Uniform and absolute convergence of improper integral 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?  
 A: 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 $v > 0$,
$$ \int_0^ \infty\left|\frac{v}{u^2 + v^2} \sin u \right| \, du \leqslant \int_0^\infty \frac{v}{u^2 + v^2} \, du  = \int_0^\infty \frac{1/v}{1 + (u/v)^2} \,du = \pi/2.$$
and 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 $v \in [0,\infty).$
