Proving Abel-Dirichlet's test for convergence of improper integrals using Integration by parts

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

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, $$\left|\lim_{b\to\infty}\int_a^bf(x)g(x)\,\mathrm{d}x\right|\le M|g(a)|\tag{3}$$

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 $\lim\limits_{b_1\to\infty}g(b_1)=0$.
• 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
• @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 2:44