Given $$\int_0^\infty \cos{x^2}\arctan{\alpha x} \,dx \ \text{on} \ \mathbb{R}$$ establish its' uniform convergence.
My thoughts are:
On
Using Fubini's theorem is enough to prove that Fresnel's integral $(b=\infty)$ converges, and since the integral clearly converges for any finite $b$, it is bounded.
Consider the quantity:
$$A=\int_0^{\pi}\cos t\sum_{n=0}^{\infty}\Big(\frac{1}{\sqrt{t+2n\pi}}-\frac{1}{\sqrt{t+(2n+1)\pi}}\Big)$$ The series in the integrand can be rewritten as:
$$\begin{align} &f(t)=\sum_{n=0}^{\infty}\Big(\frac{1}{\sqrt{t+2n\pi}}-\frac{1}{\sqrt{t+(2n+1)\pi}}\Big)\\&=\frac{\pi}{\sqrt{t}\sqrt{t+\pi}(\sqrt{t}+\sqrt{t+\pi})}+\sum_{n=1}^{\infty}\frac{\pi}{\sqrt{t+2n\pi}\sqrt{t+(2n+1)\pi}(\sqrt{t+2n\pi}+\sqrt{t+(2n+1)\pi})}\\&<\frac{1}{\sqrt{t}}+\frac{1}{4\sqrt{2\pi}}\sum_{n=1}^{\infty}\frac{1}{n^{3/2}} \end{align}$$
so it clearly converges for every $t>0$. Then we conclude that $A$ converges as well since we can write
$$A=\int_{0}^{\pi/2}dx f(x)\cos x-\int_0^{\pi/2}dxf(\pi-x)\cos x$$
and from the bound established above both integrals are bounded. By Fubini's theorem we are allowed to interchange summation and integration. Now we can show by appropriate manipulations (the steps are elementary and left to the reader) that
$$A=\int_{0}^{\infty}dx\frac{\cos x}{\sqrt{x}}=2\int_{0}^{\infty}dx\cos(x^2)$$
which proves that the Fresnel integral itself is finite.
For nonnegative integers $n$, let $$c_n = \int_{\sqrt{2n\pi}}^{\sqrt{2(n+1)\pi}} \cos(x^2)\; dx $$ By change of variables $x = \sqrt{t + 2n \pi}$, we get $$ c_n = \int_0^{2\pi} \frac{\cos(t)}{2\sqrt{2n\pi + t}}\; dt $$ Since $\int_0^{2\pi} \cos(t)\; dt = 0$ and $$\frac{1}{\sqrt{2n\pi + t}} = \frac{1}{\sqrt{2n\pi}} + O(n^{-3/2})$$ we get $c_n = O(n^{-3/2})$.