Dear Math enthusiasts,
I'm trying to see if I can find an analytical expression for the Fourier transform of a Gaussian pulse $p(\tau) = {\rm e}^{-B\tau^2}$ 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 $$G(\omega) = \int_{-\infty}^\infty {\rm e}^{-B\left(\sqrt{t^2+a^2}-a\right)^2} {\rm e}^{-\jmath \omega t} {\rm d}t.$$
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 $$G(\omega) = a\int_{-\infty}^\infty {\rm e}^{-B\left(a \cosh(x)-a\right)^2} {\rm e}^{-\jmath \omega a \sinh(x)} \cosh(x) {\rm d}x.$$
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., $$G(\omega) = \int_{-\infty}^\infty {\rm e}^{-B\left(\sqrt{t^2+a^2}-a\right)^2} \cos( \omega t) {\rm d}t = 2 \int_{0}^\infty {\rm e}^{-B\left(\sqrt{t^2+a^2}-a\right)^2} \cos( \omega t) {\rm d}t.$$ Does this make it easier? I'm not sure.
Best Answer
HINT
Firstly, $$\begin{aligned} &G(\omega,a,B)=a\int\limits_{-\infty}^\infty e^{-a^2B\left(\sqrt{\left(\frac ta\right)^2+1}-1\right)^2}e^{-ja\omega\frac ta}\,\mathrm d\dfrac ta=a\int\limits_{-\infty}^\infty e^{-a^2B(\sqrt{t^2+1}-1)^2}e^{-ja\omega t}\,\mathrm dt,\\ &G(\omega,a,B)=aH(a\omega,a^2B),\\ \end{aligned}$$ $$H(\Omega,c)=\int\limits_{-\infty}^\infty e^{-c(\sqrt{t^2+1}-1)^2}e^{-j\Omega t}\,\mathrm dt.\tag1$$
Let $$\tau(t) = \sqrt{t^2+1}-1,\quad \tau\in[0,\infty),$$ so $$t(\tau)=\pm\sqrt{(\tau+1)^2-1}=\pm\sqrt{\tau^2+2\tau},\quad t\in(-\infty,\infty),$$ $$dt=\pm\dfrac{\tau+1}{\sqrt{\tau^2+2\tau}},$$ $$\begin{aligned} &H(\Omega,c)=\int\limits_{-\infty}^{-1}e^{-c\tau^2(t)}e^{-j\Omega t}\,\mathrm dt +\int\limits_{-1}^\infty e^{-c\tau^2(t)}e^{-j\Omega t}\,\mathrm dt\\ &=-\int\limits_{\infty}^0e^{-c\tau^2}e^{j\Omega\sqrt{\tau^2+2\tau}}\dfrac{\tau+1}{\sqrt{\tau^2+2\tau}}\,\mathrm d\tau +\int\limits_{0}^\infty e^{-c\tau^2}e^{-j\Omega \sqrt{\tau^2+2\tau}}\dfrac{\tau+1}{\sqrt{\tau^2+2\tau}}\,\mathrm d\tau, \end{aligned}$$ $$H(\Omega,c)=2\int\limits_0^{\infty} e^{-c\tau^2} \cos(\Omega\sqrt{\tau^2+2\tau})\dfrac{\tau+1}{\sqrt{\tau^2+2\tau}}\,\mathrm d\tau.\tag2$$ By parts: $$\begin{aligned} &H(\omega,c) = \dfrac2\Omega\int\limits_0^{\infty} e^{-c\tau^2}\,d\sin(\Omega\sqrt{\tau^2+2\tau})\\ &=-\dfrac2\Omega e^{-c\tau^2} \sin(\Omega\sqrt{\tau^2+2\tau})\bigg|_0^\infty +\dfrac{4c}\Omega\int\limits_0^{\infty} \tau e^{-c\tau^2} \sin(\Omega\sqrt{\tau^2+2\tau})\,d\tau,\end{aligned}$$ $$H(\Omega,c)=\dfrac{4c}\Omega\int\limits_0^{\infty} \tau e^{-c\tau^2}\sin(\Omega\sqrt{\tau^2+2\tau})\,d\tau.\tag3$$ 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 $c\gtrsim\Omega,$ then the integral $(3)$ can be presented as $1D$ series, using the series expansion for sine expression in the form of $$\sin(\Omega\sqrt{x^2+2x}) = \sqrt{\dfrac x2}\Omega\left(2 + \dfrac{3-4\Omega^2}6 x + \dfrac{16 \Omega^4 - 120 \Omega^2 - 15}{240} x^2 + \dfrac{-64\Omega^6 + 1680\Omega^4 - 1260\Omega^2 + 315}{20160} x^3 + \dfrac{256\Omega^8 - 16128\Omega^6 + 90720\Omega^4 + 15120\Omega^2 - 14175}{2903040} x^4 + \dfrac{-1024\Omega^{10} + 126720\Omega^8 - 2217600\Omega^6 + 1663200\Omega^4 - 623700\Omega^2 + 1091475}{638668800}x^5 + \dots\right). $$ (the asymptotic form of $$\sin(\Omega\sqrt{x^2+2x}) = \Omega{\cos((x+1)\Omega)\left(-\dfrac1{2x} + \dfrac 1{2x^2} + \dfrac{\Omega^2-30}{48x^3} - \dfrac{\Omega^2 - 14}{16x^4} - \dfrac{\Omega^4 - 540 \Omega^2 + 5040}{3840x^5} + \dfrac{\Omega^4 - 220 \Omega^2 + 1584}{768x^6} + \dots\right)} + {\sin((x+1)\Omega)\left(1 - \dfrac{\Omega^2}{8x^2} + \dfrac{\Omega^2}{4x^3} + \dfrac{\Omega^2(\Omega^2-168)}{384 x^4} - \dfrac{\Omega^2(\Omega^2-72)}{96 x^5} - \dfrac{\Omega^2(\Omega^4 - 1320 \Omega^2 + 59400)}{46080 x^6} + \dfrac{\Omega^2(\Omega^4 - 520 \Omega^2 + 17160)}{7680 x^7} + \dots\right)} $$ is not so usable)
and the integral $$\int\limits_0^\infty x^d e^{-cx^2}\,\mathrm dx = \dfrac{\Gamma\left(\frac{d+1}2\right)}{2c^{\frac{d+1}2}}.$$ Length of the series can be selected, using the numeric experiments.