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How can estimate the following integral on $mathbb S^1$?
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For $d>0$ and $s, t in mathbb S^1=zin mathbb C mid $ the unit circle. I need help to give an estimate for the following integral
$$int_ sin mathbb S^1: , fracdsleq C_d,$$
for all fixed $t in mathbb S^1$, and $C_d$ is a constant depend of $d>0$.
For $s=e^itheta$ et $r=e^ivarphi$, using the following parameterization: $int_mathbb S^1f(s) ,ds = int_mathbb S^1f(e^itheta) ,dtheta$ for a function defined on $mathbb S^1$. Then, we have
$beginalign
int_geq d fracds &=int_ fracdthetae^itheta-e^ivarphi=int_geq d fracdtheta\
&=int_ sqrt2-2cos(theta-varphi)geq d fracdthetasqrt2-2 cos(theta-varphi) = fracsqrt22int_ 1-cos(theta-varphi)geq d^2/2 fracdthetasqrt1-cos(theta-varphi) cdots
endalign$
So how to continue if not another track?
Thank you in advance.
calculus real-analysis complex-analysis functional-analysis
asked Aug 6 at 12:38
Z. Alfata
878413
add a comment |Â
up vote
0
down vote
favorite
For $d>0$ and $s, t in mathbb S^1=zin mathbb C mid $ the unit circle. I need help to give an estimate for the following integral
$$int_ sin mathbb S^1: , fracdsleq C_d,$$
for all fixed $t in mathbb S^1$, and $C_d$ is a constant depend of $d>0$.
For $s=e^itheta$ et $r=e^ivarphi$, using the following parameterization: $int_mathbb S^1f(s) ,ds = int_mathbb S^1f(e^itheta) ,dtheta$ for a function defined on $mathbb S^1$. Then, we have
$beginalign
int_geq d fracds &=int_ fracdthetae^itheta-e^ivarphi=int_geq d fracdtheta\
&=int_ sqrt2-2cos(theta-varphi)geq d fracdthetasqrt2-2 cos(theta-varphi) = fracsqrt22int_ 1-cos(theta-varphi)geq d^2/2 fracdthetasqrt1-cos(theta-varphi) cdots
endalign$
So how to continue if not another track?
Thank you in advance.
calculus real-analysis complex-analysis functional-analysis
asked Aug 6 at 12:38
Z. Alfata
878413
1
Isn't your integral always infinite? Also, your calculations become easier if you use $1-cos x = 2sin^2frac x2$.
– Kusma
Aug 6 at 12:45
If I'm interpreting it correctly (The integral of the reciprocal of distance to a chosen point on $S_1$ over a region of $S_1$ not too far away from that point), then it would indeed diverge.
– Kajelad
Aug 6 at 12:47
The domain of integration is $ sin mathbb S^1: , $, I'm sorry!!!
– Z. Alfata
Aug 8 at 9:47
How good must be the estimate? An utterly trivial bound is $2pi/d$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:25
And you can suppose wlog that $t = 1$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:27
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For $d>0$ and $s, t in mathbb S^1=zin mathbb C mid $ the unit circle. I need help to give an estimate for the following integral
$$int_ sin mathbb S^1: , fracdsleq C_d,$$
for all fixed $t in mathbb S^1$, and $C_d$ is a constant depend of $d>0$.
For $s=e^itheta$ et $r=e^ivarphi$, using the following parameterization: $int_mathbb S^1f(s) ,ds = int_mathbb S^1f(e^itheta) ,dtheta$ for a function defined on $mathbb S^1$. Then, we have
$beginalign
int_geq d fracds &=int_ fracdthetae^itheta-e^ivarphi=int_geq d fracdtheta\
&=int_ sqrt2-2cos(theta-varphi)geq d fracdthetasqrt2-2 cos(theta-varphi) = fracsqrt22int_ 1-cos(theta-varphi)geq d^2/2 fracdthetasqrt1-cos(theta-varphi) cdots
endalign$
So how to continue if not another track?
Thank you in advance.
calculus real-analysis complex-analysis functional-analysis
asked Aug 6 at 12:38
Z. Alfata
878413
For $d>0$ and $s, t in mathbb S^1=zin mathbb C mid $ the unit circle. I need help to give an estimate for the following integral
$$int_ sin mathbb S^1: , fracdsleq C_d,$$
for all fixed $t in mathbb S^1$, and $C_d$ is a constant depend of $d>0$.
For $s=e^itheta$ et $r=e^ivarphi$, using the following parameterization: $int_mathbb S^1f(s) ,ds = int_mathbb S^1f(e^itheta) ,dtheta$ for a function defined on $mathbb S^1$. Then, we have
$beginalign
int_geq d fracds &=int_ fracdthetae^itheta-e^ivarphi=int_geq d fracdtheta\
&=int_ sqrt2-2cos(theta-varphi)geq d fracdthetasqrt2-2 cos(theta-varphi) = fracsqrt22int_ 1-cos(theta-varphi)geq d^2/2 fracdthetasqrt1-cos(theta-varphi) cdots
endalign$
So how to continue if not another track?
Thank you in advance.
calculus real-analysis complex-analysis functional-analysis
asked Aug 6 at 12:38
Z. Alfata
878413
edited Aug 8 at 9:47
asked Aug 6 at 12:38
Z. Alfata
878413
asked Aug 6 at 12:38
Z. Alfata
878413
asked Aug 6 at 12:38
Z. Alfata
878413
878413
1
Isn't your integral always infinite? Also, your calculations become easier if you use $1-cos x = 2sin^2frac x2$.
– Kusma
Aug 6 at 12:45
If I'm interpreting it correctly (The integral of the reciprocal of distance to a chosen point on $S_1$ over a region of $S_1$ not too far away from that point), then it would indeed diverge.
– Kajelad
Aug 6 at 12:47
The domain of integration is $ sin mathbb S^1: , $, I'm sorry!!!
– Z. Alfata
Aug 8 at 9:47
How good must be the estimate? An utterly trivial bound is $2pi/d$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:25
And you can suppose wlog that $t = 1$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:27
add a comment |Â
1
Isn't your integral always infinite? Also, your calculations become easier if you use $1-cos x = 2sin^2frac x2$.
– Kusma
Aug 6 at 12:45
If I'm interpreting it correctly (The integral of the reciprocal of distance to a chosen point on $S_1$ over a region of $S_1$ not too far away from that point), then it would indeed diverge.
– Kajelad
Aug 6 at 12:47
The domain of integration is $ sin mathbb S^1: , $, I'm sorry!!!
– Z. Alfata
Aug 8 at 9:47
How good must be the estimate? An utterly trivial bound is $2pi/d$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:25
And you can suppose wlog that $t = 1$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:27
1
1
Isn't your integral always infinite? Also, your calculations become easier if you use $1-cos x = 2sin^2frac x2$.
– Kusma
Aug 6 at 12:45
Isn't your integral always infinite? Also, your calculations become easier if you use $1-cos x = 2sin^2frac x2$.
– Kusma
Aug 6 at 12:45
If I'm interpreting it correctly (The integral of the reciprocal of distance to a chosen point on $S_1$ over a region of $S_1$ not too far away from that point), then it would indeed diverge.
– Kajelad
Aug 6 at 12:47
If I'm interpreting it correctly (The integral of the reciprocal of distance to a chosen point on $S_1$ over a region of $S_1$ not too far away from that point), then it would indeed diverge.
– Kajelad
Aug 6 at 12:47
The domain of integration is $ sin mathbb S^1: , $, I'm sorry!!!
– Z. Alfata
Aug 8 at 9:47
The domain of integration is $ sin mathbb S^1: , $, I'm sorry!!!
– Z. Alfata
Aug 8 at 9:47
How good must be the estimate? An utterly trivial bound is $2pi/d$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:25
How good must be the estimate? An utterly trivial bound is $2pi/d$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:25
And you can suppose wlog that $t = 1$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:27
And you can suppose wlog that $t = 1$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:27
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Draw a figure! Then you'll see that your integral is
$$J(d)=2int_alpha^pi 1over 2sinphiover2>dphi$$
whereby $alpha$ is defined by $>2sinalphaover2=d$. Replace this by
$$J(d)=2int_beta^pi/21oversinpsi>dpsi,qquad 2sinbeta=d .$$
This amounts to
$$J(d)=-2logleft(tanbetaover2right),qquad 2sinbeta=d .$$
Letting $tanbetaover2=:tau$ we have $$d=2sinbeta=4tauover 1+tau^2 ,$$
and therefore
$$tau=dover2+sqrt4-d^2 .$$
It follows that
$$J(d)=log1overtau^2=log8-d^2+4sqrt4-d^2over d^2leq3over dqquad(0<dleq2) .$$
The last estimate comes from inspecting a plot. Note that for any $alpha>0$ one has $log tleq t^alpha$ for large $t>0$.
answered Aug 8 at 18:22
Christian Blatter
164k7108306
@ Christian Blatter, with this estimate $J(d)leq log(frac16d^2)$, is what has the following estimate $J(d)leq C d^-1$, where $C$ is a constant $C>0$?
– Z. Alfata
Aug 10 at 20:32
add a comment |Â
up vote
1
down vote
WLOG you can suppose $t = 1$ ($varphi = 0$). And your integral has elementary primitive:
$$
intfracdthetasqrt1-cos(theta) =
frac2sin(theta/2)logtan(theta/4)sqrt1 - costheta + C
$$
http://www.wolframalpha.com/input/?i=int(1%2Fsqrt(1+-+cos+t),t)
answered Aug 8 at 15:23
MartÃn-Blas Pérez Pinilla
33.4k42570
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Draw a figure! Then you'll see that your integral is
$$J(d)=2int_alpha^pi 1over 2sinphiover2>dphi$$
whereby $alpha$ is defined by $>2sinalphaover2=d$. Replace this by
$$J(d)=2int_beta^pi/21oversinpsi>dpsi,qquad 2sinbeta=d .$$
This amounts to
$$J(d)=-2logleft(tanbetaover2right),qquad 2sinbeta=d .$$
Letting $tanbetaover2=:tau$ we have $$d=2sinbeta=4tauover 1+tau^2 ,$$
and therefore
$$tau=dover2+sqrt4-d^2 .$$
It follows that
$$J(d)=log1overtau^2=log8-d^2+4sqrt4-d^2over d^2leq3over dqquad(0<dleq2) .$$
The last estimate comes from inspecting a plot. Note that for any $alpha>0$ one has $log tleq t^alpha$ for large $t>0$.
answered Aug 8 at 18:22
Christian Blatter
164k7108306
@ Christian Blatter, with this estimate $J(d)leq log(frac16d^2)$, is what has the following estimate $J(d)leq C d^-1$, where $C$ is a constant $C>0$?
– Z. Alfata
Aug 10 at 20:32
add a comment |Â
up vote
1
down vote
accepted
Draw a figure! Then you'll see that your integral is
$$J(d)=2int_alpha^pi 1over 2sinphiover2>dphi$$
whereby $alpha$ is defined by $>2sinalphaover2=d$. Replace this by
$$J(d)=2int_beta^pi/21oversinpsi>dpsi,qquad 2sinbeta=d .$$
This amounts to
$$J(d)=-2logleft(tanbetaover2right),qquad 2sinbeta=d .$$
Letting $tanbetaover2=:tau$ we have $$d=2sinbeta=4tauover 1+tau^2 ,$$
and therefore
$$tau=dover2+sqrt4-d^2 .$$
It follows that
$$J(d)=log1overtau^2=log8-d^2+4sqrt4-d^2over d^2leq3over dqquad(0<dleq2) .$$
The last estimate comes from inspecting a plot. Note that for any $alpha>0$ one has $log tleq t^alpha$ for large $t>0$.
answered Aug 8 at 18:22
Christian Blatter
164k7108306
@ Christian Blatter, with this estimate $J(d)leq log(frac16d^2)$, is what has the following estimate $J(d)leq C d^-1$, where $C$ is a constant $C>0$?
– Z. Alfata
Aug 10 at 20:32
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Draw a figure! Then you'll see that your integral is
$$J(d)=2int_alpha^pi 1over 2sinphiover2>dphi$$
whereby $alpha$ is defined by $>2sinalphaover2=d$. Replace this by
$$J(d)=2int_beta^pi/21oversinpsi>dpsi,qquad 2sinbeta=d .$$
This amounts to
$$J(d)=-2logleft(tanbetaover2right),qquad 2sinbeta=d .$$
Letting $tanbetaover2=:tau$ we have $$d=2sinbeta=4tauover 1+tau^2 ,$$
and therefore
$$tau=dover2+sqrt4-d^2 .$$
It follows that
$$J(d)=log1overtau^2=log8-d^2+4sqrt4-d^2over d^2leq3over dqquad(0<dleq2) .$$
The last estimate comes from inspecting a plot. Note that for any $alpha>0$ one has $log tleq t^alpha$ for large $t>0$.
answered Aug 8 at 18:22
Christian Blatter
164k7108306
Draw a figure! Then you'll see that your integral is
$$J(d)=2int_alpha^pi 1over 2sinphiover2>dphi$$
whereby $alpha$ is defined by $>2sinalphaover2=d$. Replace this by
$$J(d)=2int_beta^pi/21oversinpsi>dpsi,qquad 2sinbeta=d .$$
This amounts to
$$J(d)=-2logleft(tanbetaover2right),qquad 2sinbeta=d .$$
Letting $tanbetaover2=:tau$ we have $$d=2sinbeta=4tauover 1+tau^2 ,$$
and therefore
$$tau=dover2+sqrt4-d^2 .$$
It follows that
$$J(d)=log1overtau^2=log8-d^2+4sqrt4-d^2over d^2leq3over dqquad(0<dleq2) .$$
The last estimate comes from inspecting a plot. Note that for any $alpha>0$ one has $log tleq t^alpha$ for large $t>0$.
answered Aug 8 at 18:22
Christian Blatter
164k7108306
edited Aug 11 at 10:17
answered Aug 8 at 18:22
Christian Blatter
164k7108306
answered Aug 8 at 18:22
Christian Blatter
164k7108306
answered Aug 8 at 18:22
Christian Blatter
164k7108306
164k7108306
@ Christian Blatter, with this estimate $J(d)leq log(frac16d^2)$, is what has the following estimate $J(d)leq C d^-1$, where $C$ is a constant $C>0$?
– Z. Alfata
Aug 10 at 20:32
add a comment |Â
@ Christian Blatter, with this estimate $J(d)leq log(frac16d^2)$, is what has the following estimate $J(d)leq C d^-1$, where $C$ is a constant $C>0$?
– Z. Alfata
Aug 10 at 20:32
@ Christian Blatter, with this estimate $J(d)leq log(frac16d^2)$, is what has the following estimate $J(d)leq C d^-1$, where $C$ is a constant $C>0$?
– Z. Alfata
Aug 10 at 20:32
@ Christian Blatter, with this estimate $J(d)leq log(frac16d^2)$, is what has the following estimate $J(d)leq C d^-1$, where $C$ is a constant $C>0$?
– Z. Alfata
Aug 10 at 20:32
add a comment |Â
up vote
1
down vote
WLOG you can suppose $t = 1$ ($varphi = 0$). And your integral has elementary primitive:
$$
intfracdthetasqrt1-cos(theta) =
frac2sin(theta/2)logtan(theta/4)sqrt1 - costheta + C
$$
http://www.wolframalpha.com/input/?i=int(1%2Fsqrt(1+-+cos+t),t)
answered Aug 8 at 15:23
MartÃn-Blas Pérez Pinilla
33.4k42570
add a comment |Â
up vote
1
down vote
WLOG you can suppose $t = 1$ ($varphi = 0$). And your integral has elementary primitive:
$$
intfracdthetasqrt1-cos(theta) =
frac2sin(theta/2)logtan(theta/4)sqrt1 - costheta + C
$$
http://www.wolframalpha.com/input/?i=int(1%2Fsqrt(1+-+cos+t),t)
answered Aug 8 at 15:23
MartÃn-Blas Pérez Pinilla
33.4k42570
add a comment |Â
up vote
1
down vote
up vote
1
down vote
WLOG you can suppose $t = 1$ ($varphi = 0$). And your integral has elementary primitive:
$$
intfracdthetasqrt1-cos(theta) =
frac2sin(theta/2)logtan(theta/4)sqrt1 - costheta + C
$$
http://www.wolframalpha.com/input/?i=int(1%2Fsqrt(1+-+cos+t),t)
answered Aug 8 at 15:23
MartÃn-Blas Pérez Pinilla
33.4k42570
WLOG you can suppose $t = 1$ ($varphi = 0$). And your integral has elementary primitive:
$$
intfracdthetasqrt1-cos(theta) =
frac2sin(theta/2)logtan(theta/4)sqrt1 - costheta + C
$$
http://www.wolframalpha.com/input/?i=int(1%2Fsqrt(1+-+cos+t),t)
answered Aug 8 at 15:23
MartÃn-Blas Pérez Pinilla
33.4k42570
answered Aug 8 at 15:23
MartÃn-Blas Pérez Pinilla
33.4k42570
answered Aug 8 at 15:23
MartÃn-Blas Pérez Pinilla
33.4k42570
answered Aug 8 at 15:23
MartÃn-Blas Pérez Pinilla
33.4k42570
33.4k42570
add a comment |Â
add a comment |Â
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1
Isn't your integral always infinite? Also, your calculations become easier if you use $1-cos x = 2sin^2frac x2$.
– Kusma
Aug 6 at 12:45
If I'm interpreting it correctly (The integral of the reciprocal of distance to a chosen point on $S_1$ over a region of $S_1$ not too far away from that point), then it would indeed diverge.
– Kajelad
Aug 6 at 12:47
The domain of integration is $ sin mathbb S^1: , $, I'm sorry!!!
– Z. Alfata
Aug 8 at 9:47
How good must be the estimate? An utterly trivial bound is $2pi/d$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:25
And you can suppose wlog that $t = 1$.
– MartÃn-Blas Pérez Pinilla
Aug 8 at 10:27