Prove Uniform integrability of $ \int_{0}^{\infty} e^{-e^{is}\cdot x^2} dx $ I want to prove Uniform Integrability of the following improper parametric integral (in the parameter $s$:
$$ \int_{0}^{\infty} e^{-e^{is}\cdot x^2} dx $$ 
for $$s \in (\frac{-\pi}{2},\frac{\pi}{2})$$
I don't want to take recourse to advanced concepts like lebesgue integration or measure theory. 
 A: Write
\begin{align}
\int_0^\infty e^{-e^{is}x^2}\,dx&=\int_0^\infty e^{-\cos(s)x^2}e^{-i\sin(s)x^2}\,dx\\
&=\int_0^\infty e^{-\cos(s)x^2}\cos(\sin(s)x^2)\,dx-i\int_0^\infty e^{-\cos(s)x^2}\sin(\sin(s)x^2)\,dx.
\end{align}
It suffices to show each of the integrands on the right hand side is uniformly integrable. I will provide the proof for the first integral; the second is almost identical in argument. By symmetry, we need only consider $0\le s<\frac\pi2$; for $0\le s<\frac\pi4$, we have
$$|e^{-\cos(s)x^2}\cos(\sin(s)x^2)|\le e^{-\frac1{\sqrt{2}}x^2}$$
which is integrable and independent of $s$, so we have uniform integrability in this interval. For $\frac\pi4\le s<\frac\pi2$, we use the substitution $u=\sin(s)x^2$ and integration by parts:
\begin{align}
\int_a^b e^{-\cos(s)x^2}\cos(\sin(s)x^2)\,dx &= \frac1{2\sqrt{\sin(s)}}\int_A^Be^{-\cot(s)u}\frac{\cos(u)}{\sqrt{u}}\,du\\
&=\frac1{2\sqrt{\sin(s)}}\left[\frac{F(s,B)}{\sqrt{B}}-\frac{F(s,A)}{\sqrt{A}}-\frac12\int_A^B\frac{F(s,u)}{u^{3/2}}\,du\right]
\end{align}
where $A=\frac{a}{\sqrt{\sin(s)}}$, $B=\frac{b}{\sqrt{\sin(s)}}$, and $\frac{\partial F}{\partial u}(s,u)=e^{-\cot(s)u}\cos(u)$. Using two more applications of integration by parts shows that we can set
$$F(s,u):=e^{-\cot(s)u}\sin^2(s)\big(\sin(x)-\cot(s)\cos(x)\big),$$
and so in particular $|F(s,u)|\le2$ for all $s\in[\frac\pi4,\frac\pi2)$ and $u\ge0$. This implies
\begin{align}
\left|\int_M^\infty e^{-\cos(s)x^2}\cos(\sin(s)x^2)\,dx\right| &\le\frac1{2\sqrt{\sin(s)}}\left[\frac{2\sqrt{\sqrt{\sin(s)}}}{\sqrt{M}}+\int_{\frac{M}{\sqrt{\sin(s)}}}^\infty\frac1{u^{3/2}}\,du\right]\\
&\le\frac{C}{\sqrt{M}}
\end{align}
for some constant $C$ independent of $s$. This completes the proof.
