Dirichlet's test for uniform convergence of improper integrals

Question

Dirichlet's test states that $F(x)=\int_a^\infty f(x,y)g(x,y)dy$ converges uniformly if primitive function of $f(x,y)$ is uniformly bounded for every value of parameter $x$, and $g(x,y)$ is monotonic and for $y\to\infty$ converges uniformly to $0$. My question is: Does $g(x,y)$ have to be monotonic for all $y\in[a,\infty)$ or is it enough if $g(x,y)$ is monotonic on an interval $[b,\infty)$, where $b>a$? For example, we know that $g(x,y)=\frac{y}{1+y^4}$ converges for $y\to\infty$ uniformly to $0$, but it is not monotonic on $[0,\infty)$. Can be Dirichlet's test used in this case when the lower bound of integration is $0$?

Answer 1

Suppose $f\cdot g$ is Riemann integrable on finite intervals $[a,c]$. There is a Cauchy criterion for uniform convergence of the improper integral over $[a,\infty)$ whereby for every $\epsilon > 0$ there exists $C(\epsilon) > 0$, independent of $x \in D$, such that for all $c_2 > c_1 > C(\epsilon)$ and for all $x\in D$ we have

$$\tag{1}\left|\int_{c_1}^{c_2} f(x,y) g(x,y) \, dy \right| < \epsilon$$

As this criterion is used to prove Dirichlet's test for uniform convergence, it is enough that $g(x,y)$ is monotone with respect to $y$ eventually.

To see this, assume that $g(x,y)$ is monotone decreasing with respect to $y$ for all $x \in D$ and only for $y \in [b, \infty)$. With $c_2 > c_1 > b$ and $x$ fixed, by the second mean value theorem for integrals there exists $\xi(x) \in (c_1,c_2)$ such that

$$\tag{1}\int_{c_1}^{c_2} f(x,y) g(x,y) \, dy = g(x,c_1) \int_{c_1}^{\xi(x)}f(x,y) \, dy$$

By the uniform boundedness condition for $f$, there exists $B > 0$ (independent of $c_1$) such that for all $x \in D$,

$$\tag{3}\left|\int_{c_1}^{\xi(x)}f(x,y) \, dy\right|< B$$

Since $g(x,y) \underset{y \to \infty}\longrightarrow 0$ uniformly for $x \in D$ there exists $C(\epsilon) > b $ such that for all $c_1> C(\epsilon)$,

$$\tag{4}|g(x,c_1)| < \epsilon/B$$

Altogether, (2), (3), and (4) imply that (1) holds and the improper integral converges uniformly.