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

Solution 1:

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 $\lim\limits_{b_1\to\infty}g(b_1)=0$.