Uniform and absolute convergence of improper integral

Question

What is an example of an improper integral , $\int_a^\infty f(u,v)du$, that converges uniformly for $v$ is some subset $S$, but where $\int_a^\infty|f(u,v)|du$ converges pointwise but NOT uniformly on $S$?

When Weierstrass’s Test shows that Riemann improper integral with a parameter $v$, $\int_a^\infty f(u,v)du,$ is uniformly convergent it also shows that $\int_a^\infty |f(u,v)|du,$ is uniformly convergent.

I can also find examples where the integral is uniformly convergent but not absolutely convergent, like $\int_a^\infty \sin(vu)/u du$ where $v \in [c,\infty)$.

Or is it always true that a uniformly and absolutely convergent improper integral must also be uniformly convergent with the absolute value of the integrand taken?

Answer 1

No -- uniform and absolute convergence does not imply uniform-absolute convergence. For a counterexample, consider the improper integral

$$\int_0^\infty \frac{v}{u^2 + v^2} \sin u \, du,$$

where $v \in [0, \infty)$ .

It is straightforward to show the integral is absolutely convergent since for $v > 0$,

$$ \int_0^ \infty\left|\frac{v}{u^2 + v^2} \sin u \right| \, du \leqslant \int_0^\infty \frac{v}{u^2 + v^2} \, du = \int_0^\infty \frac{1/v}{1 + (u/v)^2} \,du = \pi/2$$

Also, it can be shown to be uniformly convergent by the Dirichlet test.

However, we do not have uniform-absolute convergence. For all $n \in \mathbb{N}$ we have, $$\int_{n\pi}^{2n\pi} \frac{v}{u^2 + v^2} |\sin u| \, du \geqslant \frac{v}{(2n\pi)^2 + v^2}\int_{n\pi}^{2n\pi} |\sin u| \, du \\ = \frac{v}{(2n\pi)^2 + v^2}\sum_{k=0}^{n-1}\int_{n\pi + k\pi}^{n\pi + k\pi + \pi} |\sin u| \, du \\= \frac{2nv}{4\pi^2n^2 + v^2},$$

Choosing a sequence of points $v_n = n \in [0, \infty)$, we see that for all $n$

$$\int_{n \pi}^{2n\pi} \frac{n}{u^2 + n^2} |\sin u| \, du \geqslant \ \frac{2n^2}{4\pi^2n^2 + n^2} = \frac{2}{4\pi^2+1} > 0.$$

Thus, the convergence fails to be uniform for all $v \in [0,\infty).$