Asymptotic behavior of a certain oscillatory integral

Question

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:

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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 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.

Answer 1

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*}

Answer 2

$$I(x):=\int_{0}^{\infty}\frac{e^{i r}}{r^{\frac{1}{2}}}\int_{0}^{\infty}\frac{e^{-s}}{s^{\frac{1}{2}}}\frac{r}{sx+\sqrt{sxr}+r}ds dr.$$ To aid the asymptotic analysis, I regularize $I(x)$ by multiplying the integrand with $e^{-\epsilon r}$, sending $\epsilon\rightarrow 0^+$ at the end. I also change variables from $sx\mapsto s$, $$I_\epsilon(x)=\frac{1}{\sqrt{x}}\int _0^{\infty }dr\int _0^{\infty }ds\,\frac{\sqrt{r} \exp \left(i r-\epsilon r-s/x\right)}{\sqrt{s} \left(\sqrt{r s}+r+s\right)}.$$ For large $x$ I may neglect the term $-s/x$ in the exponent.$^\ast$ The integrals over $s$ and $r$ can then be evaluated in closed form, $$\lim_{\epsilon\rightarrow 0^+}I_\epsilon(x)=\frac{4 \pi i }{3 \sqrt{3x}}+{\cal O}(1/x).$$ This matches a numerical check.

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$^\ast$ The order $1/x$ error incurred by replacing $e^{-s/x}$ by 1 follows from the expansion $x^{-1/2}\int_1^\infty s^{-3/2}e^{-s/x}\,ds=2x^{-1/2}-2\sqrt{\pi}x^{-1}+2x^{-3/2}+\cdots.$