Jtdylktuy

Dear Math enthusiasts,

I'm trying to see if I can find an analytical expression for the Fourier transform of a Gaussian pulse $p(tau) = {rm e}^{-Btau^2}$ that is distorted by a hyperbolic distortion of the form $tau(t) = sqrt{t^2+a^2}-a$. What I mean is I look at the function $p(tau(t))$ and I try to transform this time $t$ to frequency.

This gives a Fourier integral of the following form $$G(omega) = int_{-infty}^infty {rm e}^{-Bleft(sqrt{t^2+a^2}-aright)^2} {rm e}^{-jmath omega t} {rm d}t.$$

Substituting the square-root doesn't help much, since it appears back when resubstituting $t$. I therefore tried to substitute $t = a sinh(x)$ instead. Assuming I didn't make a stupid mistake, this gives $$G(omega) = aint_{-infty}^infty {rm e}^{-Bleft(a cosh(x)-aright)^2} {rm e}^{-jmath omega a sinh(x)} cosh(x) {rm d}x.$$

From [*], eqn (2.3) I know that $int_{0}^infty {rm e}^{a cosh(x)} cosh(x) {rm d}x = K_0(a)$ where $K_0(a)$ is the modified Bessel function of the second kind. As the integrand is symmetric, it should be no problem to extend this to $(-infty,infty)$. But still I'm not sure this is enough to solve the integral. Mathematica has refused to give me anything, but it's possible I used it in a wrong way. I tried manipulating the exponent to bring everything to one hyperbolic trig function, but I failed.

Any hints how I can proceed?

edit: Note that since $p(tau(t))$ is even symmetric, $G(omega)$ is real-valued and even and we can get rid of complex numbers altogether, i.e., $$G(omega) = int_{-infty}^infty {rm e}^{-Bleft(sqrt{t^2+a^2}-aright)^2} cos( omega t) {rm d}t = 2 int_{0}^infty {rm e}^{-Bleft(sqrt{t^2+a^2}-aright)^2} cos( omega t) {rm d}t.$$ Does this make it easier? I'm not sure.

HINT

Firstly,
$$begin{aligned}
&G(omega,a,B)=aintlimits_{-infty}^infty e^{-a^2Bleft(sqrt{left(frac taright)^2+1}-1right)^2}e^{-jaomegafrac ta},mathrm ddfrac ta=aintlimits_{-infty}^infty e^{-a^2B(sqrt{t^2+1}-1)^2}e^{-jaomega t},mathrm dt,\
&G(omega,a,B)=aH(aomega,a^2B),\
end{aligned}$$
$$H(Omega,c)=intlimits_{-infty}^infty e^{-c(sqrt{t^2+1}-1)^2}e^{-jOmega t},mathrm dt.tag1$$

Let
$$tau(t) = sqrt{t^2+1}-1,quad tauin[0,infty),$$
so
$$t(tau)=pmsqrt{(tau+1)^2-1}=pmsqrt{tau^2+2tau},quad tin(-infty,infty),$$
$$dt=pmdfrac{tau+1}{sqrt{tau^2+2tau}},$$
$$begin{aligned}
&H(Omega,c)=intlimits_{-infty}^{-1}e^{-ctau^2(t)}e^{-jOmega t},mathrm dt
+intlimits_{-1}^infty e^{-ctau^2(t)}e^{-jOmega t},mathrm dt\
&=-intlimits_{-infty}^0e^{-ctau^2}e^{jOmegasqrt{tau^2+2tau}}dfrac{tau+1}{sqrt{tau^2+2tau}},mathrm dtau
+intlimits_{-1}^infty e^{-ctau^2}e^{-jOmega sqrt{tau^2+2tau}}dfrac{tau+1}{sqrt{tau^2+2tau}},mathrm dtau,
end{aligned}$$
$$H(Omega,c)=2intlimits_0^{infty} e^{-ctau^2} cos(Omegasqrt{tau^2+2tau})dfrac{tau+1}{sqrt{tau^2+2tau}},mathrm dtau.tag2$$
By parts:
$$begin{aligned}
&H(omega,c) = dfrac2Omegaintlimits_0^{infty} e^{-ctau^2},dsin(Omegasqrt{tau^2+2tau})\
&=-dfrac2Omega e^{-ctau^2} sin(Omegasqrt{tau^2+2tau})bigg|_0^infty
+dfrac{4c}Omegaintlimits_0^{infty} tau e^{-ctau^2}
sin(Omegasqrt{tau^2+2tau}),dtau,end{aligned}$$
$$H(Omega,c)=dfrac{4c}Omegaintlimits_0^{infty} tau e^{-ctau^2}sin(Omegasqrt{tau^2+2tau}),dtau.tag3$$
I don't see how to obtain the closed form of the integral $(3).$

At the same time, for $c<<Omega$ the exponent in the formula $(1)$ allows polynomial approximation.

If $cgtrsimOmega,$ then the integral $(3)$ can be presented as $1D$ series, using the series expansion for sine expression in the form of
$$sin(Omegasqrt{x^2+2x}) = sqrt{dfrac x2}Omegaleft(2 + dfrac{3-4Omega^2}6 x + dfrac{16 Omega^4 - 120 Omega^2 - 15}{240} x^2 + dfrac{-64Omega^6 + 1680Omega^4 - 1260Omega^2 + 315}{20160} x^3 + dfrac{256Omega^8 - 16128Omega^6 + 90720Omega^4 + 15120Omega^2 - 14175}{2903040} x^4 + dfrac{-1024Omega^{10} + 126720Omega^8 - 2217600Omega^6 + 1663200Omega^4 - 623700Omega^2 + 1091475}{638668800}x^5 + dotsright).
$$
(the asymptotic form of
$$sin(Omegasqrt{x^2+2x}) = Omega{cos((x+1)Omega)left(-dfrac1{2x} + dfrac 1{2x^2} + dfrac{Omega^2-30}{48x^3} - dfrac{Omega^2 - 14}{16x^4} - dfrac{Omega^4 - 540 Omega^2 + 5040}{3840x^5} + dfrac{Omega^4 - 220 Omega^2 + 1584}{768x^6} + dotsright)} + {sin((x+1)Omega)left(1 - dfrac{Omega^2}{8x^2} + dfrac{Omega^2}{4x^3} + dfrac{Omega^2(Omega^2-168)}{384 x^4} - dfrac{Omega^2(Omega^2-72)}{96 x^5} - dfrac{Omega^2(Omega^4 - 1320 Omega^2 + 59400)}{46080 x^6} + dfrac{Omega^2(Omega^4 - 520 Omega^2 + 17160)}{7680 x^7} + dotsright)}
$$ is not so usable)

and the integral
$$intlimits_0^infty x^d e^{-cx^2},mathrm dx = dfrac{Gammaleft(frac{d+1}2right)}{2c^{frac{d+1}2}}.$$
Length of the series can be selected, using the numeric experiments.