Asymptotic form of the integral $\int_{0}^{\infty} dx ~ \sqrt{x^2 + wx} ~ e^{-ixs}$ for $s \to \infty$

Question

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?

Answer 1

I believe that in order for the integral to make sense, we should consider $s$ to be complex with $\text{Im}\,s <0$. In fact, Fourier (or Laplace) transforms make in general the most sense in the context of complex calculus.

So let me rephrase your question:

• find the asymptotic behavior of $$\begin{equation} I(s)=\int_{0}^{\infty} \!dx \, \sqrt{x^2 + wx} \, e^{-ixs} \end{equation}$$ for $|s| \to \infty$ and the wedge in the complex plane for which this result applies (see Stokes phenomenon)

The asymptotic analysis can be performed by the method of steepest decent. In fact we can deform the integration over $x\in[0,\infty)$ into the lower half-plane of the complex $x$ plane. In particular, we have with $y=i x$ $$I(s) = -i \int_{0}^{\infty} \!dy \, \sqrt{-y^2- i w y} \, e^{-s y} .$$ For $s\to\infty$, the integral is dominated for $y$ close to 0. We can thus expand the square root and obtain $$I(s) \sim -i \sqrt{-i w} \int_0^\infty \!dy \, \Bigl[y^{1/2} + O(y^{3/2}) \Bigr] e^{-s y} \sim - \tfrac12 \pi^{1/2} e^{i\pi/4} w^{1/2} s^{-3/2} \tag{1}$$ with the next term proportional to $s^{-5/2}$. As explained above the result is valid for $\text{Im}\,s<0$.

Answer 2

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.