Mixing
How to calculate the integral $intfrac{e^{ixy}}{w^2-x^2}dx$?
$begingroup$
How to calculate the integral$$int_{-infty}^{infty}frac{e^{ixy}}{w^2-x^2}dx$$
where w,y are constants.
I tried to separate them as
$$frac{1}{2w}int_{-infty}^{infty}(frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x})dx$$but I'm not sure this is a correct way.
calculus integration complex-analysis definite-integrals fourier-analysis
edited Jan 24 at 14:20
Dylan
13.9k31127
asked Jan 24 at 2:45
kinder chenkinder chen
1434
$endgroup$
add a comment |
$begingroup$
How to calculate the integral$$int_{-infty}^{infty}frac{e^{ixy}}{w^2-x^2}dx$$
where w,y are constants.
I tried to separate them as
$$frac{1}{2w}int_{-infty}^{infty}(frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x})dx$$but I'm not sure this is a correct way.
calculus integration complex-analysis definite-integrals fourier-analysis
edited Jan 24 at 14:20
Dylan
13.9k31127
asked Jan 24 at 2:45
kinder chenkinder chen
1434
$endgroup$
4
$begingroup$
Assuming that the integral is interpreted in Cauchy principal-value sense, the answer is $$ ipi left( underset{z=-w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} + underset{z=w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} right) = frac{pi sin(wy)}{w}. $$
$endgroup$
– Sangchul Lee
Jan 24 at 2:51
2
$begingroup$
If "x,w are constants" why are you integrating with respect to $x$?
$endgroup$
– clathratus
Jan 24 at 3:41
1
$begingroup$
For $Re(w) > 0$ then $frac{1}{w^2+x^2}$ is the FT of $e^{-w |t|}$. Then the case $Re(w)= 0$ (integral interpreted in principal value) is obtained with the mean value of the FT of $e^{-(w+epsilon)|t|}$ and $e^{-(-w+epsilon)|t|}$
$endgroup$
– reuns
Jan 24 at 4:01
$begingroup$
@SangchulLee thx, but why it's $pi i$ not $2pi i$ when you use Residue theorem?
$endgroup$
– kinder chen
Jan 27 at 0:41
$begingroup$
That is essentially because we only wind the poles half-times.
$endgroup$
– Sangchul Lee
Jan 27 at 8:57
add a comment |
$begingroup$
How to calculate the integral$$int_{-infty}^{infty}frac{e^{ixy}}{w^2-x^2}dx$$
where w,y are constants.
I tried to separate them as
$$frac{1}{2w}int_{-infty}^{infty}(frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x})dx$$but I'm not sure this is a correct way.
calculus integration complex-analysis definite-integrals fourier-analysis
edited Jan 24 at 14:20
Dylan
13.9k31127
asked Jan 24 at 2:45
kinder chenkinder chen
1434
$endgroup$
How to calculate the integral$$int_{-infty}^{infty}frac{e^{ixy}}{w^2-x^2}dx$$
where w,y are constants.
I tried to separate them as
$$frac{1}{2w}int_{-infty}^{infty}(frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x})dx$$but I'm not sure this is a correct way.
calculus integration complex-analysis definite-integrals fourier-analysis
calculus integration complex-analysis definite-integrals fourier-analysis
edited Jan 24 at 14:20
Dylan
13.9k31127
asked Jan 24 at 2:45
kinder chenkinder chen
1434
edited Jan 24 at 14:20
Dylan
13.9k31127
asked Jan 24 at 2:45
kinder chenkinder chen
1434
edited Jan 24 at 14:20
Dylan
13.9k31127
edited Jan 24 at 14:20
Dylan
13.9k31127
edited Jan 24 at 14:20
Dylan
13.9k31127
13.9k31127
asked Jan 24 at 2:45
kinder chenkinder chen
1434
asked Jan 24 at 2:45
kinder chenkinder chen
1434
asked Jan 24 at 2:45
kinder chenkinder chen
1434
1434
4
$begingroup$
Assuming that the integral is interpreted in Cauchy principal-value sense, the answer is $$ ipi left( underset{z=-w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} + underset{z=w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} right) = frac{pi sin(wy)}{w}. $$
$endgroup$
– Sangchul Lee
Jan 24 at 2:51
2
$begingroup$
If "x,w are constants" why are you integrating with respect to $x$?
$endgroup$
– clathratus
Jan 24 at 3:41
1
$begingroup$
For $Re(w) > 0$ then $frac{1}{w^2+x^2}$ is the FT of $e^{-w |t|}$. Then the case $Re(w)= 0$ (integral interpreted in principal value) is obtained with the mean value of the FT of $e^{-(w+epsilon)|t|}$ and $e^{-(-w+epsilon)|t|}$
$endgroup$
– reuns
Jan 24 at 4:01
$begingroup$
@SangchulLee thx, but why it's $pi i$ not $2pi i$ when you use Residue theorem?
$endgroup$
– kinder chen
Jan 27 at 0:41
$begingroup$
That is essentially because we only wind the poles half-times.
$endgroup$
– Sangchul Lee
Jan 27 at 8:57
add a comment |
4
$begingroup$
Assuming that the integral is interpreted in Cauchy principal-value sense, the answer is $$ ipi left( underset{z=-w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} + underset{z=w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} right) = frac{pi sin(wy)}{w}. $$
$endgroup$
– Sangchul Lee
Jan 24 at 2:51
2
$begingroup$
If "x,w are constants" why are you integrating with respect to $x$?
$endgroup$
– clathratus
Jan 24 at 3:41
1
$begingroup$
For $Re(w) > 0$ then $frac{1}{w^2+x^2}$ is the FT of $e^{-w |t|}$. Then the case $Re(w)= 0$ (integral interpreted in principal value) is obtained with the mean value of the FT of $e^{-(w+epsilon)|t|}$ and $e^{-(-w+epsilon)|t|}$
$endgroup$
– reuns
Jan 24 at 4:01
$begingroup$
@SangchulLee thx, but why it's $pi i$ not $2pi i$ when you use Residue theorem?
$endgroup$
– kinder chen
Jan 27 at 0:41
$begingroup$
That is essentially because we only wind the poles half-times.
$endgroup$
– Sangchul Lee
Jan 27 at 8:57
4
4
$begingroup$
Assuming that the integral is interpreted in Cauchy principal-value sense, the answer is $$ ipi left( underset{z=-w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} + underset{z=w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} right) = frac{pi sin(wy)}{w}. $$
$endgroup$
– Sangchul Lee
Jan 24 at 2:51
$begingroup$
Assuming that the integral is interpreted in Cauchy principal-value sense, the answer is $$ ipi left( underset{z=-w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} + underset{z=w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} right) = frac{pi sin(wy)}{w}. $$
$endgroup$
– Sangchul Lee
Jan 24 at 2:51
2
2
$begingroup$
If "x,w are constants" why are you integrating with respect to $x$?
$endgroup$
– clathratus
Jan 24 at 3:41
$begingroup$
If "x,w are constants" why are you integrating with respect to $x$?
$endgroup$
– clathratus
Jan 24 at 3:41
1
1
$begingroup$
For $Re(w) > 0$ then $frac{1}{w^2+x^2}$ is the FT of $e^{-w |t|}$. Then the case $Re(w)= 0$ (integral interpreted in principal value) is obtained with the mean value of the FT of $e^{-(w+epsilon)|t|}$ and $e^{-(-w+epsilon)|t|}$
$endgroup$
– reuns
Jan 24 at 4:01
$begingroup$
For $Re(w) > 0$ then $frac{1}{w^2+x^2}$ is the FT of $e^{-w |t|}$. Then the case $Re(w)= 0$ (integral interpreted in principal value) is obtained with the mean value of the FT of $e^{-(w+epsilon)|t|}$ and $e^{-(-w+epsilon)|t|}$
$endgroup$
– reuns
Jan 24 at 4:01
$begingroup$
@SangchulLee thx, but why it's $pi i$ not $2pi i$ when you use Residue theorem?
$endgroup$
– kinder chen
Jan 27 at 0:41
$begingroup$
@SangchulLee thx, but why it's $pi i$ not $2pi i$ when you use Residue theorem?
$endgroup$
– kinder chen
Jan 27 at 0:41
$begingroup$
That is essentially because we only wind the poles half-times.
$endgroup$
– Sangchul Lee
Jan 27 at 8:57
$begingroup$
That is essentially because we only wind the poles half-times.
$endgroup$
– Sangchul Lee
Jan 27 at 8:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the following contour, where $Gamma_R$ is a simicircular arc of the radius $R$ and $gamma_{pm,epsilon}$ are semicircular arcs of radius $epsilon$ centered at $pm w$, respectively.
$hspace{5em}$ 
If we call this contour by $C$, then the Cauchy integration formula tells that
$$ int_{C} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z = 0. $$
Now note that
$displaystyle left| int_{Gamma_R} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z right|
leq frac{R}{R^2 - w^2} xrightarrow[Rtoinfty]{} 0 $,$displaystyle int_{gamma_{pm,epsilon}} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z
xrightarrow[epsilonto 0^+]{} -ipi , underset{z=pm w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} $.
So, letting $R to infty$ and $epsilon to 0^+$, we have
$$ operatorname{PV}!int_{-infty}^{infty} frac{e^{iyx}}{w^2 - x^2} , mathrm{d}x - ipi left( underset{z=- w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} + underset{z=w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} right) = 0. $$
answered Jan 27 at 8:56
Sangchul LeeSangchul Lee
95.8k12171280
$endgroup$
$begingroup$
Thx, if I separate them as $frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x}$, can I use $2pi i$?
$endgroup$
– kinder chen
Jan 28 at 19:40
$begingroup$
@Kinder, Not quite. Essentially this appears because we are considering Cauchy principal value, not because of some special property of the integrand.
$endgroup$
– Sangchul Lee
Jan 28 at 19:42
$begingroup$
I'm confused which one I'm supposed to use $pi i$ or $2pi i$ compared with wikipedia. I thought the arc integral is zero, is the whole contour integral also zero?
$endgroup$
– kinder chen
Jan 28 at 23:08
$begingroup$
@Kinder, If you integrate a holomorphic function $f$ along a closed contour that winds the point $z_0$ counter-clockwise $n$ times, then there is $$2npi i times [text{residue of }ftext{ at }z_0]$$ contribution. If one opens up the proof, one can see that this comes from the limit $$lim_{rto0^+}int_{0}^{2npi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta})=2npi i times text{[residue]}.$$ Similarly, in my computation, we end up with computing integral of the form $$lim_{rto0^+}int_{0}^{pi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta}).$$
$endgroup$
– Sangchul Lee
Jan 29 at 0:37
$begingroup$
It is like the winding number around the point $z_0$ is $frac{1}{2}$.
$endgroup$
– Sangchul Lee
Jan 29 at 0:40
|
show 2 more comments
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1 Answer
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1 Answer
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votes
$begingroup$
Consider the following contour, where $Gamma_R$ is a simicircular arc of the radius $R$ and $gamma_{pm,epsilon}$ are semicircular arcs of radius $epsilon$ centered at $pm w$, respectively.
$hspace{5em}$
If we call this contour by $C$, then the Cauchy integration formula tells that
$$ int_{C} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z = 0. $$
Now note that
$displaystyle left| int_{Gamma_R} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z right|
leq frac{R}{R^2 - w^2} xrightarrow[Rtoinfty]{} 0 $,$displaystyle int_{gamma_{pm,epsilon}} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z
xrightarrow[epsilonto 0^+]{} -ipi , underset{z=pm w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} $.
So, letting $R to infty$ and $epsilon to 0^+$, we have
$$ operatorname{PV}!int_{-infty}^{infty} frac{e^{iyx}}{w^2 - x^2} , mathrm{d}x - ipi left( underset{z=- w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} + underset{z=w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} right) = 0. $$
answered Jan 27 at 8:56
Sangchul LeeSangchul Lee
95.8k12171280
$endgroup$
$begingroup$
Thx, if I separate them as $frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x}$, can I use $2pi i$?
$endgroup$
– kinder chen
Jan 28 at 19:40
$begingroup$
@Kinder, Not quite. Essentially this appears because we are considering Cauchy principal value, not because of some special property of the integrand.
$endgroup$
– Sangchul Lee
Jan 28 at 19:42
$begingroup$
I'm confused which one I'm supposed to use $pi i$ or $2pi i$ compared with wikipedia. I thought the arc integral is zero, is the whole contour integral also zero?
$endgroup$
– kinder chen
Jan 28 at 23:08
$begingroup$
@Kinder, If you integrate a holomorphic function $f$ along a closed contour that winds the point $z_0$ counter-clockwise $n$ times, then there is $$2npi i times [text{residue of }ftext{ at }z_0]$$ contribution. If one opens up the proof, one can see that this comes from the limit $$lim_{rto0^+}int_{0}^{2npi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta})=2npi i times text{[residue]}.$$ Similarly, in my computation, we end up with computing integral of the form $$lim_{rto0^+}int_{0}^{pi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta}).$$
$endgroup$
– Sangchul Lee
Jan 29 at 0:37
$begingroup$
It is like the winding number around the point $z_0$ is $frac{1}{2}$.
$endgroup$
– Sangchul Lee
Jan 29 at 0:40
|
show 2 more comments
$begingroup$
Consider the following contour, where $Gamma_R$ is a simicircular arc of the radius $R$ and $gamma_{pm,epsilon}$ are semicircular arcs of radius $epsilon$ centered at $pm w$, respectively.
$hspace{5em}$
If we call this contour by $C$, then the Cauchy integration formula tells that
$$ int_{C} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z = 0. $$
Now note that
$displaystyle left| int_{Gamma_R} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z right|
leq frac{R}{R^2 - w^2} xrightarrow[Rtoinfty]{} 0 $,$displaystyle int_{gamma_{pm,epsilon}} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z
xrightarrow[epsilonto 0^+]{} -ipi , underset{z=pm w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} $.
So, letting $R to infty$ and $epsilon to 0^+$, we have
$$ operatorname{PV}!int_{-infty}^{infty} frac{e^{iyx}}{w^2 - x^2} , mathrm{d}x - ipi left( underset{z=- w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} + underset{z=w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} right) = 0. $$
answered Jan 27 at 8:56
Sangchul LeeSangchul Lee
95.8k12171280
$endgroup$
$begingroup$
Thx, if I separate them as $frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x}$, can I use $2pi i$?
$endgroup$
– kinder chen
Jan 28 at 19:40
$begingroup$
@Kinder, Not quite. Essentially this appears because we are considering Cauchy principal value, not because of some special property of the integrand.
$endgroup$
– Sangchul Lee
Jan 28 at 19:42
$begingroup$
I'm confused which one I'm supposed to use $pi i$ or $2pi i$ compared with wikipedia. I thought the arc integral is zero, is the whole contour integral also zero?
$endgroup$
– kinder chen
Jan 28 at 23:08
$begingroup$
@Kinder, If you integrate a holomorphic function $f$ along a closed contour that winds the point $z_0$ counter-clockwise $n$ times, then there is $$2npi i times [text{residue of }ftext{ at }z_0]$$ contribution. If one opens up the proof, one can see that this comes from the limit $$lim_{rto0^+}int_{0}^{2npi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta})=2npi i times text{[residue]}.$$ Similarly, in my computation, we end up with computing integral of the form $$lim_{rto0^+}int_{0}^{pi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta}).$$
$endgroup$
– Sangchul Lee
Jan 29 at 0:37
$begingroup$
It is like the winding number around the point $z_0$ is $frac{1}{2}$.
$endgroup$
– Sangchul Lee
Jan 29 at 0:40
|
show 2 more comments
$begingroup$
Consider the following contour, where $Gamma_R$ is a simicircular arc of the radius $R$ and $gamma_{pm,epsilon}$ are semicircular arcs of radius $epsilon$ centered at $pm w$, respectively.
$hspace{5em}$
If we call this contour by $C$, then the Cauchy integration formula tells that
$$ int_{C} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z = 0. $$
Now note that
$displaystyle left| int_{Gamma_R} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z right|
leq frac{R}{R^2 - w^2} xrightarrow[Rtoinfty]{} 0 $,$displaystyle int_{gamma_{pm,epsilon}} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z
xrightarrow[epsilonto 0^+]{} -ipi , underset{z=pm w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} $.
So, letting $R to infty$ and $epsilon to 0^+$, we have
$$ operatorname{PV}!int_{-infty}^{infty} frac{e^{iyx}}{w^2 - x^2} , mathrm{d}x - ipi left( underset{z=- w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} + underset{z=w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} right) = 0. $$
answered Jan 27 at 8:56
Sangchul LeeSangchul Lee
95.8k12171280
$endgroup$
Consider the following contour, where $Gamma_R$ is a simicircular arc of the radius $R$ and $gamma_{pm,epsilon}$ are semicircular arcs of radius $epsilon$ centered at $pm w$, respectively.
$hspace{5em}$
If we call this contour by $C$, then the Cauchy integration formula tells that
$$ int_{C} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z = 0. $$
Now note that
$displaystyle left| int_{Gamma_R} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z right|
leq frac{R}{R^2 - w^2} xrightarrow[Rtoinfty]{} 0 $,$displaystyle int_{gamma_{pm,epsilon}} frac{e^{iyz}}{w^2 - z^2} , mathrm{d}z
xrightarrow[epsilonto 0^+]{} -ipi , underset{z=pm w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} $.
So, letting $R to infty$ and $epsilon to 0^+$, we have
$$ operatorname{PV}!int_{-infty}^{infty} frac{e^{iyx}}{w^2 - x^2} , mathrm{d}x - ipi left( underset{z=- w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} + underset{z=w}{mathrm{Res}} , frac{e^{iyz}}{w^2 - z^2} right) = 0. $$
answered Jan 27 at 8:56
Sangchul LeeSangchul Lee
95.8k12171280
answered Jan 27 at 8:56
Sangchul LeeSangchul Lee
95.8k12171280
answered Jan 27 at 8:56
Sangchul LeeSangchul Lee
95.8k12171280
answered Jan 27 at 8:56
Sangchul LeeSangchul Lee
95.8k12171280
95.8k12171280
$begingroup$
Thx, if I separate them as $frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x}$, can I use $2pi i$?
$endgroup$
– kinder chen
Jan 28 at 19:40
$begingroup$
@Kinder, Not quite. Essentially this appears because we are considering Cauchy principal value, not because of some special property of the integrand.
$endgroup$
– Sangchul Lee
Jan 28 at 19:42
$begingroup$
I'm confused which one I'm supposed to use $pi i$ or $2pi i$ compared with wikipedia. I thought the arc integral is zero, is the whole contour integral also zero?
$endgroup$
– kinder chen
Jan 28 at 23:08
$begingroup$
@Kinder, If you integrate a holomorphic function $f$ along a closed contour that winds the point $z_0$ counter-clockwise $n$ times, then there is $$2npi i times [text{residue of }ftext{ at }z_0]$$ contribution. If one opens up the proof, one can see that this comes from the limit $$lim_{rto0^+}int_{0}^{2npi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta})=2npi i times text{[residue]}.$$ Similarly, in my computation, we end up with computing integral of the form $$lim_{rto0^+}int_{0}^{pi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta}).$$
$endgroup$
– Sangchul Lee
Jan 29 at 0:37
$begingroup$
It is like the winding number around the point $z_0$ is $frac{1}{2}$.
$endgroup$
– Sangchul Lee
Jan 29 at 0:40
|
show 2 more comments
$begingroup$
Thx, if I separate them as $frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x}$, can I use $2pi i$?
$endgroup$
– kinder chen
Jan 28 at 19:40
$begingroup$
@Kinder, Not quite. Essentially this appears because we are considering Cauchy principal value, not because of some special property of the integrand.
$endgroup$
– Sangchul Lee
Jan 28 at 19:42
$begingroup$
I'm confused which one I'm supposed to use $pi i$ or $2pi i$ compared with wikipedia. I thought the arc integral is zero, is the whole contour integral also zero?
$endgroup$
– kinder chen
Jan 28 at 23:08
$begingroup$
@Kinder, If you integrate a holomorphic function $f$ along a closed contour that winds the point $z_0$ counter-clockwise $n$ times, then there is $$2npi i times [text{residue of }ftext{ at }z_0]$$ contribution. If one opens up the proof, one can see that this comes from the limit $$lim_{rto0^+}int_{0}^{2npi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta})=2npi i times text{[residue]}.$$ Similarly, in my computation, we end up with computing integral of the form $$lim_{rto0^+}int_{0}^{pi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta}).$$
$endgroup$
– Sangchul Lee
Jan 29 at 0:37
$begingroup$
It is like the winding number around the point $z_0$ is $frac{1}{2}$.
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– Sangchul Lee
Jan 29 at 0:40
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Thx, if I separate them as $frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x}$, can I use $2pi i$?
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– kinder chen
Jan 28 at 19:40
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Thx, if I separate them as $frac{e^{ixy}}{w+x}+frac{e^{ixy}}{w-x}$, can I use $2pi i$?
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– kinder chen
Jan 28 at 19:40
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@Kinder, Not quite. Essentially this appears because we are considering Cauchy principal value, not because of some special property of the integrand.
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– Sangchul Lee
Jan 28 at 19:42
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@Kinder, Not quite. Essentially this appears because we are considering Cauchy principal value, not because of some special property of the integrand.
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– Sangchul Lee
Jan 28 at 19:42
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I'm confused which one I'm supposed to use $pi i$ or $2pi i$ compared with wikipedia. I thought the arc integral is zero, is the whole contour integral also zero?
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– kinder chen
Jan 28 at 23:08
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I'm confused which one I'm supposed to use $pi i$ or $2pi i$ compared with wikipedia. I thought the arc integral is zero, is the whole contour integral also zero?
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– kinder chen
Jan 28 at 23:08
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@Kinder, If you integrate a holomorphic function $f$ along a closed contour that winds the point $z_0$ counter-clockwise $n$ times, then there is $$2npi i times [text{residue of }ftext{ at }z_0]$$ contribution. If one opens up the proof, one can see that this comes from the limit $$lim_{rto0^+}int_{0}^{2npi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta})=2npi i times text{[residue]}.$$ Similarly, in my computation, we end up with computing integral of the form $$lim_{rto0^+}int_{0}^{pi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta}).$$
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– Sangchul Lee
Jan 29 at 0:37
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@Kinder, If you integrate a holomorphic function $f$ along a closed contour that winds the point $z_0$ counter-clockwise $n$ times, then there is $$2npi i times [text{residue of }ftext{ at }z_0]$$ contribution. If one opens up the proof, one can see that this comes from the limit $$lim_{rto0^+}int_{0}^{2npi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta})=2npi i times text{[residue]}.$$ Similarly, in my computation, we end up with computing integral of the form $$lim_{rto0^+}int_{0}^{pi}f(z_0+re^{itheta}),mathrm{d}(e^{itheta}).$$
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– Sangchul Lee
Jan 29 at 0:37
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It is like the winding number around the point $z_0$ is $frac{1}{2}$.
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– Sangchul Lee
Jan 29 at 0:40
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It is like the winding number around the point $z_0$ is $frac{1}{2}$.
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– Sangchul Lee
Jan 29 at 0:40
|
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Assuming that the integral is interpreted in Cauchy principal-value sense, the answer is $$ ipi left( underset{z=-w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} + underset{z=w}{mathrm{Res}},frac{e^{iyz}}{w^2-z^2} right) = frac{pi sin(wy)}{w}. $$
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– Sangchul Lee
Jan 24 at 2:51
2
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If "x,w are constants" why are you integrating with respect to $x$?
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– clathratus
Jan 24 at 3:41
1
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For $Re(w) > 0$ then $frac{1}{w^2+x^2}$ is the FT of $e^{-w |t|}$. Then the case $Re(w)= 0$ (integral interpreted in principal value) is obtained with the mean value of the FT of $e^{-(w+epsilon)|t|}$ and $e^{-(-w+epsilon)|t|}$
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– reuns
Jan 24 at 4:01
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@SangchulLee thx, but why it's $pi i$ not $2pi i$ when you use Residue theorem?
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– kinder chen
Jan 27 at 0:41
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That is essentially because we only wind the poles half-times.
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– Sangchul Lee
Jan 27 at 8:57