Can this expression be simplified? (double integral over a sphere in $mathbbR^3$.A way to split this integral/normSurface integral on sphereReference for differentiation of an integral over variable ballCan this probability expression be simplified further: $fracs_ip_i(t)sum s_jp_j(t)$?Maximize the integral over sets of a fixed volumeRelations between Fractional Sobolev spaces $H^s$ and $H^1$Does this double integral converge?Minkowski Inequality for Riemann-Stieltjes IntegralsInvestigate maxima of Gaussian integral over sphere.Show $int_partial mathbbD loglvert x-zrvert dz to int_partial mathbbD loglvert y-zrvert dz$
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Can this expression be simplified? (double integral over a sphere in $mathbbR^3$.
A way to split this integral/normSurface integral on sphereReference for differentiation of an integral over variable ballCan this probability expression be simplified further: $fracs_ip_i(t)sum s_jp_j(t)$?Maximize the integral over sets of a fixed volumeRelations between Fractional Sobolev spaces $H^s$ and $H^1$Does this double integral converge?Minkowski Inequality for Riemann-Stieltjes IntegralsInvestigate maxima of Gaussian integral over sphere.Show $int_partial mathbbD loglvert x-zrvert dz to int_partial mathbbD loglvert y-zrvert dz$
$begingroup$
Suppose $f:mathbbR^3tomathbbR$ is an integrable function (I'm open to further restricting $f$, e.g. $f$ can be assumed to be smooth).
For some constants $0<rleq R$, is there a way to simplify the expression
$$
int_lvert y rvert = Rint_lvert xrvert =r f(x+y) ,mathrmdS(x)mathrmdS(y)
$$
to an expression not involving surface integrals?
real-analysis calculus integration
asked Mar 21 at 0:08
QuokaQuoka
1,538316
$endgroup$
This question has an open bounty worth +50
reputation from rolandcyp ending ending at 2019-04-01 18:31:30Z">in 5 days.
This question has not received enough attention.
add a comment |
$begingroup$
Suppose $f:mathbbR^3tomathbbR$ is an integrable function (I'm open to further restricting $f$, e.g. $f$ can be assumed to be smooth).
For some constants $0<rleq R$, is there a way to simplify the expression
$$
int_lvert y rvert = Rint_lvert xrvert =r f(x+y) ,mathrmdS(x)mathrmdS(y)
$$
to an expression not involving surface integrals?
real-analysis calculus integration
asked Mar 21 at 0:08
QuokaQuoka
1,538316
$endgroup$
This question has an open bounty worth +50
reputation from rolandcyp ending ending at 2019-04-01 18:31:30Z">in 5 days.
This question has not received enough attention.
$begingroup$
This is a little too general for me... Maybe you could give some examples?
$endgroup$
– Yuriy S
Mar 21 at 0:54
1
$begingroup$
@YuriyS I was thinking the double integral might be equal to something of the form $$C int_S f(x) dx$$ where $Ssubseteq mathbbR^3$ and $C>0$ is a constant. I'm not entirely sure if this is possible, but my intuition tells me it might work
$endgroup$
– Quoka
Mar 21 at 1:19
$begingroup$
When $xin S_r$ and $yin S_R$ you have no control about the point $x+y$ and the value of $f$ there. I don't think there is a simple formula without further assumptions on $f$.
$endgroup$
– Christian Blatter
yesterday
add a comment |
$begingroup$
Suppose $f:mathbbR^3tomathbbR$ is an integrable function (I'm open to further restricting $f$, e.g. $f$ can be assumed to be smooth).
For some constants $0<rleq R$, is there a way to simplify the expression
$$
int_lvert y rvert = Rint_lvert xrvert =r f(x+y) ,mathrmdS(x)mathrmdS(y)
$$
to an expression not involving surface integrals?
real-analysis calculus integration
asked Mar 21 at 0:08
QuokaQuoka
1,538316
$endgroup$
Suppose $f:mathbbR^3tomathbbR$ is an integrable function (I'm open to further restricting $f$, e.g. $f$ can be assumed to be smooth).
For some constants $0<rleq R$, is there a way to simplify the expression
$$
int_lvert y rvert = Rint_lvert xrvert =r f(x+y) ,mathrmdS(x)mathrmdS(y)
$$
to an expression not involving surface integrals?
real-analysis calculus integration
real-analysis calculus integration
asked Mar 21 at 0:08
QuokaQuoka
1,538316
asked Mar 21 at 0:08
QuokaQuoka
1,538316
asked Mar 21 at 0:08
QuokaQuoka
1,538316
asked Mar 21 at 0:08
QuokaQuoka
1,538316
asked Mar 21 at 0:08
QuokaQuoka
1,538316
1,538316
This question has an open bounty worth +50
reputation from rolandcyp ending ending at 2019-04-01 18:31:30Z">in 5 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from rolandcyp ending ending at 2019-04-01 18:31:30Z">in 5 days.
This question has not received enough attention.
$begingroup$
This is a little too general for me... Maybe you could give some examples?
$endgroup$
– Yuriy S
Mar 21 at 0:54
1
$begingroup$
@YuriyS I was thinking the double integral might be equal to something of the form $$C int_S f(x) dx$$ where $Ssubseteq mathbbR^3$ and $C>0$ is a constant. I'm not entirely sure if this is possible, but my intuition tells me it might work
$endgroup$
– Quoka
Mar 21 at 1:19
$begingroup$
When $xin S_r$ and $yin S_R$ you have no control about the point $x+y$ and the value of $f$ there. I don't think there is a simple formula without further assumptions on $f$.
$endgroup$
– Christian Blatter
yesterday
add a comment |
$begingroup$
This is a little too general for me... Maybe you could give some examples?
$endgroup$
– Yuriy S
Mar 21 at 0:54
1
$begingroup$
@YuriyS I was thinking the double integral might be equal to something of the form $$C int_S f(x) dx$$ where $Ssubseteq mathbbR^3$ and $C>0$ is a constant. I'm not entirely sure if this is possible, but my intuition tells me it might work
$endgroup$
– Quoka
Mar 21 at 1:19
$begingroup$
When $xin S_r$ and $yin S_R$ you have no control about the point $x+y$ and the value of $f$ there. I don't think there is a simple formula without further assumptions on $f$.
$endgroup$
– Christian Blatter
yesterday
$begingroup$
This is a little too general for me... Maybe you could give some examples?
$endgroup$
– Yuriy S
Mar 21 at 0:54
$begingroup$
This is a little too general for me... Maybe you could give some examples?
$endgroup$
– Yuriy S
Mar 21 at 0:54
1
1
$begingroup$
@YuriyS I was thinking the double integral might be equal to something of the form $$C int_S f(x) dx$$ where $Ssubseteq mathbbR^3$ and $C>0$ is a constant. I'm not entirely sure if this is possible, but my intuition tells me it might work
$endgroup$
– Quoka
Mar 21 at 1:19
$begingroup$
@YuriyS I was thinking the double integral might be equal to something of the form $$C int_S f(x) dx$$ where $Ssubseteq mathbbR^3$ and $C>0$ is a constant. I'm not entirely sure if this is possible, but my intuition tells me it might work
$endgroup$
– Quoka
Mar 21 at 1:19
$begingroup$
When $xin S_r$ and $yin S_R$ you have no control about the point $x+y$ and the value of $f$ there. I don't think there is a simple formula without further assumptions on $f$.
$endgroup$
– Christian Blatter
yesterday
$begingroup$
When $xin S_r$ and $yin S_R$ you have no control about the point $x+y$ and the value of $f$ there. I don't think there is a simple formula without further assumptions on $f$.
$endgroup$
– Christian Blatter
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's start by writing the integral as
$$ I = int_mathbbR^3 int_mathbbR^3 f(x+y) delta(|x|-r) delta(|y|-R), rm d^3x rm d^3y$$
where $delta$ is Dirac's delta function. Changing the variables to $(z,y) = (x+y,y)$ we get
beginalign I &= int_mathbbR^3 int_mathbbR^3 f(z) delta(|z-y|-r) delta(|y|-R), rm d^3y rm d^3z \
&= int_mathbbR^3 f(z) Big(int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y Big) rm d^3zendalign
For $0le r le R$, it can be calculated that
To calculate $ int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y $ that let us use spherical coordinates for $y$ where angle $theta$ is measured from the direction of $z$, so that
$$|y| = rho $$
$$|z-y| = sqrt( = sqrtz$$
We have then
beginalign & int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = \
&quad = int_0^infty rm drhoint_0^pirm dthetaint_0^2pirm dphi,rho^2sintheta delta(sqrtz-r) delta(rho-R) = \
&quad = 2pi R^2 int_0^pirm dtheta sintheta ;delta(sqrtz-r)=^u:=sqrtz\
&quad = 2pi R^2 int_^z rm du fracu delta(u-r)= \
&quad = frac2pi R int_-infty^infty rm du ,u ,theta((R+|z|)-u)theta(u - big|R-|z|big|)delta(u-r)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - big|R-|z|big|)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - (R-|z|))theta(r - (|z|-R))=\
&quad = frac2pi R r ,theta(|z|+(R-r))theta(|z| - (R-r))theta(R+r - |z|) = ^textsince \
&quad = frac2pi R r
z thetabig(|z|-(R-r)big)thetabig(R+r-|z|big) endalign
where $theta$ is Heaviside's theta function.
In the end you get
$$ I = 2pi R r int_in[R-r, R+r] fracf(z) rm d^3z$$
answered yesterday
Adam LatosińskiAdam Latosiński
3385
$endgroup$
1
$begingroup$
I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
$endgroup$
– Quoka
yesterday
$begingroup$
It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
$endgroup$
– Adam Latosiński
yesterday
1
$begingroup$
I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
$endgroup$
– Quoka
yesterday
$begingroup$
Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
$endgroup$
– Adam Latosiński
20 hours ago
|
show 1 more comment
$begingroup$
If you work in Fourier space you can write $f(x+y)=int e^ik(x+y)hatf_k dk$ and you can separate the integral to $int dk hatf_k int e^ikx dS(x) int e^ikydS(y) $.
answered 35 mins ago
user617446user617446
4443
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's start by writing the integral as
$$ I = int_mathbbR^3 int_mathbbR^3 f(x+y) delta(|x|-r) delta(|y|-R), rm d^3x rm d^3y$$
where $delta$ is Dirac's delta function. Changing the variables to $(z,y) = (x+y,y)$ we get
beginalign I &= int_mathbbR^3 int_mathbbR^3 f(z) delta(|z-y|-r) delta(|y|-R), rm d^3y rm d^3z \
&= int_mathbbR^3 f(z) Big(int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y Big) rm d^3zendalign
For $0le r le R$, it can be calculated that
To calculate $ int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y $ that let us use spherical coordinates for $y$ where angle $theta$ is measured from the direction of $z$, so that
$$|y| = rho $$
$$|z-y| = sqrt( = sqrtz$$
We have then
beginalign & int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = \
&quad = int_0^infty rm drhoint_0^pirm dthetaint_0^2pirm dphi,rho^2sintheta delta(sqrtz-r) delta(rho-R) = \
&quad = 2pi R^2 int_0^pirm dtheta sintheta ;delta(sqrtz-r)=^u:=sqrtz\
&quad = 2pi R^2 int_^z rm du fracu delta(u-r)= \
&quad = frac2pi R int_-infty^infty rm du ,u ,theta((R+|z|)-u)theta(u - big|R-|z|big|)delta(u-r)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - big|R-|z|big|)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - (R-|z|))theta(r - (|z|-R))=\
&quad = frac2pi R r ,theta(|z|+(R-r))theta(|z| - (R-r))theta(R+r - |z|) = ^textsince \
&quad = frac2pi R r
z thetabig(|z|-(R-r)big)thetabig(R+r-|z|big) endalign
where $theta$ is Heaviside's theta function.
In the end you get
$$ I = 2pi R r int_in[R-r, R+r] fracf(z) rm d^3z$$
answered yesterday
Adam LatosińskiAdam Latosiński
3385
$endgroup$
1
$begingroup$
I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
$endgroup$
– Quoka
yesterday
$begingroup$
It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
$endgroup$
– Adam Latosiński
yesterday
1
$begingroup$
I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
$endgroup$
– Quoka
yesterday
$begingroup$
Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
$endgroup$
– Adam Latosiński
20 hours ago
|
show 1 more comment
$begingroup$
Let's start by writing the integral as
$$ I = int_mathbbR^3 int_mathbbR^3 f(x+y) delta(|x|-r) delta(|y|-R), rm d^3x rm d^3y$$
where $delta$ is Dirac's delta function. Changing the variables to $(z,y) = (x+y,y)$ we get
beginalign I &= int_mathbbR^3 int_mathbbR^3 f(z) delta(|z-y|-r) delta(|y|-R), rm d^3y rm d^3z \
&= int_mathbbR^3 f(z) Big(int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y Big) rm d^3zendalign
For $0le r le R$, it can be calculated that
To calculate $ int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y $ that let us use spherical coordinates for $y$ where angle $theta$ is measured from the direction of $z$, so that
$$|y| = rho $$
$$|z-y| = sqrt( = sqrtz$$
We have then
beginalign & int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = \
&quad = int_0^infty rm drhoint_0^pirm dthetaint_0^2pirm dphi,rho^2sintheta delta(sqrtz-r) delta(rho-R) = \
&quad = 2pi R^2 int_0^pirm dtheta sintheta ;delta(sqrtz-r)=^u:=sqrtz\
&quad = 2pi R^2 int_^z rm du fracu delta(u-r)= \
&quad = frac2pi R int_-infty^infty rm du ,u ,theta((R+|z|)-u)theta(u - big|R-|z|big|)delta(u-r)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - big|R-|z|big|)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - (R-|z|))theta(r - (|z|-R))=\
&quad = frac2pi R r ,theta(|z|+(R-r))theta(|z| - (R-r))theta(R+r - |z|) = ^textsince \
&quad = frac2pi R r
z thetabig(|z|-(R-r)big)thetabig(R+r-|z|big) endalign
where $theta$ is Heaviside's theta function.
In the end you get
$$ I = 2pi R r int_in[R-r, R+r] fracf(z) rm d^3z$$
answered yesterday
Adam LatosińskiAdam Latosiński
3385
$endgroup$
1
$begingroup$
I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
$endgroup$
– Quoka
yesterday
$begingroup$
It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
$endgroup$
– Adam Latosiński
yesterday
1
$begingroup$
I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
$endgroup$
– Quoka
yesterday
$begingroup$
Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
$endgroup$
– Adam Latosiński
20 hours ago
|
show 1 more comment
$begingroup$
Let's start by writing the integral as
$$ I = int_mathbbR^3 int_mathbbR^3 f(x+y) delta(|x|-r) delta(|y|-R), rm d^3x rm d^3y$$
where $delta$ is Dirac's delta function. Changing the variables to $(z,y) = (x+y,y)$ we get
beginalign I &= int_mathbbR^3 int_mathbbR^3 f(z) delta(|z-y|-r) delta(|y|-R), rm d^3y rm d^3z \
&= int_mathbbR^3 f(z) Big(int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y Big) rm d^3zendalign
For $0le r le R$, it can be calculated that
To calculate $ int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y $ that let us use spherical coordinates for $y$ where angle $theta$ is measured from the direction of $z$, so that
$$|y| = rho $$
$$|z-y| = sqrt( = sqrtz$$
We have then
beginalign & int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = \
&quad = int_0^infty rm drhoint_0^pirm dthetaint_0^2pirm dphi,rho^2sintheta delta(sqrtz-r) delta(rho-R) = \
&quad = 2pi R^2 int_0^pirm dtheta sintheta ;delta(sqrtz-r)=^u:=sqrtz\
&quad = 2pi R^2 int_^z rm du fracu delta(u-r)= \
&quad = frac2pi R int_-infty^infty rm du ,u ,theta((R+|z|)-u)theta(u - big|R-|z|big|)delta(u-r)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - big|R-|z|big|)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - (R-|z|))theta(r - (|z|-R))=\
&quad = frac2pi R r ,theta(|z|+(R-r))theta(|z| - (R-r))theta(R+r - |z|) = ^textsince \
&quad = frac2pi R r
z thetabig(|z|-(R-r)big)thetabig(R+r-|z|big) endalign
where $theta$ is Heaviside's theta function.
In the end you get
$$ I = 2pi R r int_in[R-r, R+r] fracf(z) rm d^3z$$
answered yesterday
Adam LatosińskiAdam Latosiński
3385
$endgroup$
Let's start by writing the integral as
$$ I = int_mathbbR^3 int_mathbbR^3 f(x+y) delta(|x|-r) delta(|y|-R), rm d^3x rm d^3y$$
where $delta$ is Dirac's delta function. Changing the variables to $(z,y) = (x+y,y)$ we get
beginalign I &= int_mathbbR^3 int_mathbbR^3 f(z) delta(|z-y|-r) delta(|y|-R), rm d^3y rm d^3z \
&= int_mathbbR^3 f(z) Big(int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y Big) rm d^3zendalign
For $0le r le R$, it can be calculated that
To calculate $ int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y $ that let us use spherical coordinates for $y$ where angle $theta$ is measured from the direction of $z$, so that
$$|y| = rho $$
$$|z-y| = sqrt( = sqrtz$$
We have then
beginalign & int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = \
&quad = int_0^infty rm drhoint_0^pirm dthetaint_0^2pirm dphi,rho^2sintheta delta(sqrtz-r) delta(rho-R) = \
&quad = 2pi R^2 int_0^pirm dtheta sintheta ;delta(sqrtz-r)=^u:=sqrtz\
&quad = 2pi R^2 int_^z rm du fracu delta(u-r)= \
&quad = frac2pi R int_-infty^infty rm du ,u ,theta((R+|z|)-u)theta(u - big|R-|z|big|)delta(u-r)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - big|R-|z|big|)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - (R-|z|))theta(r - (|z|-R))=\
&quad = frac2pi R r ,theta(|z|+(R-r))theta(|z| - (R-r))theta(R+r - |z|) = ^textsince \
&quad = frac2pi R r
z thetabig(|z|-(R-r)big)thetabig(R+r-|z|big) endalign
where $theta$ is Heaviside's theta function.
In the end you get
$$ I = 2pi R r int_in[R-r, R+r] fracf(z) rm d^3z$$
answered yesterday
Adam LatosińskiAdam Latosiński
3385
edited 20 hours ago
answered yesterday
Adam LatosińskiAdam Latosiński
3385
answered yesterday
Adam LatosińskiAdam Latosiński
3385
answered yesterday
Adam LatosińskiAdam Latosiński
3385
3385
1
$begingroup$
I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
$endgroup$
– Quoka
yesterday
$begingroup$
It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
$endgroup$
– Adam Latosiński
yesterday
1
$begingroup$
I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
$endgroup$
– Quoka
yesterday
$begingroup$
Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
$endgroup$
– Adam Latosiński
20 hours ago
|
show 1 more comment
1
$begingroup$
I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
$endgroup$
– Quoka
yesterday
$begingroup$
It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
$endgroup$
– Adam Latosiński
yesterday
1
$begingroup$
I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
$endgroup$
– Quoka
yesterday
$begingroup$
Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
$endgroup$
– Adam Latosiński
20 hours ago
1
1
$begingroup$
I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
$endgroup$
– Quoka
yesterday
$begingroup$
I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
$endgroup$
– Quoka
yesterday
$begingroup$
It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
$endgroup$
– Adam Latosiński
yesterday
$begingroup$
It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
$endgroup$
– Adam Latosiński
yesterday
1
1
$begingroup$
I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
$endgroup$
– Quoka
yesterday
$begingroup$
I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
$endgroup$
– Quoka
yesterday
$begingroup$
Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
$endgroup$
– Adam Latosiński
20 hours ago
$begingroup$
As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
$endgroup$
– Adam Latosiński
20 hours ago
|
show 1 more comment
$begingroup$
If you work in Fourier space you can write $f(x+y)=int e^ik(x+y)hatf_k dk$ and you can separate the integral to $int dk hatf_k int e^ikx dS(x) int e^ikydS(y) $.
answered 35 mins ago
user617446user617446
4443
$endgroup$
add a comment |
$begingroup$
If you work in Fourier space you can write $f(x+y)=int e^ik(x+y)hatf_k dk$ and you can separate the integral to $int dk hatf_k int e^ikx dS(x) int e^ikydS(y) $.
answered 35 mins ago
user617446user617446
4443
$endgroup$
add a comment |
$begingroup$
If you work in Fourier space you can write $f(x+y)=int e^ik(x+y)hatf_k dk$ and you can separate the integral to $int dk hatf_k int e^ikx dS(x) int e^ikydS(y) $.
answered 35 mins ago
user617446user617446
4443
$endgroup$
If you work in Fourier space you can write $f(x+y)=int e^ik(x+y)hatf_k dk$ and you can separate the integral to $int dk hatf_k int e^ikx dS(x) int e^ikydS(y) $.
answered 35 mins ago
user617446user617446
4443
answered 35 mins ago
user617446user617446
4443
answered 35 mins ago
user617446user617446
4443
answered 35 mins ago
user617446user617446
4443
4443
add a comment |
add a comment |
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$begingroup$
This is a little too general for me... Maybe you could give some examples?
$endgroup$
– Yuriy S
Mar 21 at 0:54
1
$begingroup$
@YuriyS I was thinking the double integral might be equal to something of the form $$C int_S f(x) dx$$ where $Ssubseteq mathbbR^3$ and $C>0$ is a constant. I'm not entirely sure if this is possible, but my intuition tells me it might work
$endgroup$
– Quoka
Mar 21 at 1:19
$begingroup$
When $xin S_r$ and $yin S_R$ you have no control about the point $x+y$ and the value of $f$ there. I don't think there is a simple formula without further assumptions on $f$.
$endgroup$
– Christian Blatter
yesterday