Fourier and Mellin transforms of Hilbert Transform
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I am reading Hilbert transform recently and meet two questions. The book I am reading is Debnath and Bhatta "Integral Transforms and Their Applications".
If we define the Hilbert transform on the real line is, for $x in mathbb{R}$,
$$H_x{f(t)}=frac{1}{pi}PVint_{-infty}^infty frac{f(t)}{t-x}dt=lim_{epsilon rightarrow 0}left(int_{-infty}^{t-epsilon} +int_{t+epsilon}^inftyright)frac{f(t)}{t-x}dt,$$
where PV stands for the Cauchy principal value. Then the Fourier transform of the Hilbert transform, which can be considered as a convolution, is
$$F_k{H_x{f(t)}}=i sgn(k) F_k{f(t)},$$
where $i^2=-1$, $sgn(x)=begin{cases}
1, x>0,\
-1,x<0
end{cases}$
and $F_k{f(t)}=frac{1}{sqrt{2 pi}}int_0^infty e^{-ikt}f(t)dt $, since $F_k{sqrt{frac{2}{pi}}left(-frac{1}{x}right)}=i sgn(k)$.
The first question, Eq.(9.4.3), if we let a complex variable $z=x+i y$, then why
$$F_k{H_z{f(t)}}=2 i e^{-ky} H(k) F_k{f(t)},$$
where $H(k)=frac{1}{2}(1+sgn(k))$. Which variable are we transforming here, $xrightarrow k$ or $zrightarrow k$?
The second question appears in the Mellin transform of the Hilbert transform. There is an integral I do not know how to solve it:
$$PVint_0^inftyfrac{x^{p-1}}{t-x}dx=pi Cot(pi p).$$
I understand we need to separate the integral or use the residue theorem to get rid of the singularity. But I cannot go further.
Thanks in advance.
fourier-transform integral-transforms mellin-transform
asked Dec 4 '18 at 2:06
gouwangzhangdonggouwangzhangdong
838
$endgroup$
add a comment |
$begingroup$
I am reading Hilbert transform recently and meet two questions. The book I am reading is Debnath and Bhatta "Integral Transforms and Their Applications".
If we define the Hilbert transform on the real line is, for $x in mathbb{R}$,
$$H_x{f(t)}=frac{1}{pi}PVint_{-infty}^infty frac{f(t)}{t-x}dt=lim_{epsilon rightarrow 0}left(int_{-infty}^{t-epsilon} +int_{t+epsilon}^inftyright)frac{f(t)}{t-x}dt,$$
where PV stands for the Cauchy principal value. Then the Fourier transform of the Hilbert transform, which can be considered as a convolution, is
$$F_k{H_x{f(t)}}=i sgn(k) F_k{f(t)},$$
where $i^2=-1$, $sgn(x)=begin{cases}
1, x>0,\
-1,x<0
end{cases}$
and $F_k{f(t)}=frac{1}{sqrt{2 pi}}int_0^infty e^{-ikt}f(t)dt $, since $F_k{sqrt{frac{2}{pi}}left(-frac{1}{x}right)}=i sgn(k)$.
The first question, Eq.(9.4.3), if we let a complex variable $z=x+i y$, then why
$$F_k{H_z{f(t)}}=2 i e^{-ky} H(k) F_k{f(t)},$$
where $H(k)=frac{1}{2}(1+sgn(k))$. Which variable are we transforming here, $xrightarrow k$ or $zrightarrow k$?
The second question appears in the Mellin transform of the Hilbert transform. There is an integral I do not know how to solve it:
$$PVint_0^inftyfrac{x^{p-1}}{t-x}dx=pi Cot(pi p).$$
I understand we need to separate the integral or use the residue theorem to get rid of the singularity. But I cannot go further.
Thanks in advance.
fourier-transform integral-transforms mellin-transform
asked Dec 4 '18 at 2:06
gouwangzhangdonggouwangzhangdong
838
$endgroup$
$begingroup$
For $Im(z) > 0$ let $h(z) = frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-z}dt$. If $f in L^1$ then for every $y$, $h(.+iy) in L^1$ and $h$ is complex analytic. Then $H(y,xi)=int_{-infty}^infty h(x+iy) e^{-2i pi xi (x+iy)} dx$ doesn't depend on $y > 0$. Proof : $H(y,xi)$ is complex analytic in $y$ and it doesn't depend on $Im(y)$, thus it is constant and it doesn't depend on $y$. Do the same with $h$ defined for $Im(z) < 0$ and conclude with $pv. frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-x}dt = frac12 lim_{y to 0} h(x+iy)+h(x-iy)$.
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– reuns
Dec 4 '18 at 2:38
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Sorry for my stupidness. How is your comment related to my questions?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:47
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That's what you asked, the Fourier transform of $h(z+.)$
$endgroup$
– reuns
Dec 4 '18 at 2:49
$begingroup$
I still do not understand. So how can I use your comments to get Eq.(9.4.3)?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:55
add a comment |
$begingroup$
I am reading Hilbert transform recently and meet two questions. The book I am reading is Debnath and Bhatta "Integral Transforms and Their Applications".
If we define the Hilbert transform on the real line is, for $x in mathbb{R}$,
$$H_x{f(t)}=frac{1}{pi}PVint_{-infty}^infty frac{f(t)}{t-x}dt=lim_{epsilon rightarrow 0}left(int_{-infty}^{t-epsilon} +int_{t+epsilon}^inftyright)frac{f(t)}{t-x}dt,$$
where PV stands for the Cauchy principal value. Then the Fourier transform of the Hilbert transform, which can be considered as a convolution, is
$$F_k{H_x{f(t)}}=i sgn(k) F_k{f(t)},$$
where $i^2=-1$, $sgn(x)=begin{cases}
1, x>0,\
-1,x<0
end{cases}$
and $F_k{f(t)}=frac{1}{sqrt{2 pi}}int_0^infty e^{-ikt}f(t)dt $, since $F_k{sqrt{frac{2}{pi}}left(-frac{1}{x}right)}=i sgn(k)$.
The first question, Eq.(9.4.3), if we let a complex variable $z=x+i y$, then why
$$F_k{H_z{f(t)}}=2 i e^{-ky} H(k) F_k{f(t)},$$
where $H(k)=frac{1}{2}(1+sgn(k))$. Which variable are we transforming here, $xrightarrow k$ or $zrightarrow k$?
The second question appears in the Mellin transform of the Hilbert transform. There is an integral I do not know how to solve it:
$$PVint_0^inftyfrac{x^{p-1}}{t-x}dx=pi Cot(pi p).$$
I understand we need to separate the integral or use the residue theorem to get rid of the singularity. But I cannot go further.
Thanks in advance.
fourier-transform integral-transforms mellin-transform
asked Dec 4 '18 at 2:06
gouwangzhangdonggouwangzhangdong
838
$endgroup$
I am reading Hilbert transform recently and meet two questions. The book I am reading is Debnath and Bhatta "Integral Transforms and Their Applications".
If we define the Hilbert transform on the real line is, for $x in mathbb{R}$,
$$H_x{f(t)}=frac{1}{pi}PVint_{-infty}^infty frac{f(t)}{t-x}dt=lim_{epsilon rightarrow 0}left(int_{-infty}^{t-epsilon} +int_{t+epsilon}^inftyright)frac{f(t)}{t-x}dt,$$
where PV stands for the Cauchy principal value. Then the Fourier transform of the Hilbert transform, which can be considered as a convolution, is
$$F_k{H_x{f(t)}}=i sgn(k) F_k{f(t)},$$
where $i^2=-1$, $sgn(x)=begin{cases}
1, x>0,\
-1,x<0
end{cases}$
and $F_k{f(t)}=frac{1}{sqrt{2 pi}}int_0^infty e^{-ikt}f(t)dt $, since $F_k{sqrt{frac{2}{pi}}left(-frac{1}{x}right)}=i sgn(k)$.
The first question, Eq.(9.4.3), if we let a complex variable $z=x+i y$, then why
$$F_k{H_z{f(t)}}=2 i e^{-ky} H(k) F_k{f(t)},$$
where $H(k)=frac{1}{2}(1+sgn(k))$. Which variable are we transforming here, $xrightarrow k$ or $zrightarrow k$?
The second question appears in the Mellin transform of the Hilbert transform. There is an integral I do not know how to solve it:
$$PVint_0^inftyfrac{x^{p-1}}{t-x}dx=pi Cot(pi p).$$
I understand we need to separate the integral or use the residue theorem to get rid of the singularity. But I cannot go further.
Thanks in advance.
fourier-transform integral-transforms mellin-transform
fourier-transform integral-transforms mellin-transform
asked Dec 4 '18 at 2:06
gouwangzhangdonggouwangzhangdong
838
asked Dec 4 '18 at 2:06
gouwangzhangdonggouwangzhangdong
838
edited Dec 15 '18 at 4:25
gouwangzhangdong
asked Dec 4 '18 at 2:06
gouwangzhangdonggouwangzhangdong
838
asked Dec 4 '18 at 2:06
gouwangzhangdonggouwangzhangdong
838
asked Dec 4 '18 at 2:06
gouwangzhangdonggouwangzhangdong
838
838
$begingroup$
For $Im(z) > 0$ let $h(z) = frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-z}dt$. If $f in L^1$ then for every $y$, $h(.+iy) in L^1$ and $h$ is complex analytic. Then $H(y,xi)=int_{-infty}^infty h(x+iy) e^{-2i pi xi (x+iy)} dx$ doesn't depend on $y > 0$. Proof : $H(y,xi)$ is complex analytic in $y$ and it doesn't depend on $Im(y)$, thus it is constant and it doesn't depend on $y$. Do the same with $h$ defined for $Im(z) < 0$ and conclude with $pv. frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-x}dt = frac12 lim_{y to 0} h(x+iy)+h(x-iy)$.
$endgroup$
– reuns
Dec 4 '18 at 2:38
$begingroup$
Sorry for my stupidness. How is your comment related to my questions?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:47
$begingroup$
That's what you asked, the Fourier transform of $h(z+.)$
$endgroup$
– reuns
Dec 4 '18 at 2:49
$begingroup$
I still do not understand. So how can I use your comments to get Eq.(9.4.3)?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:55
add a comment |
$begingroup$
For $Im(z) > 0$ let $h(z) = frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-z}dt$. If $f in L^1$ then for every $y$, $h(.+iy) in L^1$ and $h$ is complex analytic. Then $H(y,xi)=int_{-infty}^infty h(x+iy) e^{-2i pi xi (x+iy)} dx$ doesn't depend on $y > 0$. Proof : $H(y,xi)$ is complex analytic in $y$ and it doesn't depend on $Im(y)$, thus it is constant and it doesn't depend on $y$. Do the same with $h$ defined for $Im(z) < 0$ and conclude with $pv. frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-x}dt = frac12 lim_{y to 0} h(x+iy)+h(x-iy)$.
$endgroup$
– reuns
Dec 4 '18 at 2:38
$begingroup$
Sorry for my stupidness. How is your comment related to my questions?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:47
$begingroup$
That's what you asked, the Fourier transform of $h(z+.)$
$endgroup$
– reuns
Dec 4 '18 at 2:49
$begingroup$
I still do not understand. So how can I use your comments to get Eq.(9.4.3)?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:55
$begingroup$
For $Im(z) > 0$ let $h(z) = frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-z}dt$. If $f in L^1$ then for every $y$, $h(.+iy) in L^1$ and $h$ is complex analytic. Then $H(y,xi)=int_{-infty}^infty h(x+iy) e^{-2i pi xi (x+iy)} dx$ doesn't depend on $y > 0$. Proof : $H(y,xi)$ is complex analytic in $y$ and it doesn't depend on $Im(y)$, thus it is constant and it doesn't depend on $y$. Do the same with $h$ defined for $Im(z) < 0$ and conclude with $pv. frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-x}dt = frac12 lim_{y to 0} h(x+iy)+h(x-iy)$.
$endgroup$
– reuns
Dec 4 '18 at 2:38
$begingroup$
For $Im(z) > 0$ let $h(z) = frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-z}dt$. If $f in L^1$ then for every $y$, $h(.+iy) in L^1$ and $h$ is complex analytic. Then $H(y,xi)=int_{-infty}^infty h(x+iy) e^{-2i pi xi (x+iy)} dx$ doesn't depend on $y > 0$. Proof : $H(y,xi)$ is complex analytic in $y$ and it doesn't depend on $Im(y)$, thus it is constant and it doesn't depend on $y$. Do the same with $h$ defined for $Im(z) < 0$ and conclude with $pv. frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-x}dt = frac12 lim_{y to 0} h(x+iy)+h(x-iy)$.
$endgroup$
– reuns
Dec 4 '18 at 2:38
$begingroup$
Sorry for my stupidness. How is your comment related to my questions?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:47
$begingroup$
Sorry for my stupidness. How is your comment related to my questions?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:47
$begingroup$
That's what you asked, the Fourier transform of $h(z+.)$
$endgroup$
– reuns
Dec 4 '18 at 2:49
$begingroup$
That's what you asked, the Fourier transform of $h(z+.)$
$endgroup$
– reuns
Dec 4 '18 at 2:49
$begingroup$
I still do not understand. So how can I use your comments to get Eq.(9.4.3)?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:55
$begingroup$
I still do not understand. So how can I use your comments to get Eq.(9.4.3)?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:55
add a comment |
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$begingroup$
For $Im(z) > 0$ let $h(z) = frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-z}dt$. If $f in L^1$ then for every $y$, $h(.+iy) in L^1$ and $h$ is complex analytic. Then $H(y,xi)=int_{-infty}^infty h(x+iy) e^{-2i pi xi (x+iy)} dx$ doesn't depend on $y > 0$. Proof : $H(y,xi)$ is complex analytic in $y$ and it doesn't depend on $Im(y)$, thus it is constant and it doesn't depend on $y$. Do the same with $h$ defined for $Im(z) < 0$ and conclude with $pv. frac{1}{pi} int_{-infty}^infty frac{f(t)}{t-x}dt = frac12 lim_{y to 0} h(x+iy)+h(x-iy)$.
$endgroup$
– reuns
Dec 4 '18 at 2:38
$begingroup$
Sorry for my stupidness. How is your comment related to my questions?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:47
$begingroup$
That's what you asked, the Fourier transform of $h(z+.)$
$endgroup$
– reuns
Dec 4 '18 at 2:49
$begingroup$
I still do not understand. So how can I use your comments to get Eq.(9.4.3)?
$endgroup$
– gouwangzhangdong
Dec 4 '18 at 2:55