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How to estimate the following integral?
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I would like to estimate the integral $I_s,t$ by a constant $C$ independent of $s, t in mathbb S^1$:
$$I_s,t=int_r-s fract-s^2 , dr,$$
where $r, s , mboxand, t$ are in the unit circle $mathbb S^1=zetain mathbb C: , $. More precisely, I want to prove that, there exist a constant $C$ independent of $s, t in mathbb S^1$ such that
$$I_s,t=int_r-s fract-s^2 , dr< C.$$
Thank you in advance
calculus real-analysis integration complex-analysis inequality
asked Jul 20 at 19:06
Z. Alfata
873413
 |Â
show 1 more comment
up vote
3
down vote
favorite
I would like to estimate the integral $I_s,t$ by a constant $C$ independent of $s, t in mathbb S^1$:
$$I_s,t=int_r-s fract-s^2 , dr,$$
where $r, s , mboxand, t$ are in the unit circle $mathbb S^1=zetain mathbb C: , $. More precisely, I want to prove that, there exist a constant $C$ independent of $s, t in mathbb S^1$ such that
$$I_s,t=int_r-s fract-s^2 , dr< C.$$
Thank you in advance
calculus real-analysis integration complex-analysis inequality
asked Jul 20 at 19:06
Z. Alfata
873413
Did you mean "I would like to estimate the integral..."?
– Adrian Keister
Jul 20 at 19:31
Also, when you say $r,s,t$ are in the unit circle, do you mean $|s|<1, ; |t|<1,$ and $|r|<1?$ Or do you mean $|s|=1, ; |t|=1, ; |r|=1?$
– Adrian Keister
Jul 20 at 19:32
@ Adrian Keister: yes that what I would like to say and $zetain mathbb S^1$ i.e., $|zeta|=1$. Thanks
– Z. Alfata
Jul 20 at 19:41
Great! This is an interesting problem, +1. I'm going to think about it some.
– Adrian Keister
Jul 20 at 19:42
Do you know if $t$ can equal $s?$
– Adrian Keister
Jul 20 at 19:43
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I would like to estimate the integral $I_s,t$ by a constant $C$ independent of $s, t in mathbb S^1$:
$$I_s,t=int_r-s fract-s^2 , dr,$$
where $r, s , mboxand, t$ are in the unit circle $mathbb S^1=zetain mathbb C: , $. More precisely, I want to prove that, there exist a constant $C$ independent of $s, t in mathbb S^1$ such that
$$I_s,t=int_r-s fract-s^2 , dr< C.$$
Thank you in advance
calculus real-analysis integration complex-analysis inequality
asked Jul 20 at 19:06
Z. Alfata
873413
I would like to estimate the integral $I_s,t$ by a constant $C$ independent of $s, t in mathbb S^1$:
$$I_s,t=int_r-s fract-s^2 , dr,$$
where $r, s , mboxand, t$ are in the unit circle $mathbb S^1=zetain mathbb C: , $. More precisely, I want to prove that, there exist a constant $C$ independent of $s, t in mathbb S^1$ such that
$$I_s,t=int_r-s fract-s^2 , dr< C.$$
Thank you in advance
calculus real-analysis integration complex-analysis inequality
asked Jul 20 at 19:06
Z. Alfata
873413
edited Jul 20 at 19:34
asked Jul 20 at 19:06
Z. Alfata
873413
asked Jul 20 at 19:06
Z. Alfata
873413
asked Jul 20 at 19:06
Z. Alfata
873413
873413
Did you mean "I would like to estimate the integral..."?
– Adrian Keister
Jul 20 at 19:31
Also, when you say $r,s,t$ are in the unit circle, do you mean $|s|<1, ; |t|<1,$ and $|r|<1?$ Or do you mean $|s|=1, ; |t|=1, ; |r|=1?$
– Adrian Keister
Jul 20 at 19:32
@ Adrian Keister: yes that what I would like to say and $zetain mathbb S^1$ i.e., $|zeta|=1$. Thanks
– Z. Alfata
Jul 20 at 19:41
Great! This is an interesting problem, +1. I'm going to think about it some.
– Adrian Keister
Jul 20 at 19:42
Do you know if $t$ can equal $s?$
– Adrian Keister
Jul 20 at 19:43
 |Â
show 1 more comment
Did you mean "I would like to estimate the integral..."?
– Adrian Keister
Jul 20 at 19:31
Also, when you say $r,s,t$ are in the unit circle, do you mean $|s|<1, ; |t|<1,$ and $|r|<1?$ Or do you mean $|s|=1, ; |t|=1, ; |r|=1?$
– Adrian Keister
Jul 20 at 19:32
@ Adrian Keister: yes that what I would like to say and $zetain mathbb S^1$ i.e., $|zeta|=1$. Thanks
– Z. Alfata
Jul 20 at 19:41
Great! This is an interesting problem, +1. I'm going to think about it some.
– Adrian Keister
Jul 20 at 19:42
Do you know if $t$ can equal $s?$
– Adrian Keister
Jul 20 at 19:43
Did you mean "I would like to estimate the integral..."?
– Adrian Keister
Jul 20 at 19:31
Did you mean "I would like to estimate the integral..."?
– Adrian Keister
Jul 20 at 19:31
Also, when you say $r,s,t$ are in the unit circle, do you mean $|s|<1, ; |t|<1,$ and $|r|<1?$ Or do you mean $|s|=1, ; |t|=1, ; |r|=1?$
– Adrian Keister
Jul 20 at 19:32
Also, when you say $r,s,t$ are in the unit circle, do you mean $|s|<1, ; |t|<1,$ and $|r|<1?$ Or do you mean $|s|=1, ; |t|=1, ; |r|=1?$
– Adrian Keister
Jul 20 at 19:32
@ Adrian Keister: yes that what I would like to say and $zetain mathbb S^1$ i.e., $|zeta|=1$. Thanks
– Z. Alfata
Jul 20 at 19:41
@ Adrian Keister: yes that what I would like to say and $zetain mathbb S^1$ i.e., $|zeta|=1$. Thanks
– Z. Alfata
Jul 20 at 19:41
Great! This is an interesting problem, +1. I'm going to think about it some.
– Adrian Keister
Jul 20 at 19:42
Great! This is an interesting problem, +1. I'm going to think about it some.
– Adrian Keister
Jul 20 at 19:42
Do you know if $t$ can equal $s?$
– Adrian Keister
Jul 20 at 19:43
Do you know if $t$ can equal $s?$
– Adrian Keister
Jul 20 at 19:43
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
3
down vote
I don't think this is possible. The problem is, if $tnot=s,$ then the integrand is bounded in size, but it's bounded by something that is dependent on $t$ and $s$, which is itself not bounded. Let $D=t-s.$ From the domain $D$ over which we are integrating, we have that
beginalign*
|r-s|&ge 2|t-s| \
frac1&lefrac1t-s \
frac1^2&lefrac1t-s \
fract-s^2&lefrac1.
endalign*
The $ML$ estimate would then say that
beginalign*
left|int_Dfract-s^2,drright|&le left|int_Dfrac1,drright| \
&=frac1left|int_D,drright| \
&le frac2pi \
&=fracpit-s.
endalign*
This bound can get quite large if $t$ gets close to $s.$
I'm curious to try a few numerical calculations. Since $r,s,tinmathbbS^1,$ we can write
beginalign*
r&=e^itheta \
s&=e^ivarphi \
t&=e^ixi.
endalign*
Then $dr=i,e^itheta,dtheta.$ Note a curious fact: this problem is symmetric under any rotation. Moreover, we have that $|r-s|ge 2|t-s|$ if and only if $|theta-varphi|ge 2|xi-varphi|,$ modulo $2pi$. This is at least true when $tapprox s,$ which is what we're concerned about. So, let us set $varphi-xi=0,$ (from rotational symmetry) so that the limits on the $theta$ integral will be $xi$ to $2pi.$ The integral becomes
$$int_Dfract-s^2,dr=iint_xi^2pifrace^itheta-e^ivarphi,e^itheta,dtheta. $$
This we can set up in Mathematica, and the results show, indeed, that the magnitude of this integral can be quite large, depending on how close $varphi$ is to $xi$.
answered Jul 20 at 20:36
Adrian Keister
3,61721533
@ Adrian Keister, If $s=1$, can not hope to estimate this integral $iint_xi^2pifrac^2,e^itheta,dtheta$ independent of $xi$ ?
– Z. Alfata
Jul 25 at 10:53
add a comment |Â
up vote
1
down vote
Hint:
If I understand correctly, we can write
$$r=e^iw,s=e^iu,t=e^iv$$
and the integral is
$$int_t-sfract-s(cos w-cos u)^2+(sin w-sin u)^2e^iwdw
\=|t-s|int_t-sfrace^iwdw2-2cos wcos u-2sin wsin u
\=|t-s|frace^iu2int_t-sfrace^i(w-u)dw1-cos(w-u).$$
With $z:=w-u$, the indefinite integral has an analytical form
$$intfraccos z+isin z1-cos zdz=fracsin z+2sin^2dfrac z2left(z-2ilogsindfrac z2right)cos z-1.$$
Finally, the integration domain is the portion of the unit circle which is outside a circle centered at $s$, with radius $2|t-s|$. Hence, $w$ varies in a single interval delimited by the intersections of the two circles.
This allows to write the integral in a (complicated) closed-form.
answered Jul 20 at 20:22
Yves Daoust
111k665204
@ Yves Daoust, we can take for example $s=1$ i.e, $I_1,t=int_1-r frac^2 , dr$, my question is: can not hope to estimate this integral independent of of $t$?
– Z. Alfata
Jul 25 at 10:57
@Z.Alfata: actually, there is no dependency on $t$ but just on the distance $|t-s|$, which determines the extent of the arc. There should be a way to approxmate the integrand along the arc.
– Yves Daoust
Jul 25 at 12:28
$t=e^itheta$ and $r=e^ia$ (with the condition $2a<pi$) and we define the measure $dt$ ($=|dt|$) as the Lebesgue on $mathbb R$, then, $$int_1-r frac^2 , dr= 2int_2a^pi fracatheta^2 , dtheta = 2a(frac12a-frac1pi) leq 1$$ it is working now?
– Z. Alfata
Jul 25 at 13:04
@Z.Alfata: I doubt it, this looks too simple.
– Yves Daoust
Jul 25 at 13:15
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I don't think this is possible. The problem is, if $tnot=s,$ then the integrand is bounded in size, but it's bounded by something that is dependent on $t$ and $s$, which is itself not bounded. Let $D=t-s.$ From the domain $D$ over which we are integrating, we have that
beginalign*
|r-s|&ge 2|t-s| \
frac1&lefrac1t-s \
frac1^2&lefrac1t-s \
fract-s^2&lefrac1.
endalign*
The $ML$ estimate would then say that
beginalign*
left|int_Dfract-s^2,drright|&le left|int_Dfrac1,drright| \
&=frac1left|int_D,drright| \
&le frac2pi \
&=fracpit-s.
endalign*
This bound can get quite large if $t$ gets close to $s.$
I'm curious to try a few numerical calculations. Since $r,s,tinmathbbS^1,$ we can write
beginalign*
r&=e^itheta \
s&=e^ivarphi \
t&=e^ixi.
endalign*
Then $dr=i,e^itheta,dtheta.$ Note a curious fact: this problem is symmetric under any rotation. Moreover, we have that $|r-s|ge 2|t-s|$ if and only if $|theta-varphi|ge 2|xi-varphi|,$ modulo $2pi$. This is at least true when $tapprox s,$ which is what we're concerned about. So, let us set $varphi-xi=0,$ (from rotational symmetry) so that the limits on the $theta$ integral will be $xi$ to $2pi.$ The integral becomes
$$int_Dfract-s^2,dr=iint_xi^2pifrace^itheta-e^ivarphi,e^itheta,dtheta. $$
This we can set up in Mathematica, and the results show, indeed, that the magnitude of this integral can be quite large, depending on how close $varphi$ is to $xi$.
answered Jul 20 at 20:36
Adrian Keister
3,61721533
@ Adrian Keister, If $s=1$, can not hope to estimate this integral $iint_xi^2pifrac^2,e^itheta,dtheta$ independent of $xi$ ?
– Z. Alfata
Jul 25 at 10:53
add a comment |Â
up vote
3
down vote
I don't think this is possible. The problem is, if $tnot=s,$ then the integrand is bounded in size, but it's bounded by something that is dependent on $t$ and $s$, which is itself not bounded. Let $D=t-s.$ From the domain $D$ over which we are integrating, we have that
beginalign*
|r-s|&ge 2|t-s| \
frac1&lefrac1t-s \
frac1^2&lefrac1t-s \
fract-s^2&lefrac1.
endalign*
The $ML$ estimate would then say that
beginalign*
left|int_Dfract-s^2,drright|&le left|int_Dfrac1,drright| \
&=frac1left|int_D,drright| \
&le frac2pi \
&=fracpit-s.
endalign*
This bound can get quite large if $t$ gets close to $s.$
I'm curious to try a few numerical calculations. Since $r,s,tinmathbbS^1,$ we can write
beginalign*
r&=e^itheta \
s&=e^ivarphi \
t&=e^ixi.
endalign*
Then $dr=i,e^itheta,dtheta.$ Note a curious fact: this problem is symmetric under any rotation. Moreover, we have that $|r-s|ge 2|t-s|$ if and only if $|theta-varphi|ge 2|xi-varphi|,$ modulo $2pi$. This is at least true when $tapprox s,$ which is what we're concerned about. So, let us set $varphi-xi=0,$ (from rotational symmetry) so that the limits on the $theta$ integral will be $xi$ to $2pi.$ The integral becomes
$$int_Dfract-s^2,dr=iint_xi^2pifrace^itheta-e^ivarphi,e^itheta,dtheta. $$
This we can set up in Mathematica, and the results show, indeed, that the magnitude of this integral can be quite large, depending on how close $varphi$ is to $xi$.
answered Jul 20 at 20:36
Adrian Keister
3,61721533
@ Adrian Keister, If $s=1$, can not hope to estimate this integral $iint_xi^2pifrac^2,e^itheta,dtheta$ independent of $xi$ ?
– Z. Alfata
Jul 25 at 10:53
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I don't think this is possible. The problem is, if $tnot=s,$ then the integrand is bounded in size, but it's bounded by something that is dependent on $t$ and $s$, which is itself not bounded. Let $D=t-s.$ From the domain $D$ over which we are integrating, we have that
beginalign*
|r-s|&ge 2|t-s| \
frac1&lefrac1t-s \
frac1^2&lefrac1t-s \
fract-s^2&lefrac1.
endalign*
The $ML$ estimate would then say that
beginalign*
left|int_Dfract-s^2,drright|&le left|int_Dfrac1,drright| \
&=frac1left|int_D,drright| \
&le frac2pi \
&=fracpit-s.
endalign*
This bound can get quite large if $t$ gets close to $s.$
I'm curious to try a few numerical calculations. Since $r,s,tinmathbbS^1,$ we can write
beginalign*
r&=e^itheta \
s&=e^ivarphi \
t&=e^ixi.
endalign*
Then $dr=i,e^itheta,dtheta.$ Note a curious fact: this problem is symmetric under any rotation. Moreover, we have that $|r-s|ge 2|t-s|$ if and only if $|theta-varphi|ge 2|xi-varphi|,$ modulo $2pi$. This is at least true when $tapprox s,$ which is what we're concerned about. So, let us set $varphi-xi=0,$ (from rotational symmetry) so that the limits on the $theta$ integral will be $xi$ to $2pi.$ The integral becomes
$$int_Dfract-s^2,dr=iint_xi^2pifrace^itheta-e^ivarphi,e^itheta,dtheta. $$
This we can set up in Mathematica, and the results show, indeed, that the magnitude of this integral can be quite large, depending on how close $varphi$ is to $xi$.
answered Jul 20 at 20:36
Adrian Keister
3,61721533
I don't think this is possible. The problem is, if $tnot=s,$ then the integrand is bounded in size, but it's bounded by something that is dependent on $t$ and $s$, which is itself not bounded. Let $D=t-s.$ From the domain $D$ over which we are integrating, we have that
beginalign*
|r-s|&ge 2|t-s| \
frac1&lefrac1t-s \
frac1^2&lefrac1t-s \
fract-s^2&lefrac1.
endalign*
The $ML$ estimate would then say that
beginalign*
left|int_Dfract-s^2,drright|&le left|int_Dfrac1,drright| \
&=frac1left|int_D,drright| \
&le frac2pi \
&=fracpit-s.
endalign*
This bound can get quite large if $t$ gets close to $s.$
I'm curious to try a few numerical calculations. Since $r,s,tinmathbbS^1,$ we can write
beginalign*
r&=e^itheta \
s&=e^ivarphi \
t&=e^ixi.
endalign*
Then $dr=i,e^itheta,dtheta.$ Note a curious fact: this problem is symmetric under any rotation. Moreover, we have that $|r-s|ge 2|t-s|$ if and only if $|theta-varphi|ge 2|xi-varphi|,$ modulo $2pi$. This is at least true when $tapprox s,$ which is what we're concerned about. So, let us set $varphi-xi=0,$ (from rotational symmetry) so that the limits on the $theta$ integral will be $xi$ to $2pi.$ The integral becomes
$$int_Dfract-s^2,dr=iint_xi^2pifrace^itheta-e^ivarphi,e^itheta,dtheta. $$
This we can set up in Mathematica, and the results show, indeed, that the magnitude of this integral can be quite large, depending on how close $varphi$ is to $xi$.
answered Jul 20 at 20:36
Adrian Keister
3,61721533
answered Jul 20 at 20:36
Adrian Keister
3,61721533
answered Jul 20 at 20:36
Adrian Keister
3,61721533
answered Jul 20 at 20:36
Adrian Keister
3,61721533
3,61721533
@ Adrian Keister, If $s=1$, can not hope to estimate this integral $iint_xi^2pifrac^2,e^itheta,dtheta$ independent of $xi$ ?
– Z. Alfata
Jul 25 at 10:53
add a comment |Â
@ Adrian Keister, If $s=1$, can not hope to estimate this integral $iint_xi^2pifrac^2,e^itheta,dtheta$ independent of $xi$ ?
– Z. Alfata
Jul 25 at 10:53
@ Adrian Keister, If $s=1$, can not hope to estimate this integral $iint_xi^2pifrac^2,e^itheta,dtheta$ independent of $xi$ ?
– Z. Alfata
Jul 25 at 10:53
@ Adrian Keister, If $s=1$, can not hope to estimate this integral $iint_xi^2pifrac^2,e^itheta,dtheta$ independent of $xi$ ?
– Z. Alfata
Jul 25 at 10:53
add a comment |Â
up vote
1
down vote
Hint:
If I understand correctly, we can write
$$r=e^iw,s=e^iu,t=e^iv$$
and the integral is
$$int_t-sfract-s(cos w-cos u)^2+(sin w-sin u)^2e^iwdw
\=|t-s|int_t-sfrace^iwdw2-2cos wcos u-2sin wsin u
\=|t-s|frace^iu2int_t-sfrace^i(w-u)dw1-cos(w-u).$$
With $z:=w-u$, the indefinite integral has an analytical form
$$intfraccos z+isin z1-cos zdz=fracsin z+2sin^2dfrac z2left(z-2ilogsindfrac z2right)cos z-1.$$
Finally, the integration domain is the portion of the unit circle which is outside a circle centered at $s$, with radius $2|t-s|$. Hence, $w$ varies in a single interval delimited by the intersections of the two circles.
This allows to write the integral in a (complicated) closed-form.
answered Jul 20 at 20:22
Yves Daoust
111k665204
@ Yves Daoust, we can take for example $s=1$ i.e, $I_1,t=int_1-r frac^2 , dr$, my question is: can not hope to estimate this integral independent of of $t$?
– Z. Alfata
Jul 25 at 10:57
@Z.Alfata: actually, there is no dependency on $t$ but just on the distance $|t-s|$, which determines the extent of the arc. There should be a way to approxmate the integrand along the arc.
– Yves Daoust
Jul 25 at 12:28
$t=e^itheta$ and $r=e^ia$ (with the condition $2a<pi$) and we define the measure $dt$ ($=|dt|$) as the Lebesgue on $mathbb R$, then, $$int_1-r frac^2 , dr= 2int_2a^pi fracatheta^2 , dtheta = 2a(frac12a-frac1pi) leq 1$$ it is working now?
– Z. Alfata
Jul 25 at 13:04
@Z.Alfata: I doubt it, this looks too simple.
– Yves Daoust
Jul 25 at 13:15
add a comment |Â
up vote
1
down vote
Hint:
If I understand correctly, we can write
$$r=e^iw,s=e^iu,t=e^iv$$
and the integral is
$$int_t-sfract-s(cos w-cos u)^2+(sin w-sin u)^2e^iwdw
\=|t-s|int_t-sfrace^iwdw2-2cos wcos u-2sin wsin u
\=|t-s|frace^iu2int_t-sfrace^i(w-u)dw1-cos(w-u).$$
With $z:=w-u$, the indefinite integral has an analytical form
$$intfraccos z+isin z1-cos zdz=fracsin z+2sin^2dfrac z2left(z-2ilogsindfrac z2right)cos z-1.$$
Finally, the integration domain is the portion of the unit circle which is outside a circle centered at $s$, with radius $2|t-s|$. Hence, $w$ varies in a single interval delimited by the intersections of the two circles.
This allows to write the integral in a (complicated) closed-form.
answered Jul 20 at 20:22
Yves Daoust
111k665204
@ Yves Daoust, we can take for example $s=1$ i.e, $I_1,t=int_1-r frac^2 , dr$, my question is: can not hope to estimate this integral independent of of $t$?
– Z. Alfata
Jul 25 at 10:57
@Z.Alfata: actually, there is no dependency on $t$ but just on the distance $|t-s|$, which determines the extent of the arc. There should be a way to approxmate the integrand along the arc.
– Yves Daoust
Jul 25 at 12:28
$t=e^itheta$ and $r=e^ia$ (with the condition $2a<pi$) and we define the measure $dt$ ($=|dt|$) as the Lebesgue on $mathbb R$, then, $$int_1-r frac^2 , dr= 2int_2a^pi fracatheta^2 , dtheta = 2a(frac12a-frac1pi) leq 1$$ it is working now?
– Z. Alfata
Jul 25 at 13:04
@Z.Alfata: I doubt it, this looks too simple.
– Yves Daoust
Jul 25 at 13:15
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint:
If I understand correctly, we can write
$$r=e^iw,s=e^iu,t=e^iv$$
and the integral is
$$int_t-sfract-s(cos w-cos u)^2+(sin w-sin u)^2e^iwdw
\=|t-s|int_t-sfrace^iwdw2-2cos wcos u-2sin wsin u
\=|t-s|frace^iu2int_t-sfrace^i(w-u)dw1-cos(w-u).$$
With $z:=w-u$, the indefinite integral has an analytical form
$$intfraccos z+isin z1-cos zdz=fracsin z+2sin^2dfrac z2left(z-2ilogsindfrac z2right)cos z-1.$$
Finally, the integration domain is the portion of the unit circle which is outside a circle centered at $s$, with radius $2|t-s|$. Hence, $w$ varies in a single interval delimited by the intersections of the two circles.
This allows to write the integral in a (complicated) closed-form.
answered Jul 20 at 20:22
Yves Daoust
111k665204
Hint:
If I understand correctly, we can write
$$r=e^iw,s=e^iu,t=e^iv$$
and the integral is
$$int_t-sfract-s(cos w-cos u)^2+(sin w-sin u)^2e^iwdw
\=|t-s|int_t-sfrace^iwdw2-2cos wcos u-2sin wsin u
\=|t-s|frace^iu2int_t-sfrace^i(w-u)dw1-cos(w-u).$$
With $z:=w-u$, the indefinite integral has an analytical form
$$intfraccos z+isin z1-cos zdz=fracsin z+2sin^2dfrac z2left(z-2ilogsindfrac z2right)cos z-1.$$
Finally, the integration domain is the portion of the unit circle which is outside a circle centered at $s$, with radius $2|t-s|$. Hence, $w$ varies in a single interval delimited by the intersections of the two circles.
This allows to write the integral in a (complicated) closed-form.
answered Jul 20 at 20:22
Yves Daoust
111k665204
answered Jul 20 at 20:22
Yves Daoust
111k665204
answered Jul 20 at 20:22
Yves Daoust
111k665204
answered Jul 20 at 20:22
Yves Daoust
111k665204
111k665204
@ Yves Daoust, we can take for example $s=1$ i.e, $I_1,t=int_1-r frac^2 , dr$, my question is: can not hope to estimate this integral independent of of $t$?
– Z. Alfata
Jul 25 at 10:57
@Z.Alfata: actually, there is no dependency on $t$ but just on the distance $|t-s|$, which determines the extent of the arc. There should be a way to approxmate the integrand along the arc.
– Yves Daoust
Jul 25 at 12:28
$t=e^itheta$ and $r=e^ia$ (with the condition $2a<pi$) and we define the measure $dt$ ($=|dt|$) as the Lebesgue on $mathbb R$, then, $$int_1-r frac^2 , dr= 2int_2a^pi fracatheta^2 , dtheta = 2a(frac12a-frac1pi) leq 1$$ it is working now?
– Z. Alfata
Jul 25 at 13:04
@Z.Alfata: I doubt it, this looks too simple.
– Yves Daoust
Jul 25 at 13:15
add a comment |Â
@ Yves Daoust, we can take for example $s=1$ i.e, $I_1,t=int_1-r frac^2 , dr$, my question is: can not hope to estimate this integral independent of of $t$?
– Z. Alfata
Jul 25 at 10:57
@Z.Alfata: actually, there is no dependency on $t$ but just on the distance $|t-s|$, which determines the extent of the arc. There should be a way to approxmate the integrand along the arc.
– Yves Daoust
Jul 25 at 12:28
$t=e^itheta$ and $r=e^ia$ (with the condition $2a<pi$) and we define the measure $dt$ ($=|dt|$) as the Lebesgue on $mathbb R$, then, $$int_1-r frac^2 , dr= 2int_2a^pi fracatheta^2 , dtheta = 2a(frac12a-frac1pi) leq 1$$ it is working now?
– Z. Alfata
Jul 25 at 13:04
@Z.Alfata: I doubt it, this looks too simple.
– Yves Daoust
Jul 25 at 13:15
@ Yves Daoust, we can take for example $s=1$ i.e, $I_1,t=int_1-r frac^2 , dr$, my question is: can not hope to estimate this integral independent of of $t$?
– Z. Alfata
Jul 25 at 10:57
@ Yves Daoust, we can take for example $s=1$ i.e, $I_1,t=int_1-r frac^2 , dr$, my question is: can not hope to estimate this integral independent of of $t$?
– Z. Alfata
Jul 25 at 10:57
@Z.Alfata: actually, there is no dependency on $t$ but just on the distance $|t-s|$, which determines the extent of the arc. There should be a way to approxmate the integrand along the arc.
– Yves Daoust
Jul 25 at 12:28
@Z.Alfata: actually, there is no dependency on $t$ but just on the distance $|t-s|$, which determines the extent of the arc. There should be a way to approxmate the integrand along the arc.
– Yves Daoust
Jul 25 at 12:28
$t=e^itheta$ and $r=e^ia$ (with the condition $2a<pi$) and we define the measure $dt$ ($=|dt|$) as the Lebesgue on $mathbb R$, then, $$int_1-r frac^2 , dr= 2int_2a^pi fracatheta^2 , dtheta = 2a(frac12a-frac1pi) leq 1$$ it is working now?
– Z. Alfata
Jul 25 at 13:04
$t=e^itheta$ and $r=e^ia$ (with the condition $2a<pi$) and we define the measure $dt$ ($=|dt|$) as the Lebesgue on $mathbb R$, then, $$int_1-r frac^2 , dr= 2int_2a^pi fracatheta^2 , dtheta = 2a(frac12a-frac1pi) leq 1$$ it is working now?
– Z. Alfata
Jul 25 at 13:04
@Z.Alfata: I doubt it, this looks too simple.
– Yves Daoust
Jul 25 at 13:15
@Z.Alfata: I doubt it, this looks too simple.
– Yves Daoust
Jul 25 at 13:15
add a comment |Â
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Did you mean "I would like to estimate the integral..."?
– Adrian Keister
Jul 20 at 19:31
Also, when you say $r,s,t$ are in the unit circle, do you mean $|s|<1, ; |t|<1,$ and $|r|<1?$ Or do you mean $|s|=1, ; |t|=1, ; |r|=1?$
– Adrian Keister
Jul 20 at 19:32
@ Adrian Keister: yes that what I would like to say and $zetain mathbb S^1$ i.e., $|zeta|=1$. Thanks
– Z. Alfata
Jul 20 at 19:41
Great! This is an interesting problem, +1. I'm going to think about it some.
– Adrian Keister
Jul 20 at 19:42
Do you know if $t$ can equal $s?$
– Adrian Keister
Jul 20 at 19:43