Asymptotic behaviour from Fourier transform Given the Fourier transform $G(\vec{k}) = \frac{1}{A + B k^4}$ of a function $G(\vec{x})$, I wish to infer the asymptotic behaviour of $G(\vec{x})$ as $|{\vec{x}}| \to \infty$. Note that $\vec{x} \in \mathbb{R}^2$ and $A,B > 0$. I thought of writing $G(\vec{x}/a)$, doing u-substitutions in the integrals and eventually expanding in a series for $a$ close to $0$, but this got me nowhere. Does anyone have any hints on how to proceed?
Thanks.
 A: First of all, let's define Fourier Transformation as
$$F(\vec k)=\int_{-\infty}^\infty f(\vec x) e^{-i(\vec k,\vec x)} d^2x$$
Thus,
$$f(\vec x)=\frac{1}{(2\pi)^2}\int_{-\infty}^\infty F(\vec k) e^{i(\vec k,\vec x)} d^2k=\frac{1}{(2\pi)^2B}\int_0^\infty \int_0^{2\pi}\frac{e^{ikx\cos\phi}}{a+k^4} d\phi$$
where $a=\frac{A}{B}$.
But
$$\int_0^{2\pi}e^{ikx\cos\phi}d\phi=2\pi J_0(kx)$$
We can also express the Bessel function (please, look here - formula 10.9.8) in the form
$$J_0(z)=\frac{2}{\pi}\int_0^\infty\sin(z\cosh t)dt\,\,(z>0)$$
The integrand in our case is even, and we can write
$$f(\vec x)=f(x)=\frac{1}{4\pi^2 B}\int_{-\infty}^\infty dt\int_{-\infty}^\infty\frac{\sin(kx\cosh t)}{a+k^4}kdk=\frac{1}{4\pi^2 B}\Im\int_{-\infty}^\infty dt\int_{-\infty}^\infty\frac{e^{ikx\cosh t}}{a+k^4}kdk$$
To evaluate $I=\int_{-\infty}^\infty\frac{e^{ikx\cosh t}}{a+k^4}k\,dk$ we integrate in the complex plane, adding a big half-circle in the upper half-plane. The integrand declines rapidly enough, so the integral along this half-circle tends to zero as the radius tends to $\infty$. The integrand has two poles inside the contour.
$$I=\int_{-\infty}^\infty\frac{e^{ikx\cosh t}}{a+k^4}kdk=2\pi i Res_{\binom{k=a^{1/4}e^{\pi i/4}}{k=a^{1/4}e^{3\pi i/4}}}\frac{e^{ikx\cosh t}}{a+k^4}$$
Residues evaluation is straightforward:
$$I=\frac{\pi}{2\sqrt a}\big(e^{-a^{1/4}x\cosh t\frac{1-i}{\sqrt2}}-e^{-a^{1/4}x\cosh t\frac{1+i}{\sqrt2}}\big)$$
$$f(x)=\frac{1}{4\pi B\sqrt a}\Im\int_{-\infty}^\infty e^{-c\cosh t}dt$$
where we denoted $c=a^{1/4}xe^{-\frac{\pi i}{4}}$.
To find the asymptotics of the integral at $x\to\infty$ we use the steepest descent method. The saddle point is $t=0$. Near this point $\cosh t\approx 1+\frac{t^2}{2}$
$$\int_{-\infty}^\infty e^{-c\cosh t}dt\sim\int_{-\infty}^\infty e^{-c(1+\frac{t^2}{2})}dt=e^{-c}\sqrt\frac{2\pi}{c}$$
$$\Im \,e^{-c}\sqrt\frac{2\pi}{c}=\Im \Bigg(e^{-a^{1/4}x\frac{1-i}{\sqrt 2}}\sqrt\frac{2\pi}{a^{1/4}xe^{-\frac{\pi i}{4}}}\Bigg)=\sqrt\frac{2\pi}{a^{1/4}x}e^{-\frac{a^{1/4}x}{\sqrt 2}}\sin\Big(\frac{a^{1/4}x}{\sqrt 2}-\frac{\pi}{8}\Big)$$
$$f(x)=\frac{1}{2\sqrt{2\pi}}\frac{1}{Ba^{5/8}}\frac{e^{-\frac{a^{1/4}x}{\sqrt2}}}{\sqrt x}\sin\Big(\frac{a^{1/4}x}{\sqrt 2}-\frac{\pi}{8}\Big)\,, \,\, a=\frac{A}{B}, \,\,x\gg1$$
I would recommend to check the calculations.
