TITLE: Asymptotic form of the integral $\int_{0}^{\infty} dx ~ \sqrt{x^2 + wx} ~ e^{-ixs}$ for $s \to \infty$
QUESTION [6 upvotes]: I would like to find an asymptotic form of the following integral when $s \to \infty$ ($s$ and $w$ are positive)
\begin{equation}
\int_{0}^{\infty} dx ~ \sqrt{x^2 + wx} ~ e^{-ixs}
\end{equation}
I have tried a change of variable first $xs = y$ to get
\begin{equation}
\int_{0}^{\infty} \frac{dy}{s}~ \sqrt{\frac{y^2}{s^2} + \frac{wy}{s}} e^{-iy}
\end{equation}
I suspect also that the main contribution will come from around $y=0$, and I know that the rapid oscillations of the integrand will cancel most of the integral outside that region, so I could perform the integral only up to $L$ instead of $\infty$ (although I am not sure what $L$ should be chosen). This, I think, would lead to the asymptotics (first order is enough) for the integral being
$$
\sqrt{w/s} \int_0^L dy ~ \sqrt{y} e^{-iy}
$$
Is this correct and or can be improved?

REPLY [3 votes]: In the quadrant $[0,R]\cup Re^{-i\pi[0,1/2]}\cup[-iR,0]$ there are no singularities, so we can adjust the contour anywhere in that quadrant.
Furthermore, the part of the integral past any fixed $x$ decays exponentially in $s$, so we only need pay attention to the part near $0$:
$$
\begin{align}
&\lim_{t\to0^-}\int_0^\infty\sqrt{x^2+wx}\,e^{-ix(s+it)}\,\mathrm{d}x\\
&=-\int_0^\infty\sqrt{x^2+iwx}\,e^{-xs}\,\mathrm{d}x\\
&=-\frac{1+i}{\sqrt2}\sqrt{w}\int_0^\infty\sqrt{x}\sqrt{1-ix/w}\,e^{-xs}\,\mathrm{d}x\\
&=-\frac{1+i}{\sqrt2}w^2\int_0^\infty\sqrt{x}\sqrt{1-ix}\,e^{-xsw}\,\mathrm{d}x\\
&\sim-\frac{1+i}{\sqrt2}w^2\int_0^\infty\sqrt{x}{\small\left(1+\frac{\frac12}1(-ix)+\frac{\frac12(-\frac12)}{1\cdot2}(-ix)^2+\frac{\frac12(-\frac12)(-\frac32)}{1\cdot2\cdot3}(-ix)^3+\dots\right)}\,e^{-xsw}\,\mathrm{d}x\\
&=-\frac{1+i}{\sqrt2}w^2{\small\left(\frac{\sqrt\pi}2(sw)^{-3/2}-\frac i2\frac{3\sqrt\pi}4(sw)^{-5/2}+\frac18\frac{15\sqrt\pi}8(sw)^{-7/2}+\frac i{16}\frac{105\sqrt\pi}{16}(sw)^{-9/2}+\dots\right)}\\
&=-\frac{1+i}{2\sqrt2}\sqrt{\frac{\pi w}{s^3}}{\small\left(1-i\frac3{4sw}+\frac{15}{32(sw)^2}+i\frac{105}{128(sw)^3}+\dots\right)}\\
\end{align}
$$
The first term agrees with Fabian's formula.