Fourier transform of $e^{-i\lambda\sqrt{1+x^2}}$ - asymptotics for $\lambda$?

Question

As the title says: I want to compute the Fourier transform (in the distributional sense) of $f(x)=e^{-i\sqrt{1+x^2}}$, $x\in \mathbb{R}^n$ - say $n=1$ for the moment. I have no idea how to get it done: I have tried with the usual machinery ("ODE approach" like for the Gaussian, explicit computations, estimates for oscillatory integrals, complex analysis) but I am not able to come up with a useful suggestion. Any hints or partial results? Is there any result in order to connect this transform with the one of $e^{-i|x|}$, at least in asymptotic terms? In fact, I am interested in an estimate for the parameter $\lambda >0$ in the Fourier transform of $f_{\lambda}(x)=e^{-i\lambda\sqrt{1+x^2}}$. Given the analogy with $$ \mathcal{F}[e^{-i\lambda|\cdot|}](y)=\frac{\lambda}{y^2+\lambda^2},$$I expect a similar decay for $\lambda$ in $\mathcal{F}[f_{\lambda}]$. Is there any way to prove that?

Answer 1

We have $$ F(k;\lambda)=\int_{-\infty}^{+\infty}e^{-ikx-i\lambda\sqrt{1+x^2}}dx=\int_{-\infty}^{+\infty}e^{-ikx}e^{-i\lambda\varphi(x)}dx\,, $$ with $$ \varphi(x)=\sqrt{1+x^2}\,. $$ Since $$ \varphi(x)=1+\frac{x^2}{2}+\cdots $$ for small $x$, i.e. near the stationary point $x_0=0$ of $\varphi(x)$. The stationary phase approximation method gives, as $\lambda\to\infty$, $$ F(k;\lambda)\sim e^{-ik x_0}e^{-i\lambda\varphi(x_0)}\sqrt{\frac{2\pi}{i\varphi''(x_0)}}=e^{-i\lambda}\sqrt{\frac{2\pi}{i\lambda}}\,, $$ up to corrections $o(\lambda^{-1/2})$. So, indeed the decay as $\lambda\to\infty$ is a power-like behavior $|F(k,\lambda)|\approx 2\pi/\sqrt{\lambda}$ as OP seems to suggests. Comparing with the Fourier transform of $e^{-\lambda|x|}$, which decays as $1/\lambda$ instead, we see that the different behavior of the function near $x=0$ indeed influences the specific power appearing in the asymptotics, but not the qualitative power-law behavior.

For $\lambda=0$ instead $$ F(k;0)=\int_{-\infty}^{+\infty}e^{-ikx} dx=2\pi \delta(k)\,. $$

Answer 2

OP's integrand in the title for $\lambda>0$ is not integrable, not even as an improper integral. For this reason we will in this answer assume that $${\rm Im}(\lambda)~<~0$$ to make OP's integral convergent$^1$. OP is interested in the asymptotics for ${\rm Re}(\lambda)\to \infty$. OP's integral then reads:

$$F(k;\lambda)~:=~\int_{\mathbb{R}} \mathrm{d}x ~e^{ -i (\lambda\sqrt{x^2+1}+kx)} ~=~\int_{\mathbb{R}} \mathrm{d}x ~e^{ -i\lambda\sqrt{x^2+1}}\cos(kx) ~=~2\int_{\mathbb{R_+}} \mathrm{d}x ~e^{-i\lambda\sqrt{x^2+1}}\cos(kx)$$ $$~\stackrel{w=\sqrt{x^2+1}-1}{=}~2e^{-i\lambda}\int_{\mathbb{R_+}} \mathrm{d}w\frac{1+w}{\sqrt{w(2+w)}} ~e^{-i\lambda w}\cos(k\sqrt{w(2+w)})$$ $$~\stackrel{w=\frac{v}{i\lambda}}{=}~\frac{2e^{-i\lambda}}{\sqrt{i\lambda}}\int_{\mathbb{R_+}} \mathrm{d}v\frac{1+\frac{v}{i\lambda}}{\sqrt{v(2+\frac{v}{i\lambda})}} e^{-v}\cos(k\sqrt{\frac{v}{i\lambda}(2+\frac{v}{i\lambda})})$$ $$~=~\sqrt{\frac{2}{i\lambda}}e^{-i\lambda}\int_{\mathbb{R_+}} \frac{\mathrm{d}v}{\sqrt{v}}~e^{-v}\left( 1+ (\frac{3}{4}-k^2)\frac{v}{i\lambda}+O(\lambda^{-2})\right)$$ $$ ~=~\sqrt{\frac{2\pi}{i\lambda}}e^{-i\lambda} \left( 1+ (\frac{3}{4}-k^2)\frac{1}{2i\lambda}+O(\lambda^{-2})\right). $$

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$^1$ A similar regularization seems needed to salvage OP's last formula. If we assume ${\rm Im}(\lambda)<0$, then the LHS of OP's last formula reads $$ \int_{\mathbb{R}} \mathrm{d}x ~e^{ -i (\lambda|x|+kx)} ~=~\int_{\mathbb{R}_+} \mathrm{d}x ~e^{ -i (\lambda+k)x} ~+~(k\leftrightarrow -k) ~=~\frac{i}{\lambda+k} ~+~(k\leftrightarrow -k) ~=~\frac{2i\lambda}{\lambda^2-k^2}. $$ This may be further rewritten with the help of the Sokhotski–Plemelj theorem: $$\begin{align}\int_{\mathbb{R}} \mathrm{d}x ~e^{ -i \{({\rm Re}(\lambda) - i0^+)|x|+kx\}} &~=~ \frac{i}{{\rm Re}(\lambda) \!-\! i0^+\!+\!k} ~+~(k\leftrightarrow -k)\cr &~=~P\frac{i}{{\rm Re}(\lambda)\!+\!k} -\pi\delta({\rm Re}(\lambda)\!+\!k) ~+~(k\leftrightarrow -k).\end{align}$$

Answer 3

The Fourier transform of $e^{-i \lambda |x|}$ is not $\lambda/(\omega^2 + \lambda^2)$, it's $$(e^{-i \lambda |x|}, e^{i \omega x}) = \frac {2 i \lambda} {\omega^2 - \lambda^2} + 2 \pi |\lambda| \hspace {1px} \delta(\omega^2 - \lambda^2),$$ where $1/(\omega^2 - \lambda^2)$ is the p.v. functional.

Let $f(x) = e^{-i \lambda \sqrt {x^2 + 1}}$. Since $f(x) - e^{-i \lambda |x|} \to 0$ when $x \to \pm \infty$, it follows that $\mathcal F[f]$ is also a singular functional. The value it gives when applied to a test function can be estimated as $$(\mathcal F[f], \phi) = (f, \mathcal F[\phi]) = \\ \int_{\mathbb R} f(x) \, \mathcal F[\phi](x) \, dx = e^{-i (\lambda + \pi/4)} \sqrt {\frac {2 \pi} \lambda} \, \mathcal F[\phi](0) + o(\lambda^{-1/2}) = \\ e^{-i (\lambda + \pi/4)} \sqrt {\frac {2 \pi} \lambda} \int_{\mathbb R} \phi(x) \, dx + o(\lambda^{-1/2}), \quad \lambda \to \infty.$$