What is Fourier transform of $|(x,t)|^{-alpha}$?












1












$begingroup$


Let the $xin mathbb{R}^d, tin mathbb{R}$, i.e. $(x,t)in {mathbb{R^{d+1}}}$.



I already know the Fourier transform of $|x|^{-alpha}$ is $|xi|^{-d+alpha}$.



How do I get the Fourier transform of $|(x,t)|^{-alpha} = left(sqrt{|x|^2 +t^2}right)^{-alpha}$ with respect to $x$ ?






fourier-analysis harmonic-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    @reuns Isn't there the more general form of that?
    $endgroup$
    – Idkwhat
    Jan 15 at 9:24










  • $begingroup$
    I meant the inverse Fourier transform of $|(y,u)|^{alpha-d-1}$ in $u in mathbb{R}$ (and $y in mathbb{R}^d$ fixed).
    $endgroup$
    – reuns
    Jan 15 at 10:46
















1












$begingroup$


Let the $xin mathbb{R}^d, tin mathbb{R}$, i.e. $(x,t)in {mathbb{R^{d+1}}}$.



I already know the Fourier transform of $|x|^{-alpha}$ is $|xi|^{-d+alpha}$.



How do I get the Fourier transform of $|(x,t)|^{-alpha} = left(sqrt{|x|^2 +t^2}right)^{-alpha}$ with respect to $x$ ?






fourier-analysis harmonic-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    @reuns Isn't there the more general form of that?
    $endgroup$
    – Idkwhat
    Jan 15 at 9:24










  • $begingroup$
    I meant the inverse Fourier transform of $|(y,u)|^{alpha-d-1}$ in $u in mathbb{R}$ (and $y in mathbb{R}^d$ fixed).
    $endgroup$
    – reuns
    Jan 15 at 10:46














1












1








1





$begingroup$


Let the $xin mathbb{R}^d, tin mathbb{R}$, i.e. $(x,t)in {mathbb{R^{d+1}}}$.



I already know the Fourier transform of $|x|^{-alpha}$ is $|xi|^{-d+alpha}$.



How do I get the Fourier transform of $|(x,t)|^{-alpha} = left(sqrt{|x|^2 +t^2}right)^{-alpha}$ with respect to $x$ ?






fourier-analysis harmonic-analysis







share|cite|improve this question











$endgroup$




Let the $xin mathbb{R}^d, tin mathbb{R}$, i.e. $(x,t)in {mathbb{R^{d+1}}}$.



I already know the Fourier transform of $|x|^{-alpha}$ is $|xi|^{-d+alpha}$.



How do I get the Fourier transform of $|(x,t)|^{-alpha} = left(sqrt{|x|^2 +t^2}right)^{-alpha}$ with respect to $x$ ?






fourier-analysis harmonic-analysis




fourier-analysis harmonic-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 15 at 9:16







Idkwhat

















asked Jan 15 at 8:55









IdkwhatIdkwhat

236




236












  • $begingroup$
    @reuns Isn't there the more general form of that?
    $endgroup$
    – Idkwhat
    Jan 15 at 9:24










  • $begingroup$
    I meant the inverse Fourier transform of $|(y,u)|^{alpha-d-1}$ in $u in mathbb{R}$ (and $y in mathbb{R}^d$ fixed).
    $endgroup$
    – reuns
    Jan 15 at 10:46


















  • $begingroup$
    @reuns Isn't there the more general form of that?
    $endgroup$
    – Idkwhat
    Jan 15 at 9:24










  • $begingroup$
    I meant the inverse Fourier transform of $|(y,u)|^{alpha-d-1}$ in $u in mathbb{R}$ (and $y in mathbb{R}^d$ fixed).
    $endgroup$
    – reuns
    Jan 15 at 10:46
















$begingroup$
@reuns Isn't there the more general form of that?
$endgroup$
– Idkwhat
Jan 15 at 9:24




$begingroup$
@reuns Isn't there the more general form of that?
$endgroup$
– Idkwhat
Jan 15 at 9:24












$begingroup$
I meant the inverse Fourier transform of $|(y,u)|^{alpha-d-1}$ in $u in mathbb{R}$ (and $y in mathbb{R}^d$ fixed).
$endgroup$
– reuns
Jan 15 at 10:46




$begingroup$
I meant the inverse Fourier transform of $|(y,u)|^{alpha-d-1}$ in $u in mathbb{R}$ (and $y in mathbb{R}^d$ fixed).
$endgroup$
– reuns
Jan 15 at 10:46










1 Answer
1






active

oldest

votes


















1












$begingroup$

You are interested in obtaining an explicit expression for
$$I(k,t) =frac{1}{(2pi)^{n/2}} intleft(sqrt{|x|^2 +t^2}right)^{-alpha} e^{i xcdot k} d^n x,.$$
Let us introduce spherical coordinates with the first coordinate of $x$ with component $r cos phi$ pointing along $k$. We thus reduce the integral to the form
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty int_{0}^pi r^{n-1}left(sqrt{r^2 +t^2}right)^{-alpha} e^{i r |k| cos(phi)} dphi dr $$
where $S_n$ is the surface area of the $n$-sphere given by
$$ S_{n-1} =frac{npi^frac{n}{2}}{Gammaleft(frac{n}{2}+1 right)}.$$



The integral over $phi$ can be easily executed with the result
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty frac{pi r^{n-1} J_0(r |k|)}{(r^2+t^2)^{alpha/2}} dr,. $$
I am not sure about the remaining integral. Maybe somebody has an idea...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is also $frac{1}{pi}int_0^infty (r^2+u^2)^c cos(ut)du, c =( alpha-d-1)/2,r=|x|$
    $endgroup$
    – reuns
    Jan 15 at 11:02













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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









1












$begingroup$

You are interested in obtaining an explicit expression for
$$I(k,t) =frac{1}{(2pi)^{n/2}} intleft(sqrt{|x|^2 +t^2}right)^{-alpha} e^{i xcdot k} d^n x,.$$
Let us introduce spherical coordinates with the first coordinate of $x$ with component $r cos phi$ pointing along $k$. We thus reduce the integral to the form
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty int_{0}^pi r^{n-1}left(sqrt{r^2 +t^2}right)^{-alpha} e^{i r |k| cos(phi)} dphi dr $$
where $S_n$ is the surface area of the $n$-sphere given by
$$ S_{n-1} =frac{npi^frac{n}{2}}{Gammaleft(frac{n}{2}+1 right)}.$$



The integral over $phi$ can be easily executed with the result
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty frac{pi r^{n-1} J_0(r |k|)}{(r^2+t^2)^{alpha/2}} dr,. $$
I am not sure about the remaining integral. Maybe somebody has an idea...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is also $frac{1}{pi}int_0^infty (r^2+u^2)^c cos(ut)du, c =( alpha-d-1)/2,r=|x|$
    $endgroup$
    – reuns
    Jan 15 at 11:02


















1












$begingroup$

You are interested in obtaining an explicit expression for
$$I(k,t) =frac{1}{(2pi)^{n/2}} intleft(sqrt{|x|^2 +t^2}right)^{-alpha} e^{i xcdot k} d^n x,.$$
Let us introduce spherical coordinates with the first coordinate of $x$ with component $r cos phi$ pointing along $k$. We thus reduce the integral to the form
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty int_{0}^pi r^{n-1}left(sqrt{r^2 +t^2}right)^{-alpha} e^{i r |k| cos(phi)} dphi dr $$
where $S_n$ is the surface area of the $n$-sphere given by
$$ S_{n-1} =frac{npi^frac{n}{2}}{Gammaleft(frac{n}{2}+1 right)}.$$



The integral over $phi$ can be easily executed with the result
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty frac{pi r^{n-1} J_0(r |k|)}{(r^2+t^2)^{alpha/2}} dr,. $$
I am not sure about the remaining integral. Maybe somebody has an idea...






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It is also $frac{1}{pi}int_0^infty (r^2+u^2)^c cos(ut)du, c =( alpha-d-1)/2,r=|x|$
    $endgroup$
    – reuns
    Jan 15 at 11:02
















1












1








1





$begingroup$

You are interested in obtaining an explicit expression for
$$I(k,t) =frac{1}{(2pi)^{n/2}} intleft(sqrt{|x|^2 +t^2}right)^{-alpha} e^{i xcdot k} d^n x,.$$
Let us introduce spherical coordinates with the first coordinate of $x$ with component $r cos phi$ pointing along $k$. We thus reduce the integral to the form
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty int_{0}^pi r^{n-1}left(sqrt{r^2 +t^2}right)^{-alpha} e^{i r |k| cos(phi)} dphi dr $$
where $S_n$ is the surface area of the $n$-sphere given by
$$ S_{n-1} =frac{npi^frac{n}{2}}{Gammaleft(frac{n}{2}+1 right)}.$$



The integral over $phi$ can be easily executed with the result
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty frac{pi r^{n-1} J_0(r |k|)}{(r^2+t^2)^{alpha/2}} dr,. $$
I am not sure about the remaining integral. Maybe somebody has an idea...






share|cite|improve this answer









$endgroup$



You are interested in obtaining an explicit expression for
$$I(k,t) =frac{1}{(2pi)^{n/2}} intleft(sqrt{|x|^2 +t^2}right)^{-alpha} e^{i xcdot k} d^n x,.$$
Let us introduce spherical coordinates with the first coordinate of $x$ with component $r cos phi$ pointing along $k$. We thus reduce the integral to the form
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty int_{0}^pi r^{n-1}left(sqrt{r^2 +t^2}right)^{-alpha} e^{i r |k| cos(phi)} dphi dr $$
where $S_n$ is the surface area of the $n$-sphere given by
$$ S_{n-1} =frac{npi^frac{n}{2}}{Gammaleft(frac{n}{2}+1 right)}.$$



The integral over $phi$ can be easily executed with the result
$$I(k,t) =frac{S_{n-2}}{(2pi)^{n/2}} int_0^infty frac{pi r^{n-1} J_0(r |k|)}{(r^2+t^2)^{alpha/2}} dr,. $$
I am not sure about the remaining integral. Maybe somebody has an idea...







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 15 at 9:51









FabianFabian

19.7k3674




19.7k3674












  • $begingroup$
    It is also $frac{1}{pi}int_0^infty (r^2+u^2)^c cos(ut)du, c =( alpha-d-1)/2,r=|x|$
    $endgroup$
    – reuns
    Jan 15 at 11:02




















  • $begingroup$
    It is also $frac{1}{pi}int_0^infty (r^2+u^2)^c cos(ut)du, c =( alpha-d-1)/2,r=|x|$
    $endgroup$
    – reuns
    Jan 15 at 11:02


















$begingroup$
It is also $frac{1}{pi}int_0^infty (r^2+u^2)^c cos(ut)du, c =( alpha-d-1)/2,r=|x|$
$endgroup$
– reuns
Jan 15 at 11:02






$begingroup$
It is also $frac{1}{pi}int_0^infty (r^2+u^2)^c cos(ut)du, c =( alpha-d-1)/2,r=|x|$
$endgroup$
– reuns
Jan 15 at 11:02




















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