Mixing
Demonstrating $int_{pi}^{0}{ie^{-i2pi s epsilon e^{itheta}}dtheta} = ipi space mathrm{sgn}(s)$
$begingroup$
In his derivation of the Fourier Transform of $dfrac{1}{x}$, Bracewell starts with
$$mathscr{F}left{dfrac{1}{x}right} = P.V. int_{-infty}^{infty}{dfrac{e^{-i2pi sx}}{x}}dx$$
And goes on to consider the contour integral with the expected semicircular contour (but Bracewell omits specifying which half-plane, upper or lower) with a small semicircular arc excursion around the pole at the origin:
$$int_{C}{dfrac{e^{-i2pi sz}}{z}}dz$$
When evaluating the integral for the semi-circular arc around the origin, with $epsilon$ a vanishingly small positive number, Bracewell states the integral "equals $pm ipi$ according to the sign of s", so we have
$$int_{pi}^{0}{ie^{-i2pi s epsilon e^{itheta}}dtheta} = ipi spacemathrm{sgn}(s)$$
Could someone please show me in more detail how the $mathrm{sgn}(s)$ arises in the evaluation of this integral? You can assume $epsilon rightarrow 0$ in the limit.
Is it that the sign of $s$ dictates which half plane, upper or lower, the contour should be in, and hence affects this integral around this small semi-circular arc around $0$?
definite-integrals fourier-analysis contour-integration fourier-transform
asked Jan 24 at 14:10
Andy WallsAndy Walls
1,764139
$endgroup$
add a comment |
$begingroup$
In his derivation of the Fourier Transform of $dfrac{1}{x}$, Bracewell starts with
$$mathscr{F}left{dfrac{1}{x}right} = P.V. int_{-infty}^{infty}{dfrac{e^{-i2pi sx}}{x}}dx$$
And goes on to consider the contour integral with the expected semicircular contour (but Bracewell omits specifying which half-plane, upper or lower) with a small semicircular arc excursion around the pole at the origin:
$$int_{C}{dfrac{e^{-i2pi sz}}{z}}dz$$
When evaluating the integral for the semi-circular arc around the origin, with $epsilon$ a vanishingly small positive number, Bracewell states the integral "equals $pm ipi$ according to the sign of s", so we have
$$int_{pi}^{0}{ie^{-i2pi s epsilon e^{itheta}}dtheta} = ipi spacemathrm{sgn}(s)$$
Could someone please show me in more detail how the $mathrm{sgn}(s)$ arises in the evaluation of this integral? You can assume $epsilon rightarrow 0$ in the limit.
Is it that the sign of $s$ dictates which half plane, upper or lower, the contour should be in, and hence affects this integral around this small semi-circular arc around $0$?
definite-integrals fourier-analysis contour-integration fourier-transform
asked Jan 24 at 14:10
Andy WallsAndy Walls
1,764139
$endgroup$
1
$begingroup$
As you have noticed, the claim is wrong. The integrand converges to $i$ uniformly in $theta$, therefore the integral tends to $int_pi^0 i ,dtheta$.
$endgroup$
– Maxim
Jan 25 at 18:27
add a comment |
$begingroup$
In his derivation of the Fourier Transform of $dfrac{1}{x}$, Bracewell starts with
$$mathscr{F}left{dfrac{1}{x}right} = P.V. int_{-infty}^{infty}{dfrac{e^{-i2pi sx}}{x}}dx$$
And goes on to consider the contour integral with the expected semicircular contour (but Bracewell omits specifying which half-plane, upper or lower) with a small semicircular arc excursion around the pole at the origin:
$$int_{C}{dfrac{e^{-i2pi sz}}{z}}dz$$
When evaluating the integral for the semi-circular arc around the origin, with $epsilon$ a vanishingly small positive number, Bracewell states the integral "equals $pm ipi$ according to the sign of s", so we have
$$int_{pi}^{0}{ie^{-i2pi s epsilon e^{itheta}}dtheta} = ipi spacemathrm{sgn}(s)$$
Could someone please show me in more detail how the $mathrm{sgn}(s)$ arises in the evaluation of this integral? You can assume $epsilon rightarrow 0$ in the limit.
Is it that the sign of $s$ dictates which half plane, upper or lower, the contour should be in, and hence affects this integral around this small semi-circular arc around $0$?
definite-integrals fourier-analysis contour-integration fourier-transform
asked Jan 24 at 14:10
Andy WallsAndy Walls
1,764139
$endgroup$
In his derivation of the Fourier Transform of $dfrac{1}{x}$, Bracewell starts with
$$mathscr{F}left{dfrac{1}{x}right} = P.V. int_{-infty}^{infty}{dfrac{e^{-i2pi sx}}{x}}dx$$
And goes on to consider the contour integral with the expected semicircular contour (but Bracewell omits specifying which half-plane, upper or lower) with a small semicircular arc excursion around the pole at the origin:
$$int_{C}{dfrac{e^{-i2pi sz}}{z}}dz$$
When evaluating the integral for the semi-circular arc around the origin, with $epsilon$ a vanishingly small positive number, Bracewell states the integral "equals $pm ipi$ according to the sign of s", so we have
$$int_{pi}^{0}{ie^{-i2pi s epsilon e^{itheta}}dtheta} = ipi spacemathrm{sgn}(s)$$
Could someone please show me in more detail how the $mathrm{sgn}(s)$ arises in the evaluation of this integral? You can assume $epsilon rightarrow 0$ in the limit.
Is it that the sign of $s$ dictates which half plane, upper or lower, the contour should be in, and hence affects this integral around this small semi-circular arc around $0$?
definite-integrals fourier-analysis contour-integration fourier-transform
definite-integrals fourier-analysis contour-integration fourier-transform
asked Jan 24 at 14:10
Andy WallsAndy Walls
1,764139
asked Jan 24 at 14:10
Andy WallsAndy Walls
1,764139
edited Jan 24 at 14:28
Andy Walls
asked Jan 24 at 14:10
Andy WallsAndy Walls
1,764139
asked Jan 24 at 14:10
Andy WallsAndy Walls
1,764139
asked Jan 24 at 14:10
Andy WallsAndy Walls
1,764139
1,764139
1
$begingroup$
As you have noticed, the claim is wrong. The integrand converges to $i$ uniformly in $theta$, therefore the integral tends to $int_pi^0 i ,dtheta$.
$endgroup$
– Maxim
Jan 25 at 18:27
add a comment |
1
$begingroup$
As you have noticed, the claim is wrong. The integrand converges to $i$ uniformly in $theta$, therefore the integral tends to $int_pi^0 i ,dtheta$.
$endgroup$
– Maxim
Jan 25 at 18:27
1
1
$begingroup$
As you have noticed, the claim is wrong. The integrand converges to $i$ uniformly in $theta$, therefore the integral tends to $int_pi^0 i ,dtheta$.
$endgroup$
– Maxim
Jan 25 at 18:27
$begingroup$
As you have noticed, the claim is wrong. The integrand converges to $i$ uniformly in $theta$, therefore the integral tends to $int_pi^0 i ,dtheta$.
$endgroup$
– Maxim
Jan 25 at 18:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Edited: Your guess is right, we should take upper semi-circle contour when $text{sgn}(s)<0$ and lower one when $text{sgn}(s)>0$. It is related to the behavior of the integral
$$
int_{C_R^+} frac{e^{isz}}{z}dz,quad s=pm 1
$$ as $Rtoinfty$ where $C_R^+: Re^{it}, 0le tle pi$ is an upper semi-circle contour. When we choose this contour, we expect that the limit of the above integral over $C_R^+$ tends to $0$ so that $int_{-infty}^infty frac{e^{pm iz}}{z}dz$ can be computed using the singularity at $z=0$. When $s=1$, we can do it since
$$
int_{C_R^+} frac{e^{iz}}{z}dz=iint_0^pi e^{iRcos t-Rsin t}dtstackrel{Rtoinfty}longrightarrow 0.
$$ This happens essentially because $|e^{iz}|=e^{-text{Im}(z)}=e^{-Rsin t}to 0$ as $Rtoinfty$. However, when $s=-1$, it is not the case anymore because $|e^{-iz}|=e^{text{Im}(z)}=e^{Rsin t}to infty$ for $tne 0,pi$, making us unable to control $int_{C_R^+} frac{e^{-iz}}{z}dz$ as $Rto infty$. In this case, it is natural to take the lower semi-circle contour $C_R^-$to have $int_{C_R^-} frac{e^{isz}}{z}dz to 0$. I hope this will help you.
We have
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=text{p.v.}int frac{cos (2pi sz)-isin(2pi sz)}{z}dz=-itext{p.v.}int frac{sin(2pi sz)}{z}dz
$$ since $zmapsto frac{cos(2pi sz)}{z}$ is an odd function. Note that if $a>0$, then by making change of variables $t=ax$,
$$
int_{-infty}^infty frac{sin ax}{x}dx=int_{-infty}^infty frac{sin t}{t}dt=pi
$$ and if $a=-|a|<0$, then
$$
int_{-infty}^infty frac{sin ax}{x}dx=-int_{-infty}^infty frac{sin |a|x}{x}dx=-pi.
$$ Therefore it follows
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=-ipi cdottext{sgn}(s)
$$
answered Jan 24 at 15:35
SongSong
18.1k21449
$endgroup$
$begingroup$
Have an upvote for a clear and concise derivation of the particular Fourier Transform given in the background information. However, the question is how sgn() could emerge from the integral in the title. I now believe that it can't. It is likely Bracewell was hand-waving over the detail of having to use two different closed contours depending on the sign of $s$, which is required to get the contour integral along the infinitely large semi-circular arc to vanish to $0$.
$endgroup$
– Andy Walls
Jan 25 at 13:53
$begingroup$
@AndyWalls I've edited my answer explaining what you have asked. I hope this will help ...
$endgroup$
– Song
Jan 25 at 14:37
add a comment |
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$begingroup$
Edited: Your guess is right, we should take upper semi-circle contour when $text{sgn}(s)<0$ and lower one when $text{sgn}(s)>0$. It is related to the behavior of the integral
$$
int_{C_R^+} frac{e^{isz}}{z}dz,quad s=pm 1
$$ as $Rtoinfty$ where $C_R^+: Re^{it}, 0le tle pi$ is an upper semi-circle contour. When we choose this contour, we expect that the limit of the above integral over $C_R^+$ tends to $0$ so that $int_{-infty}^infty frac{e^{pm iz}}{z}dz$ can be computed using the singularity at $z=0$. When $s=1$, we can do it since
$$
int_{C_R^+} frac{e^{iz}}{z}dz=iint_0^pi e^{iRcos t-Rsin t}dtstackrel{Rtoinfty}longrightarrow 0.
$$ This happens essentially because $|e^{iz}|=e^{-text{Im}(z)}=e^{-Rsin t}to 0$ as $Rtoinfty$. However, when $s=-1$, it is not the case anymore because $|e^{-iz}|=e^{text{Im}(z)}=e^{Rsin t}to infty$ for $tne 0,pi$, making us unable to control $int_{C_R^+} frac{e^{-iz}}{z}dz$ as $Rto infty$. In this case, it is natural to take the lower semi-circle contour $C_R^-$to have $int_{C_R^-} frac{e^{isz}}{z}dz to 0$. I hope this will help you.
We have
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=text{p.v.}int frac{cos (2pi sz)-isin(2pi sz)}{z}dz=-itext{p.v.}int frac{sin(2pi sz)}{z}dz
$$ since $zmapsto frac{cos(2pi sz)}{z}$ is an odd function. Note that if $a>0$, then by making change of variables $t=ax$,
$$
int_{-infty}^infty frac{sin ax}{x}dx=int_{-infty}^infty frac{sin t}{t}dt=pi
$$ and if $a=-|a|<0$, then
$$
int_{-infty}^infty frac{sin ax}{x}dx=-int_{-infty}^infty frac{sin |a|x}{x}dx=-pi.
$$ Therefore it follows
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=-ipi cdottext{sgn}(s)
$$
answered Jan 24 at 15:35
SongSong
18.1k21449
$endgroup$
$begingroup$
Have an upvote for a clear and concise derivation of the particular Fourier Transform given in the background information. However, the question is how sgn() could emerge from the integral in the title. I now believe that it can't. It is likely Bracewell was hand-waving over the detail of having to use two different closed contours depending on the sign of $s$, which is required to get the contour integral along the infinitely large semi-circular arc to vanish to $0$.
$endgroup$
– Andy Walls
Jan 25 at 13:53
$begingroup$
@AndyWalls I've edited my answer explaining what you have asked. I hope this will help ...
$endgroup$
– Song
Jan 25 at 14:37
add a comment |
$begingroup$
Edited: Your guess is right, we should take upper semi-circle contour when $text{sgn}(s)<0$ and lower one when $text{sgn}(s)>0$. It is related to the behavior of the integral
$$
int_{C_R^+} frac{e^{isz}}{z}dz,quad s=pm 1
$$ as $Rtoinfty$ where $C_R^+: Re^{it}, 0le tle pi$ is an upper semi-circle contour. When we choose this contour, we expect that the limit of the above integral over $C_R^+$ tends to $0$ so that $int_{-infty}^infty frac{e^{pm iz}}{z}dz$ can be computed using the singularity at $z=0$. When $s=1$, we can do it since
$$
int_{C_R^+} frac{e^{iz}}{z}dz=iint_0^pi e^{iRcos t-Rsin t}dtstackrel{Rtoinfty}longrightarrow 0.
$$ This happens essentially because $|e^{iz}|=e^{-text{Im}(z)}=e^{-Rsin t}to 0$ as $Rtoinfty$. However, when $s=-1$, it is not the case anymore because $|e^{-iz}|=e^{text{Im}(z)}=e^{Rsin t}to infty$ for $tne 0,pi$, making us unable to control $int_{C_R^+} frac{e^{-iz}}{z}dz$ as $Rto infty$. In this case, it is natural to take the lower semi-circle contour $C_R^-$to have $int_{C_R^-} frac{e^{isz}}{z}dz to 0$. I hope this will help you.
We have
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=text{p.v.}int frac{cos (2pi sz)-isin(2pi sz)}{z}dz=-itext{p.v.}int frac{sin(2pi sz)}{z}dz
$$ since $zmapsto frac{cos(2pi sz)}{z}$ is an odd function. Note that if $a>0$, then by making change of variables $t=ax$,
$$
int_{-infty}^infty frac{sin ax}{x}dx=int_{-infty}^infty frac{sin t}{t}dt=pi
$$ and if $a=-|a|<0$, then
$$
int_{-infty}^infty frac{sin ax}{x}dx=-int_{-infty}^infty frac{sin |a|x}{x}dx=-pi.
$$ Therefore it follows
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=-ipi cdottext{sgn}(s)
$$
answered Jan 24 at 15:35
SongSong
18.1k21449
$endgroup$
$begingroup$
Have an upvote for a clear and concise derivation of the particular Fourier Transform given in the background information. However, the question is how sgn() could emerge from the integral in the title. I now believe that it can't. It is likely Bracewell was hand-waving over the detail of having to use two different closed contours depending on the sign of $s$, which is required to get the contour integral along the infinitely large semi-circular arc to vanish to $0$.
$endgroup$
– Andy Walls
Jan 25 at 13:53
$begingroup$
@AndyWalls I've edited my answer explaining what you have asked. I hope this will help ...
$endgroup$
– Song
Jan 25 at 14:37
add a comment |
$begingroup$
Edited: Your guess is right, we should take upper semi-circle contour when $text{sgn}(s)<0$ and lower one when $text{sgn}(s)>0$. It is related to the behavior of the integral
$$
int_{C_R^+} frac{e^{isz}}{z}dz,quad s=pm 1
$$ as $Rtoinfty$ where $C_R^+: Re^{it}, 0le tle pi$ is an upper semi-circle contour. When we choose this contour, we expect that the limit of the above integral over $C_R^+$ tends to $0$ so that $int_{-infty}^infty frac{e^{pm iz}}{z}dz$ can be computed using the singularity at $z=0$. When $s=1$, we can do it since
$$
int_{C_R^+} frac{e^{iz}}{z}dz=iint_0^pi e^{iRcos t-Rsin t}dtstackrel{Rtoinfty}longrightarrow 0.
$$ This happens essentially because $|e^{iz}|=e^{-text{Im}(z)}=e^{-Rsin t}to 0$ as $Rtoinfty$. However, when $s=-1$, it is not the case anymore because $|e^{-iz}|=e^{text{Im}(z)}=e^{Rsin t}to infty$ for $tne 0,pi$, making us unable to control $int_{C_R^+} frac{e^{-iz}}{z}dz$ as $Rto infty$. In this case, it is natural to take the lower semi-circle contour $C_R^-$to have $int_{C_R^-} frac{e^{isz}}{z}dz to 0$. I hope this will help you.
We have
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=text{p.v.}int frac{cos (2pi sz)-isin(2pi sz)}{z}dz=-itext{p.v.}int frac{sin(2pi sz)}{z}dz
$$ since $zmapsto frac{cos(2pi sz)}{z}$ is an odd function. Note that if $a>0$, then by making change of variables $t=ax$,
$$
int_{-infty}^infty frac{sin ax}{x}dx=int_{-infty}^infty frac{sin t}{t}dt=pi
$$ and if $a=-|a|<0$, then
$$
int_{-infty}^infty frac{sin ax}{x}dx=-int_{-infty}^infty frac{sin |a|x}{x}dx=-pi.
$$ Therefore it follows
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=-ipi cdottext{sgn}(s)
$$
answered Jan 24 at 15:35
SongSong
18.1k21449
$endgroup$
Edited: Your guess is right, we should take upper semi-circle contour when $text{sgn}(s)<0$ and lower one when $text{sgn}(s)>0$. It is related to the behavior of the integral
$$
int_{C_R^+} frac{e^{isz}}{z}dz,quad s=pm 1
$$ as $Rtoinfty$ where $C_R^+: Re^{it}, 0le tle pi$ is an upper semi-circle contour. When we choose this contour, we expect that the limit of the above integral over $C_R^+$ tends to $0$ so that $int_{-infty}^infty frac{e^{pm iz}}{z}dz$ can be computed using the singularity at $z=0$. When $s=1$, we can do it since
$$
int_{C_R^+} frac{e^{iz}}{z}dz=iint_0^pi e^{iRcos t-Rsin t}dtstackrel{Rtoinfty}longrightarrow 0.
$$ This happens essentially because $|e^{iz}|=e^{-text{Im}(z)}=e^{-Rsin t}to 0$ as $Rtoinfty$. However, when $s=-1$, it is not the case anymore because $|e^{-iz}|=e^{text{Im}(z)}=e^{Rsin t}to infty$ for $tne 0,pi$, making us unable to control $int_{C_R^+} frac{e^{-iz}}{z}dz$ as $Rto infty$. In this case, it is natural to take the lower semi-circle contour $C_R^-$to have $int_{C_R^-} frac{e^{isz}}{z}dz to 0$. I hope this will help you.
We have
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=text{p.v.}int frac{cos (2pi sz)-isin(2pi sz)}{z}dz=-itext{p.v.}int frac{sin(2pi sz)}{z}dz
$$ since $zmapsto frac{cos(2pi sz)}{z}$ is an odd function. Note that if $a>0$, then by making change of variables $t=ax$,
$$
int_{-infty}^infty frac{sin ax}{x}dx=int_{-infty}^infty frac{sin t}{t}dt=pi
$$ and if $a=-|a|<0$, then
$$
int_{-infty}^infty frac{sin ax}{x}dx=-int_{-infty}^infty frac{sin |a|x}{x}dx=-pi.
$$ Therefore it follows
$$
text{p.v.}int frac{e^{-2pi isz}}{z}dz=-ipi cdottext{sgn}(s)
$$
answered Jan 24 at 15:35
SongSong
18.1k21449
edited Jan 25 at 14:21
answered Jan 24 at 15:35
SongSong
18.1k21449
answered Jan 24 at 15:35
SongSong
18.1k21449
answered Jan 24 at 15:35
SongSong
18.1k21449
18.1k21449
$begingroup$
Have an upvote for a clear and concise derivation of the particular Fourier Transform given in the background information. However, the question is how sgn() could emerge from the integral in the title. I now believe that it can't. It is likely Bracewell was hand-waving over the detail of having to use two different closed contours depending on the sign of $s$, which is required to get the contour integral along the infinitely large semi-circular arc to vanish to $0$.
$endgroup$
– Andy Walls
Jan 25 at 13:53
$begingroup$
@AndyWalls I've edited my answer explaining what you have asked. I hope this will help ...
$endgroup$
– Song
Jan 25 at 14:37
add a comment |
$begingroup$
Have an upvote for a clear and concise derivation of the particular Fourier Transform given in the background information. However, the question is how sgn() could emerge from the integral in the title. I now believe that it can't. It is likely Bracewell was hand-waving over the detail of having to use two different closed contours depending on the sign of $s$, which is required to get the contour integral along the infinitely large semi-circular arc to vanish to $0$.
$endgroup$
– Andy Walls
Jan 25 at 13:53
$begingroup$
@AndyWalls I've edited my answer explaining what you have asked. I hope this will help ...
$endgroup$
– Song
Jan 25 at 14:37
$begingroup$
Have an upvote for a clear and concise derivation of the particular Fourier Transform given in the background information. However, the question is how sgn() could emerge from the integral in the title. I now believe that it can't. It is likely Bracewell was hand-waving over the detail of having to use two different closed contours depending on the sign of $s$, which is required to get the contour integral along the infinitely large semi-circular arc to vanish to $0$.
$endgroup$
– Andy Walls
Jan 25 at 13:53
$begingroup$
Have an upvote for a clear and concise derivation of the particular Fourier Transform given in the background information. However, the question is how sgn() could emerge from the integral in the title. I now believe that it can't. It is likely Bracewell was hand-waving over the detail of having to use two different closed contours depending on the sign of $s$, which is required to get the contour integral along the infinitely large semi-circular arc to vanish to $0$.
$endgroup$
– Andy Walls
Jan 25 at 13:53
$begingroup$
@AndyWalls I've edited my answer explaining what you have asked. I hope this will help ...
$endgroup$
– Song
Jan 25 at 14:37
$begingroup$
@AndyWalls I've edited my answer explaining what you have asked. I hope this will help ...
$endgroup$
– Song
Jan 25 at 14:37
add a comment |
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As you have noticed, the claim is wrong. The integrand converges to $i$ uniformly in $theta$, therefore the integral tends to $int_pi^0 i ,dtheta$.
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– Maxim
Jan 25 at 18:27