Consider following integral $\displaystyle \int_{0}^{\infty}\sin(yx^2)\arctan(yx) dx$. What can we say about uniformly convergent of it and about continuity?
I've tried to maximize it by $\arctan(yx) \le \frac{\pi}{2}$ but then i've got an integral , which doesn't converge (but bounded).
Any hints?
There is a generalization of the Abel test for series to improper integrals that states that
$$\int_a^\infty f(x,y) g(x,y) \, dx$$
is uniformly convergent if (1) $g$ is monotonic in $x$ and bounded and (2) if $\int_a^\infty f(x,y) \, dx $ is uniformly convergent. The proof uses the second mean value theorem for integrals and the boundedness of $g$ to show that the Cauchy criterion
$$\left|\int_c^d f(x,y)g(x,y) \, dx\right| < \epsilon $$
is satisfied for sufficiently large $d > c$.
Condition (1) is satisfied with $g(x,y) = \arctan(yx)$ for all $y$.
Try to confirm condition (2). Hint: This requires restriction of the set of values for $y$ to obtain uniform convergence. Note that $\int_0^\infty \sin (y x^2) \, dx = O(1/\sqrt{|y|})$ for $y \neq 0.$
