Let $x>0$ and consider the integral
$$I(x):=\int\_0^\infty \frac{e^{i r}}{r^{\frac{1}{2}}} \int\_0^\infty \frac{e^{-s}}{s^{\frac{1}{2}}} \frac{r}{sx+\sqrt{sxr}+r} \, ds \, dr.$$


I am trying to determine the asymptotic behavior of $I(x)$ as $x\rightarrow+\infty$.


Note that $\lim\_{x\rightarrow+\infty}I(x)=0$.
Here is why:




---


Notice that  

$(r,s)\mapsto{e^{i r}}{r^{-\frac{1}{2}}}{e^{-s}}{s^{-\frac{1}{2}}}\frac{r^{2}}{s^2 x^2+sxr+r^2}$
is dominated by ${r^{-\frac{1}{2}}}{e^{-s}}{s^{-\frac{1}{2}}}$ which is an $L^{1}([0,1]\times [0,+\infty[)$ function.


On $[1,+\infty[\times [0,+\infty[$ we need an integration by parts w.r.t. $r$ to justify the passing of the limit as $x\rightarrow+\infty$. Doing so, we get
\begin{align}
& \int\_1^\infty\frac{e^{i r}}{r^{\frac{1}{2}}} \int\_0^\infty \frac{e^{-s}}{s^{\frac{1}{2}}} \frac{r}{sx+\sqrt{sxr}+r} \, ds \, dr \\[6pt]
= {} & {i e^{i }}\int\_{0}^{\infty}\frac{e^{-s}}{s^{\frac{1}{2}}}\frac{1}{sx+\sqrt{sx}+1} \, ds \\[6pt]
& {}+\frac{1}{2}\int\_1^\infty \frac{e^{i r}}{r^{\frac{3}{2}}} \int\_0^\infty \frac{e^{-s}}{s^{\frac{1}{2}}} \frac{r}{sx+\sqrt{sxr}+r} \, ds \, dr \\[6pt]
& {}-\int\_1^\infty \frac{e^{i r}}{r^{\frac{3}{2}}} \int\_0^\infty \frac{e^{-s}}{s^{\frac{1}{2}}} \left(\frac{r}{sx+\sqrt{sxr}+r} -\frac{\frac{1}{2}\sqrt{sxr}+r}{(sx+\sqrt{sxr}+r)^2} \right)\, ds \, dr \tag{$\*\*$}
\end{align}


Simple remarks on the formula $(\*\*)$:


(1) The limit (taken at computing the boundary terms) and derivative under the $s$-integral are both justified by
the fact that $e^{-s}/\sqrt{s}\in L^1 ([0,+\infty[)$.


(2) A factor $1/r$ has been pulled outside the $s$-integral.


More importantly,


(3) The integrand in this formula, as a function of $(r,s)$, is absolutely dominated by
$r^{-3/2}s^{-1/2}e^{-s}$ which is an $L^1( \left[1,+\infty\right[\times [0,+\infty[ )$ function.




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What is the precise the asymptotic behavior of $I(x)$ as $x\to+\infty\text{ ?}$


I am probably wrong, but I tried rescaling/changing order of integration/Taylor-expanding the rational factor in any variable. Nothing seems to work.



Asymptotic behavior of a certain oscillatory integral
We can evaluate $I(x)$ explicitly, and then asymptotically.


Indeed, using the substitution $s=ru/x$, we get
\begin{equation\*}
 I(x)=\frac1{\sqrt x}\lim\_{R\to\infty}J\_R(x), \tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
 J\_R(x)&:=\int\_0^R dr\,e^{ir}\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}e^{-ru/x} \\ 
 &=\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\int\_0^R dr\,e^{(i-u/x)r} \\ 
 &=\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\frac{1-e^{(i-u/x)R}}{u/x-i}. 
\end{aligned}
\end{equation\*}
Next, (i) for any real $u,x>0$ we have $\dfrac{1-e^{(i-u/x)R}}{u/x-i}\to\dfrac1{u/x-i}$ as $R\to\infty$, (ii) for any real $u,x,R>0$ we have $\Big|\dfrac{1-e^{(i-u/x)R}}{u/x-i}\Big|\le\dfrac2{|u/x-i|}\le2$, and (iii) for any real $x>0$ we have $\displaystyle{\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\,2<\infty}$.


So, by dominated convergence,
\begin{equation\*}
 \lim\_{R\to\infty}J\_R(x)=J(x), \tag{2}\label{2}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
 &J(x):=\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\frac1{u/x-i} \\ 
 &=\frac{x \left(-18 \ln x+\pi \left(8 i \sqrt{3} x+(9-9 i) \sqrt{2} \sqrt{x}-\frac{18
 \sqrt[4]{-1}}{\sqrt{x}}+4 \sqrt{3}+9 i\right)\right)}{18 (-1+x (x-i))} 
\end{aligned}
\tag{3}\label{3}
\end{equation\*}
(note that the integrand in \eqref{3} is rational in $\sqrt u\,$, so that one can use partial fraction decomposition to get \eqref{3}).
So, as $x\to\infty$,
\begin{equation\*}
 J(x)\to c:=\frac{4 i \pi }{3 \sqrt{3}},
\end{equation\*}
so that, by \eqref{1},
\begin{equation\*}
 I(x)\sim \frac c{\sqrt x}.
\end{equation\*}


