I'm struggling with the following calculus question.
Let there be two functions $f,g : [a, \infty) \to \mathbb R$ such that:
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$g$ is monotonic, differentiable and has a limit at zero
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$f$ 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
Best Answer
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 $\lim\limits_{b_1\to\infty}g(b_1)=0$.