean Value Theorem, I've got no idea how to prove it using integration by parts. How can this be done?

Any hints or leads will be greatly appreciated.

Thank you

davyjones
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vondip
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1 Answers1

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I assume you mean that $\lim\limits_{x\to\infty}g(x)=0$.

Let $F(x)=\int_a^xf(t)\,\mathrm{d}t$. Then, using the Riemann-Stieltjes integral $$ \begin{align} \lim_{b\to\infty}\int_a^bf(x)g(x)\,\mathrm{d}x &=\lim_{b\to\infty}\int_a^bg(x)\,\mathrm{d}F(x)\\ &=\lim_{b\to\infty}g(b)F(b)-g(a)F(a)-\lim_{b\to\infty}\int_a^bF(x)\,\mathrm{d}g(x)\\ &=0-0-\lim_{b\to\infty}\int_a^bF(x)\,\mathrm{d}g(x)\tag{1} \end{align} $$ Since $|F(x)|\le M$ and $g'$ doesn't change signs $$ \left|\int_a^bF(x)\,\mathrm{d}g(x)\right|\le M|g(a)-g(b)|\tag{2} $$ Thus, $$ \left|\lim_{b\to\infty}\int_a^bf(x)g(x)\,\mathrm{d}x\right|\le M|g(a)|\tag{3} $$

Addition from Comments

It was asked in a comment how we know that the limit in $(1)$ exists, since $(3)$ actually shows only that the integral is bounded in $b$. Using $(1)$, we need to show that $$ \lim\limits_{b\to\infty}\int_a^bF(x)\,\mathrm{d}g(x)\tag{4} $$ exists. The limit in $(4)$ exists because $$ \left|\int_{b_1}^{b_2}F(x)\,\mathrm{d}g(x)\right|\le M|g(b_1)|\tag{5} $$ and $\lim\limits_{b_1\to\infty}g(b_1)=0$.

robjohn
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  • Why does the limit $\lim_{b\to\infty}\int_a^bf(x)g(x),\mathrm{d}x$ exist? – user72870 Jul 20 '14 at 17:27
  • Integration by parts: $\int_a^bf(x)g(x),\mathrm{d}x=F(b)g(b)-F(a)g(a)-\int_a^bF(x)g'(x),\mathrm{d}x$ and $\lim\limits_{b\to\infty}F(b)g(b)=0$ and $\left|\int_{b_1}^{b_2}F(x)g'(x),\mathrm{d}x\right|\le M|g(b_1)|$. – robjohn Jul 20 '14 at 19:21
  • Sorry to insist, but it seems to me that you are just proving that $\int_a^bf(x)g(x),dx$ is bounded. – user72870 Jul 20 '14 at 20:07
  • No, I am showing that the limit exists. There are three parts on the right hand side: $$\tag*{}$$ 1. $$\lim_{b\to\infty}F(b)g(b)=0$$ 2. $$\lim\limits_{b\to\infty}F(a)g(a)=F(a)g(a)=0$$ 3. $$\lim_{b\to\infty}\int_a^bF(x)g'(x),\mathrm{d}x$$ which exists because $$\left|\int_{b_1}^{b_2}F(x)g'(x),\mathrm{d}x\right|\le M|g(b_1)|$$ – robjohn Jul 20 '14 at 20:25
  • Now it's fine +1 ;) – user72870 Jul 20 '14 at 20:58
  • Regarding the last line: Why do you let the lower bound go to $\infty$? Has it something to do with Cauchy? Didn't you show with $(2)$ and $(3)$ that the limit exists by showing it is bounded? – mdcq Jan 25 '18 at 22:52
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    @philmcole: The definition of a Cauchy sequence talks about $|a_k-a_n|$ going to $0$ as $k$ and $n$ go to $\infty$. This is the same thing; we're showing the integral converges by showing $$\left|\int_a^{b_2}f(x),\mathrm{d}x-\int_a^{b_1}f(x),\mathrm{d}x\right|=\left|\int_{b_1}^{b_2}f(x),\mathrm{d}x\right|\to0$$ as $b_1,b_2\to\infty$. – robjohn Jan 26 '18 at 02:44
  • @philmcole: Just because a sequence is bounded, does not mean it converges. – robjohn Jan 26 '18 at 02:47
  • @robjohn Thanks! But then we don't really need $(2)$ and $(3)$ anymore, when we can just show this because we then know it converges by Cauchy. – mdcq Jan 26 '18 at 09:09
  • @philmcole: Cauchy sequences are sequences. We are duplicating the idea since this is not a sequence. – robjohn Jan 26 '18 at 09:54
  • How do you know that $g'$ is Riemann integrable over $[a, b]$, and if you don't claim it is, how do you know that $\int \limits_a^b F(x) g'(x) , \mathrm{d} x$ exists? – Adayah Jul 20 '18 at 12:35
  • @Adayah: If $g'$ is not Riemann integrable, then $\int_a^bF(x)g'(x),\mathrm{d}x$ should be replaced by the Riemann-Stieltjes integral: $\int_a^bF(x),\mathrm{d}g(x)$. Each of the estimates above are valid with this replacement. – robjohn Jul 28 '18 at 00:52
  • Can I ask, f(x) is not continuous, so how does this integration by parts formula hold? – alan23273850 Feb 09 '22 at 14:57
  • @alan23273850: take $f=v'$ and $g=u$ in this. – robjohn Feb 09 '22 at 15:55