Help with Seemingly Hopeless Double Integral
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I hate to be that guy to just post an integration problem and ask how to solve it so I'll give a little relevant info
Okay, so I'm working on a physics project and my professor proposed that the following double integral could potentially solve a problem that I've used an alternative method to solve:
$$I=int_0^piint_0^rhofrac{t^2sinphileft(tcosphi-dright)}{left[t^2sin^2phi+left(tcosphi-dright)^2right]^{3/2}};dt dphi$$
$rho$ is an arbitrary, strictly positive real constant
$d$ is a real constant that satisfies $d>rho$
This integral's value could provide immense insight into fields of uniform, solid spherical objects, so it's actually pretty important for my work.
After some quick attempts to simplify, I decided to try some integral calculators with set values. Needless to say, the result after the first integral seemed so hopeless that I couldn't imagine simplifying and integrating again--not to mention then generalising constant inputs to their original variable form.
However, there is a strong likelihood that $I$ simplifies to one of the following two solutions:
$$text{1.This solution comes from inverse square laws}$$
$$I=frac{1}{d^2}$$
$$text{2. This solution comes from a separate computation that I did (integrals below)}$$
$$I=left(1-frac{rho^2}{5d^2}right)left[frac{3}{2rho^2}+frac{3(rho^2-d^2)}{4drho^3}lnleft(frac{d+rho}{d-rho}right)right]$$
Although it looks like these are vastly different answers, given $rho=1$ and $d=10$, you get the following outputs from $(1)$ and $(2)$:
$$1.; I=0.01$$
$$2.; Iapprox 0.01000046$$
Here's the ratio of solution (2) over (1) for $rhoin(0,1),;din(0,50)$
I tried to tackle this problem differently than my professor, and set up the following integrals to solve the problem that lead to solution $(2)$:
$$frac{9}{4rho^6}left[;intlimits_{d-rho}^{d+rho}xleft[x-frac{x^2+d^2-rho^2}{2d}right]left[frac{(x+d)^2-rho^2}{4dcdot x}right];dxright]cdotleft[;intlimits_{d-rho}^{d+rho}frac{rho^2-(x-d)^2}{2dcdot x};dxright]$$
Where you come in
If the double integral is correctly composed (which my professor felt confident with), I need someone skilled in integration to solve said double integral. I've given two possible solutions and it's probable that the answer will be one of those. If it's solution $(1)$, I know that mine will have an error and you will essentially have proved the inverse square law for gravitational and electric fields. If it's solution $(2)$, then this will be far more exciting to me but less likely. If it's neither, then there are several possible implications
BOUNTY
I'm willing to award the following bounties for solving the double integral at the beginning. Since certain solutions have stronger implications (as explained above), I'm rewarding the following bounties:
- +200 rep if you verify solution $(1)$
- +500 rep if you verify solution $(2)$
- +75 rep for any other solutions (note they'll have to be verified by a second user first)
QUESTIONS
If you have any additional questions feel free to ask, and thanks for reading all this!
calculus integration multivariable-calculus physics
$endgroup$
|
show 6 more comments
$begingroup$
I hate to be that guy to just post an integration problem and ask how to solve it so I'll give a little relevant info
Okay, so I'm working on a physics project and my professor proposed that the following double integral could potentially solve a problem that I've used an alternative method to solve:
$$I=int_0^piint_0^rhofrac{t^2sinphileft(tcosphi-dright)}{left[t^2sin^2phi+left(tcosphi-dright)^2right]^{3/2}};dt dphi$$
$rho$ is an arbitrary, strictly positive real constant
$d$ is a real constant that satisfies $d>rho$
This integral's value could provide immense insight into fields of uniform, solid spherical objects, so it's actually pretty important for my work.
After some quick attempts to simplify, I decided to try some integral calculators with set values. Needless to say, the result after the first integral seemed so hopeless that I couldn't imagine simplifying and integrating again--not to mention then generalising constant inputs to their original variable form.
However, there is a strong likelihood that $I$ simplifies to one of the following two solutions:
$$text{1.This solution comes from inverse square laws}$$
$$I=frac{1}{d^2}$$
$$text{2. This solution comes from a separate computation that I did (integrals below)}$$
$$I=left(1-frac{rho^2}{5d^2}right)left[frac{3}{2rho^2}+frac{3(rho^2-d^2)}{4drho^3}lnleft(frac{d+rho}{d-rho}right)right]$$
Although it looks like these are vastly different answers, given $rho=1$ and $d=10$, you get the following outputs from $(1)$ and $(2)$:
$$1.; I=0.01$$
$$2.; Iapprox 0.01000046$$
Here's the ratio of solution (2) over (1) for $rhoin(0,1),;din(0,50)$
I tried to tackle this problem differently than my professor, and set up the following integrals to solve the problem that lead to solution $(2)$:
$$frac{9}{4rho^6}left[;intlimits_{d-rho}^{d+rho}xleft[x-frac{x^2+d^2-rho^2}{2d}right]left[frac{(x+d)^2-rho^2}{4dcdot x}right];dxright]cdotleft[;intlimits_{d-rho}^{d+rho}frac{rho^2-(x-d)^2}{2dcdot x};dxright]$$
Where you come in
If the double integral is correctly composed (which my professor felt confident with), I need someone skilled in integration to solve said double integral. I've given two possible solutions and it's probable that the answer will be one of those. If it's solution $(1)$, I know that mine will have an error and you will essentially have proved the inverse square law for gravitational and electric fields. If it's solution $(2)$, then this will be far more exciting to me but less likely. If it's neither, then there are several possible implications
BOUNTY
I'm willing to award the following bounties for solving the double integral at the beginning. Since certain solutions have stronger implications (as explained above), I'm rewarding the following bounties:
- +200 rep if you verify solution $(1)$
- +500 rep if you verify solution $(2)$
- +75 rep for any other solutions (note they'll have to be verified by a second user first)
QUESTIONS
If you have any additional questions feel free to ask, and thanks for reading all this!
calculus integration multivariable-calculus physics
$endgroup$
$begingroup$
Edited: Typo in the original post
$endgroup$
– Lanier Freeman
yesterday
1
$begingroup$
So $phiin[0,pi]$, and $tin[0,rho]$ for some ?fixed? constant $rho$? (And the first solution seems to not depend on $rho$. Unexpected, since i can take $rho=0$.) Please fix some framework for all used constants. Things seem to be important, please just fix these details for the eye of a first reader... Help will come in some seconds... (At least numerically, this is the easiest (experimental) validation when explicit choices are given.)
$endgroup$
– dan_fulea
yesterday
1
$begingroup$
@LanierFreeman I'm not sure if this is too helpful, but I think the result must depend on rho. If you call the integral $I= I(rho,d)$, then differentiate with rho, I got $I'= -2(frac{rho}{d})^{2}$, again, differentiated in rho. But this means that the original integral can't depend only on $d$, right?
$endgroup$
– Ryan Goulden
yesterday
1
$begingroup$
Yes this is the type of integration found in some older textbooks on electromagnetism; the aim would be to eventually calculate the magnetic moment of a "classical" spinning electron, say if its charge is uniformly distributed over a spherical volume of radius $d$.
$endgroup$
– James Arathoon
yesterday
1
$begingroup$
If it still matters, I also found the answer to be $frac{-2rho^3}{3d^2}$ by converting to rectangular.
$endgroup$
– Tom Himler
yesterday
|
show 6 more comments
$begingroup$
I hate to be that guy to just post an integration problem and ask how to solve it so I'll give a little relevant info
Okay, so I'm working on a physics project and my professor proposed that the following double integral could potentially solve a problem that I've used an alternative method to solve:
$$I=int_0^piint_0^rhofrac{t^2sinphileft(tcosphi-dright)}{left[t^2sin^2phi+left(tcosphi-dright)^2right]^{3/2}};dt dphi$$
$rho$ is an arbitrary, strictly positive real constant
$d$ is a real constant that satisfies $d>rho$
This integral's value could provide immense insight into fields of uniform, solid spherical objects, so it's actually pretty important for my work.
After some quick attempts to simplify, I decided to try some integral calculators with set values. Needless to say, the result after the first integral seemed so hopeless that I couldn't imagine simplifying and integrating again--not to mention then generalising constant inputs to their original variable form.
However, there is a strong likelihood that $I$ simplifies to one of the following two solutions:
$$text{1.This solution comes from inverse square laws}$$
$$I=frac{1}{d^2}$$
$$text{2. This solution comes from a separate computation that I did (integrals below)}$$
$$I=left(1-frac{rho^2}{5d^2}right)left[frac{3}{2rho^2}+frac{3(rho^2-d^2)}{4drho^3}lnleft(frac{d+rho}{d-rho}right)right]$$
Although it looks like these are vastly different answers, given $rho=1$ and $d=10$, you get the following outputs from $(1)$ and $(2)$:
$$1.; I=0.01$$
$$2.; Iapprox 0.01000046$$
Here's the ratio of solution (2) over (1) for $rhoin(0,1),;din(0,50)$
I tried to tackle this problem differently than my professor, and set up the following integrals to solve the problem that lead to solution $(2)$:
$$frac{9}{4rho^6}left[;intlimits_{d-rho}^{d+rho}xleft[x-frac{x^2+d^2-rho^2}{2d}right]left[frac{(x+d)^2-rho^2}{4dcdot x}right];dxright]cdotleft[;intlimits_{d-rho}^{d+rho}frac{rho^2-(x-d)^2}{2dcdot x};dxright]$$
Where you come in
If the double integral is correctly composed (which my professor felt confident with), I need someone skilled in integration to solve said double integral. I've given two possible solutions and it's probable that the answer will be one of those. If it's solution $(1)$, I know that mine will have an error and you will essentially have proved the inverse square law for gravitational and electric fields. If it's solution $(2)$, then this will be far more exciting to me but less likely. If it's neither, then there are several possible implications
BOUNTY
I'm willing to award the following bounties for solving the double integral at the beginning. Since certain solutions have stronger implications (as explained above), I'm rewarding the following bounties:
- +200 rep if you verify solution $(1)$
- +500 rep if you verify solution $(2)$
- +75 rep for any other solutions (note they'll have to be verified by a second user first)
QUESTIONS
If you have any additional questions feel free to ask, and thanks for reading all this!
calculus integration multivariable-calculus physics
$endgroup$
I hate to be that guy to just post an integration problem and ask how to solve it so I'll give a little relevant info
Okay, so I'm working on a physics project and my professor proposed that the following double integral could potentially solve a problem that I've used an alternative method to solve:
$$I=int_0^piint_0^rhofrac{t^2sinphileft(tcosphi-dright)}{left[t^2sin^2phi+left(tcosphi-dright)^2right]^{3/2}};dt dphi$$
$rho$ is an arbitrary, strictly positive real constant
$d$ is a real constant that satisfies $d>rho$
This integral's value could provide immense insight into fields of uniform, solid spherical objects, so it's actually pretty important for my work.
After some quick attempts to simplify, I decided to try some integral calculators with set values. Needless to say, the result after the first integral seemed so hopeless that I couldn't imagine simplifying and integrating again--not to mention then generalising constant inputs to their original variable form.
However, there is a strong likelihood that $I$ simplifies to one of the following two solutions:
$$text{1.This solution comes from inverse square laws}$$
$$I=frac{1}{d^2}$$
$$text{2. This solution comes from a separate computation that I did (integrals below)}$$
$$I=left(1-frac{rho^2}{5d^2}right)left[frac{3}{2rho^2}+frac{3(rho^2-d^2)}{4drho^3}lnleft(frac{d+rho}{d-rho}right)right]$$
Although it looks like these are vastly different answers, given $rho=1$ and $d=10$, you get the following outputs from $(1)$ and $(2)$:
$$1.; I=0.01$$
$$2.; Iapprox 0.01000046$$
Here's the ratio of solution (2) over (1) for $rhoin(0,1),;din(0,50)$
I tried to tackle this problem differently than my professor, and set up the following integrals to solve the problem that lead to solution $(2)$:
$$frac{9}{4rho^6}left[;intlimits_{d-rho}^{d+rho}xleft[x-frac{x^2+d^2-rho^2}{2d}right]left[frac{(x+d)^2-rho^2}{4dcdot x}right];dxright]cdotleft[;intlimits_{d-rho}^{d+rho}frac{rho^2-(x-d)^2}{2dcdot x};dxright]$$
Where you come in
If the double integral is correctly composed (which my professor felt confident with), I need someone skilled in integration to solve said double integral. I've given two possible solutions and it's probable that the answer will be one of those. If it's solution $(1)$, I know that mine will have an error and you will essentially have proved the inverse square law for gravitational and electric fields. If it's solution $(2)$, then this will be far more exciting to me but less likely. If it's neither, then there are several possible implications
BOUNTY
I'm willing to award the following bounties for solving the double integral at the beginning. Since certain solutions have stronger implications (as explained above), I'm rewarding the following bounties:
- +200 rep if you verify solution $(1)$
- +500 rep if you verify solution $(2)$
- +75 rep for any other solutions (note they'll have to be verified by a second user first)
QUESTIONS
If you have any additional questions feel free to ask, and thanks for reading all this!
calculus integration multivariable-calculus physics
calculus integration multivariable-calculus physics
edited 23 hours ago
Lanier Freeman
asked yesterday
Lanier FreemanLanier Freeman
2,907929
2,907929
$begingroup$
Edited: Typo in the original post
$endgroup$
– Lanier Freeman
yesterday
1
$begingroup$
So $phiin[0,pi]$, and $tin[0,rho]$ for some ?fixed? constant $rho$? (And the first solution seems to not depend on $rho$. Unexpected, since i can take $rho=0$.) Please fix some framework for all used constants. Things seem to be important, please just fix these details for the eye of a first reader... Help will come in some seconds... (At least numerically, this is the easiest (experimental) validation when explicit choices are given.)
$endgroup$
– dan_fulea
yesterday
1
$begingroup$
@LanierFreeman I'm not sure if this is too helpful, but I think the result must depend on rho. If you call the integral $I= I(rho,d)$, then differentiate with rho, I got $I'= -2(frac{rho}{d})^{2}$, again, differentiated in rho. But this means that the original integral can't depend only on $d$, right?
$endgroup$
– Ryan Goulden
yesterday
1
$begingroup$
Yes this is the type of integration found in some older textbooks on electromagnetism; the aim would be to eventually calculate the magnetic moment of a "classical" spinning electron, say if its charge is uniformly distributed over a spherical volume of radius $d$.
$endgroup$
– James Arathoon
yesterday
1
$begingroup$
If it still matters, I also found the answer to be $frac{-2rho^3}{3d^2}$ by converting to rectangular.
$endgroup$
– Tom Himler
yesterday
|
show 6 more comments
$begingroup$
Edited: Typo in the original post
$endgroup$
– Lanier Freeman
yesterday
1
$begingroup$
So $phiin[0,pi]$, and $tin[0,rho]$ for some ?fixed? constant $rho$? (And the first solution seems to not depend on $rho$. Unexpected, since i can take $rho=0$.) Please fix some framework for all used constants. Things seem to be important, please just fix these details for the eye of a first reader... Help will come in some seconds... (At least numerically, this is the easiest (experimental) validation when explicit choices are given.)
$endgroup$
– dan_fulea
yesterday
1
$begingroup$
@LanierFreeman I'm not sure if this is too helpful, but I think the result must depend on rho. If you call the integral $I= I(rho,d)$, then differentiate with rho, I got $I'= -2(frac{rho}{d})^{2}$, again, differentiated in rho. But this means that the original integral can't depend only on $d$, right?
$endgroup$
– Ryan Goulden
yesterday
1
$begingroup$
Yes this is the type of integration found in some older textbooks on electromagnetism; the aim would be to eventually calculate the magnetic moment of a "classical" spinning electron, say if its charge is uniformly distributed over a spherical volume of radius $d$.
$endgroup$
– James Arathoon
yesterday
1
$begingroup$
If it still matters, I also found the answer to be $frac{-2rho^3}{3d^2}$ by converting to rectangular.
$endgroup$
– Tom Himler
yesterday
$begingroup$
Edited: Typo in the original post
$endgroup$
– Lanier Freeman
yesterday
$begingroup$
Edited: Typo in the original post
$endgroup$
– Lanier Freeman
yesterday
1
1
$begingroup$
So $phiin[0,pi]$, and $tin[0,rho]$ for some ?fixed? constant $rho$? (And the first solution seems to not depend on $rho$. Unexpected, since i can take $rho=0$.) Please fix some framework for all used constants. Things seem to be important, please just fix these details for the eye of a first reader... Help will come in some seconds... (At least numerically, this is the easiest (experimental) validation when explicit choices are given.)
$endgroup$
– dan_fulea
yesterday
$begingroup$
So $phiin[0,pi]$, and $tin[0,rho]$ for some ?fixed? constant $rho$? (And the first solution seems to not depend on $rho$. Unexpected, since i can take $rho=0$.) Please fix some framework for all used constants. Things seem to be important, please just fix these details for the eye of a first reader... Help will come in some seconds... (At least numerically, this is the easiest (experimental) validation when explicit choices are given.)
$endgroup$
– dan_fulea
yesterday
1
1
$begingroup$
@LanierFreeman I'm not sure if this is too helpful, but I think the result must depend on rho. If you call the integral $I= I(rho,d)$, then differentiate with rho, I got $I'= -2(frac{rho}{d})^{2}$, again, differentiated in rho. But this means that the original integral can't depend only on $d$, right?
$endgroup$
– Ryan Goulden
yesterday
$begingroup$
@LanierFreeman I'm not sure if this is too helpful, but I think the result must depend on rho. If you call the integral $I= I(rho,d)$, then differentiate with rho, I got $I'= -2(frac{rho}{d})^{2}$, again, differentiated in rho. But this means that the original integral can't depend only on $d$, right?
$endgroup$
– Ryan Goulden
yesterday
1
1
$begingroup$
Yes this is the type of integration found in some older textbooks on electromagnetism; the aim would be to eventually calculate the magnetic moment of a "classical" spinning electron, say if its charge is uniformly distributed over a spherical volume of radius $d$.
$endgroup$
– James Arathoon
yesterday
$begingroup$
Yes this is the type of integration found in some older textbooks on electromagnetism; the aim would be to eventually calculate the magnetic moment of a "classical" spinning electron, say if its charge is uniformly distributed over a spherical volume of radius $d$.
$endgroup$
– James Arathoon
yesterday
1
1
$begingroup$
If it still matters, I also found the answer to be $frac{-2rho^3}{3d^2}$ by converting to rectangular.
$endgroup$
– Tom Himler
yesterday
$begingroup$
If it still matters, I also found the answer to be $frac{-2rho^3}{3d^2}$ by converting to rectangular.
$endgroup$
– Tom Himler
yesterday
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
As a mathematician, I would divide by force in the numerator and denominator by $d^3$, substitute $t/d$ by something, thus reducing to the case $d=1$. But here, let it be, we conserve the homogeneous setting as a control of the computations.
We split the numerator, compute first
$$
begin{aligned}
J_1
&=
int_0^rho
dt
int_0^pi
frac
{t^2sinphicdot tcosphi}
{left[t^2sin^2phi+left(tcosphi-Dright)^2right]^{3/2}}; dphi
\
&=
int_0^rho
dt
int_0^pi
frac
{t^2(-cosphi)'cdot tcosphi}
{left[t^2-2Dtcosphi+D^2right]^{3/2}}; dphi
\
&qquadtext{ Substitution: }u=cos phi ,
\
&=
int_0^rho
dt
int_{-1}^1
frac
{t^3; u}
{left[t^2-2Dt;u+D^2right]^{3/2}}; du
\
&qquadtext{ Substitution (for $u$, fixed $t$) of the radical }v=sqrt{t^2-2Dt;u+D^2} ,
\
&qquad u=frac 1{2Dt}(t^2+D^2-v^2) , du=-frac v{Dt}; dv\ ,
\
&=
-
int_0^rho
dt
int_{sqrt{t^2+2Dt+D^2}}^{sqrt{t^2-2Dt+D^2}}
frac
{t^3; frac 1{2Dt}(t^2+D^2-v^2)}
{v^3}; frac v{Dt}; dv
\
&=
int_0^rho
t;dt
int_{D-t}^{D+t}
frac 1{2D^2}
cdot
frac {t^2+D^2-v^2}
{v^2}; dv
\
&=
int_0^rho
t;dt
;frac 1{2D^2}
left[
-(t^2+D^2)frac 1v
-1
right]_{v=D-t}^{v=D+t}
\
&=
int_0^rho
dt
;frac t{2D^2}
left[
(t^2+D^2)left(frac 1{D-t}-frac 1{D+t}right)
-
2t
right]
\
&=
int_0^rho
dt
left[
frac D{D+t}
+frac D{D-t}
-2frac{D^2+t^2}{D^2}
right]
\
&=
Dlnfrac {D+t}{D-t}
-
2rholeft(1+frac {rho^2}{3D^2}right)
.
end{aligned}
$$
Computer check for $D=2$, $rho=1$ (pari/gp code):
? D=2; r=1;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%19 = 0.030557910669552716123823807178384744388
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%20 = 0.030557910669552716123823807178384742634
?
? D=223; r=101;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%22 = 1.9969022076015148346071622544965636670
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%23 = 1.9969022076015148346071622544965636629
The other integral. I will integrate here first w.r.t. $t$.
$$
begin{aligned}
J_2
&=
-D
int_0^pi
dphi
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&qquadtext{ and we consider separately (without the factor $-D$)}
\
J_2(phi)
&=
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&=
int_{0-Dcosphi}^{rho-Dcosphi}
frac
{(u+Dcosphi)^2}
{(u^2+a^2)^{3/2}}
; du ,qquad a:= Dsinphi
.
\
&qquad
text{ Now the integrals can be computed}
\
int frac{u^2}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac t{(u^2+a^2)^{1/2}}+operatorname{arcsinh} frac ta+C ,
\
int frac{u}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac 1{(u^2+a^2)^{1/2}}+C ,
\
int frac{1}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac {a^2;u}{(u^2+a^2)^{1/2}}+C ,
end{aligned}
$$
and the computation goes on.
If my calculus is ok, then
$$
begin{aligned}
J_2(phi)
&=
int_0^pi
dphi;
Bigg[
operatorname{arcsinh} frac{t-Dcos phi}{Dsinphi}
\&qquadqquadqquad+
frac{t-Dcosphi}{(t^2-2Dtcosphi+D^2)^{1/2}sin^2phi}
\&qquadqquadqquadqquadqquadqquad
+frac2{(t^2-2Dtcosphi+D^2)^{1/2}}
Bigg]_0^rho .
end{aligned}
$$
I have to submit, hope this is helpful to check with the own computations.
I'll be back, but typing kills a lot of time.
$endgroup$
add a comment |
$begingroup$
Hint:
With the change of variable $u=cosphi$, the integral on $phi$ becomes
$$int_{-1}^1frac{t^2(tu-d)}{sqrt{(u-dt)^2+d^2(1-t^2)}}du.$$
By decomposition of the numerator, you will get a term
$$c(t)log((u-dt)^2+d^2(1-t^2))$$
and another
$$c'(t)arctanfrac{u-dt}{dsqrt{1-t^2}}.$$
These terms do not simplify at the bounds of the integration interval.
The integral on $t$ (cubic in $t$ at the denominator) is worse. I am not optimisitc about existence of a closed-form.
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add a comment |
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$begingroup$
As a mathematician, I would divide by force in the numerator and denominator by $d^3$, substitute $t/d$ by something, thus reducing to the case $d=1$. But here, let it be, we conserve the homogeneous setting as a control of the computations.
We split the numerator, compute first
$$
begin{aligned}
J_1
&=
int_0^rho
dt
int_0^pi
frac
{t^2sinphicdot tcosphi}
{left[t^2sin^2phi+left(tcosphi-Dright)^2right]^{3/2}}; dphi
\
&=
int_0^rho
dt
int_0^pi
frac
{t^2(-cosphi)'cdot tcosphi}
{left[t^2-2Dtcosphi+D^2right]^{3/2}}; dphi
\
&qquadtext{ Substitution: }u=cos phi ,
\
&=
int_0^rho
dt
int_{-1}^1
frac
{t^3; u}
{left[t^2-2Dt;u+D^2right]^{3/2}}; du
\
&qquadtext{ Substitution (for $u$, fixed $t$) of the radical }v=sqrt{t^2-2Dt;u+D^2} ,
\
&qquad u=frac 1{2Dt}(t^2+D^2-v^2) , du=-frac v{Dt}; dv\ ,
\
&=
-
int_0^rho
dt
int_{sqrt{t^2+2Dt+D^2}}^{sqrt{t^2-2Dt+D^2}}
frac
{t^3; frac 1{2Dt}(t^2+D^2-v^2)}
{v^3}; frac v{Dt}; dv
\
&=
int_0^rho
t;dt
int_{D-t}^{D+t}
frac 1{2D^2}
cdot
frac {t^2+D^2-v^2}
{v^2}; dv
\
&=
int_0^rho
t;dt
;frac 1{2D^2}
left[
-(t^2+D^2)frac 1v
-1
right]_{v=D-t}^{v=D+t}
\
&=
int_0^rho
dt
;frac t{2D^2}
left[
(t^2+D^2)left(frac 1{D-t}-frac 1{D+t}right)
-
2t
right]
\
&=
int_0^rho
dt
left[
frac D{D+t}
+frac D{D-t}
-2frac{D^2+t^2}{D^2}
right]
\
&=
Dlnfrac {D+t}{D-t}
-
2rholeft(1+frac {rho^2}{3D^2}right)
.
end{aligned}
$$
Computer check for $D=2$, $rho=1$ (pari/gp code):
? D=2; r=1;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%19 = 0.030557910669552716123823807178384744388
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%20 = 0.030557910669552716123823807178384742634
?
? D=223; r=101;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%22 = 1.9969022076015148346071622544965636670
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%23 = 1.9969022076015148346071622544965636629
The other integral. I will integrate here first w.r.t. $t$.
$$
begin{aligned}
J_2
&=
-D
int_0^pi
dphi
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&qquadtext{ and we consider separately (without the factor $-D$)}
\
J_2(phi)
&=
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&=
int_{0-Dcosphi}^{rho-Dcosphi}
frac
{(u+Dcosphi)^2}
{(u^2+a^2)^{3/2}}
; du ,qquad a:= Dsinphi
.
\
&qquad
text{ Now the integrals can be computed}
\
int frac{u^2}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac t{(u^2+a^2)^{1/2}}+operatorname{arcsinh} frac ta+C ,
\
int frac{u}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac 1{(u^2+a^2)^{1/2}}+C ,
\
int frac{1}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac {a^2;u}{(u^2+a^2)^{1/2}}+C ,
end{aligned}
$$
and the computation goes on.
If my calculus is ok, then
$$
begin{aligned}
J_2(phi)
&=
int_0^pi
dphi;
Bigg[
operatorname{arcsinh} frac{t-Dcos phi}{Dsinphi}
\&qquadqquadqquad+
frac{t-Dcosphi}{(t^2-2Dtcosphi+D^2)^{1/2}sin^2phi}
\&qquadqquadqquadqquadqquadqquad
+frac2{(t^2-2Dtcosphi+D^2)^{1/2}}
Bigg]_0^rho .
end{aligned}
$$
I have to submit, hope this is helpful to check with the own computations.
I'll be back, but typing kills a lot of time.
$endgroup$
add a comment |
$begingroup$
As a mathematician, I would divide by force in the numerator and denominator by $d^3$, substitute $t/d$ by something, thus reducing to the case $d=1$. But here, let it be, we conserve the homogeneous setting as a control of the computations.
We split the numerator, compute first
$$
begin{aligned}
J_1
&=
int_0^rho
dt
int_0^pi
frac
{t^2sinphicdot tcosphi}
{left[t^2sin^2phi+left(tcosphi-Dright)^2right]^{3/2}}; dphi
\
&=
int_0^rho
dt
int_0^pi
frac
{t^2(-cosphi)'cdot tcosphi}
{left[t^2-2Dtcosphi+D^2right]^{3/2}}; dphi
\
&qquadtext{ Substitution: }u=cos phi ,
\
&=
int_0^rho
dt
int_{-1}^1
frac
{t^3; u}
{left[t^2-2Dt;u+D^2right]^{3/2}}; du
\
&qquadtext{ Substitution (for $u$, fixed $t$) of the radical }v=sqrt{t^2-2Dt;u+D^2} ,
\
&qquad u=frac 1{2Dt}(t^2+D^2-v^2) , du=-frac v{Dt}; dv\ ,
\
&=
-
int_0^rho
dt
int_{sqrt{t^2+2Dt+D^2}}^{sqrt{t^2-2Dt+D^2}}
frac
{t^3; frac 1{2Dt}(t^2+D^2-v^2)}
{v^3}; frac v{Dt}; dv
\
&=
int_0^rho
t;dt
int_{D-t}^{D+t}
frac 1{2D^2}
cdot
frac {t^2+D^2-v^2}
{v^2}; dv
\
&=
int_0^rho
t;dt
;frac 1{2D^2}
left[
-(t^2+D^2)frac 1v
-1
right]_{v=D-t}^{v=D+t}
\
&=
int_0^rho
dt
;frac t{2D^2}
left[
(t^2+D^2)left(frac 1{D-t}-frac 1{D+t}right)
-
2t
right]
\
&=
int_0^rho
dt
left[
frac D{D+t}
+frac D{D-t}
-2frac{D^2+t^2}{D^2}
right]
\
&=
Dlnfrac {D+t}{D-t}
-
2rholeft(1+frac {rho^2}{3D^2}right)
.
end{aligned}
$$
Computer check for $D=2$, $rho=1$ (pari/gp code):
? D=2; r=1;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%19 = 0.030557910669552716123823807178384744388
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%20 = 0.030557910669552716123823807178384742634
?
? D=223; r=101;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%22 = 1.9969022076015148346071622544965636670
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%23 = 1.9969022076015148346071622544965636629
The other integral. I will integrate here first w.r.t. $t$.
$$
begin{aligned}
J_2
&=
-D
int_0^pi
dphi
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&qquadtext{ and we consider separately (without the factor $-D$)}
\
J_2(phi)
&=
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&=
int_{0-Dcosphi}^{rho-Dcosphi}
frac
{(u+Dcosphi)^2}
{(u^2+a^2)^{3/2}}
; du ,qquad a:= Dsinphi
.
\
&qquad
text{ Now the integrals can be computed}
\
int frac{u^2}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac t{(u^2+a^2)^{1/2}}+operatorname{arcsinh} frac ta+C ,
\
int frac{u}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac 1{(u^2+a^2)^{1/2}}+C ,
\
int frac{1}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac {a^2;u}{(u^2+a^2)^{1/2}}+C ,
end{aligned}
$$
and the computation goes on.
If my calculus is ok, then
$$
begin{aligned}
J_2(phi)
&=
int_0^pi
dphi;
Bigg[
operatorname{arcsinh} frac{t-Dcos phi}{Dsinphi}
\&qquadqquadqquad+
frac{t-Dcosphi}{(t^2-2Dtcosphi+D^2)^{1/2}sin^2phi}
\&qquadqquadqquadqquadqquadqquad
+frac2{(t^2-2Dtcosphi+D^2)^{1/2}}
Bigg]_0^rho .
end{aligned}
$$
I have to submit, hope this is helpful to check with the own computations.
I'll be back, but typing kills a lot of time.
$endgroup$
add a comment |
$begingroup$
As a mathematician, I would divide by force in the numerator and denominator by $d^3$, substitute $t/d$ by something, thus reducing to the case $d=1$. But here, let it be, we conserve the homogeneous setting as a control of the computations.
We split the numerator, compute first
$$
begin{aligned}
J_1
&=
int_0^rho
dt
int_0^pi
frac
{t^2sinphicdot tcosphi}
{left[t^2sin^2phi+left(tcosphi-Dright)^2right]^{3/2}}; dphi
\
&=
int_0^rho
dt
int_0^pi
frac
{t^2(-cosphi)'cdot tcosphi}
{left[t^2-2Dtcosphi+D^2right]^{3/2}}; dphi
\
&qquadtext{ Substitution: }u=cos phi ,
\
&=
int_0^rho
dt
int_{-1}^1
frac
{t^3; u}
{left[t^2-2Dt;u+D^2right]^{3/2}}; du
\
&qquadtext{ Substitution (for $u$, fixed $t$) of the radical }v=sqrt{t^2-2Dt;u+D^2} ,
\
&qquad u=frac 1{2Dt}(t^2+D^2-v^2) , du=-frac v{Dt}; dv\ ,
\
&=
-
int_0^rho
dt
int_{sqrt{t^2+2Dt+D^2}}^{sqrt{t^2-2Dt+D^2}}
frac
{t^3; frac 1{2Dt}(t^2+D^2-v^2)}
{v^3}; frac v{Dt}; dv
\
&=
int_0^rho
t;dt
int_{D-t}^{D+t}
frac 1{2D^2}
cdot
frac {t^2+D^2-v^2}
{v^2}; dv
\
&=
int_0^rho
t;dt
;frac 1{2D^2}
left[
-(t^2+D^2)frac 1v
-1
right]_{v=D-t}^{v=D+t}
\
&=
int_0^rho
dt
;frac t{2D^2}
left[
(t^2+D^2)left(frac 1{D-t}-frac 1{D+t}right)
-
2t
right]
\
&=
int_0^rho
dt
left[
frac D{D+t}
+frac D{D-t}
-2frac{D^2+t^2}{D^2}
right]
\
&=
Dlnfrac {D+t}{D-t}
-
2rholeft(1+frac {rho^2}{3D^2}right)
.
end{aligned}
$$
Computer check for $D=2$, $rho=1$ (pari/gp code):
? D=2; r=1;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%19 = 0.030557910669552716123823807178384744388
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%20 = 0.030557910669552716123823807178384742634
?
? D=223; r=101;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%22 = 1.9969022076015148346071622544965636670
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%23 = 1.9969022076015148346071622544965636629
The other integral. I will integrate here first w.r.t. $t$.
$$
begin{aligned}
J_2
&=
-D
int_0^pi
dphi
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&qquadtext{ and we consider separately (without the factor $-D$)}
\
J_2(phi)
&=
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&=
int_{0-Dcosphi}^{rho-Dcosphi}
frac
{(u+Dcosphi)^2}
{(u^2+a^2)^{3/2}}
; du ,qquad a:= Dsinphi
.
\
&qquad
text{ Now the integrals can be computed}
\
int frac{u^2}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac t{(u^2+a^2)^{1/2}}+operatorname{arcsinh} frac ta+C ,
\
int frac{u}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac 1{(u^2+a^2)^{1/2}}+C ,
\
int frac{1}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac {a^2;u}{(u^2+a^2)^{1/2}}+C ,
end{aligned}
$$
and the computation goes on.
If my calculus is ok, then
$$
begin{aligned}
J_2(phi)
&=
int_0^pi
dphi;
Bigg[
operatorname{arcsinh} frac{t-Dcos phi}{Dsinphi}
\&qquadqquadqquad+
frac{t-Dcosphi}{(t^2-2Dtcosphi+D^2)^{1/2}sin^2phi}
\&qquadqquadqquadqquadqquadqquad
+frac2{(t^2-2Dtcosphi+D^2)^{1/2}}
Bigg]_0^rho .
end{aligned}
$$
I have to submit, hope this is helpful to check with the own computations.
I'll be back, but typing kills a lot of time.
$endgroup$
As a mathematician, I would divide by force in the numerator and denominator by $d^3$, substitute $t/d$ by something, thus reducing to the case $d=1$. But here, let it be, we conserve the homogeneous setting as a control of the computations.
We split the numerator, compute first
$$
begin{aligned}
J_1
&=
int_0^rho
dt
int_0^pi
frac
{t^2sinphicdot tcosphi}
{left[t^2sin^2phi+left(tcosphi-Dright)^2right]^{3/2}}; dphi
\
&=
int_0^rho
dt
int_0^pi
frac
{t^2(-cosphi)'cdot tcosphi}
{left[t^2-2Dtcosphi+D^2right]^{3/2}}; dphi
\
&qquadtext{ Substitution: }u=cos phi ,
\
&=
int_0^rho
dt
int_{-1}^1
frac
{t^3; u}
{left[t^2-2Dt;u+D^2right]^{3/2}}; du
\
&qquadtext{ Substitution (for $u$, fixed $t$) of the radical }v=sqrt{t^2-2Dt;u+D^2} ,
\
&qquad u=frac 1{2Dt}(t^2+D^2-v^2) , du=-frac v{Dt}; dv\ ,
\
&=
-
int_0^rho
dt
int_{sqrt{t^2+2Dt+D^2}}^{sqrt{t^2-2Dt+D^2}}
frac
{t^3; frac 1{2Dt}(t^2+D^2-v^2)}
{v^3}; frac v{Dt}; dv
\
&=
int_0^rho
t;dt
int_{D-t}^{D+t}
frac 1{2D^2}
cdot
frac {t^2+D^2-v^2}
{v^2}; dv
\
&=
int_0^rho
t;dt
;frac 1{2D^2}
left[
-(t^2+D^2)frac 1v
-1
right]_{v=D-t}^{v=D+t}
\
&=
int_0^rho
dt
;frac t{2D^2}
left[
(t^2+D^2)left(frac 1{D-t}-frac 1{D+t}right)
-
2t
right]
\
&=
int_0^rho
dt
left[
frac D{D+t}
+frac D{D-t}
-2frac{D^2+t^2}{D^2}
right]
\
&=
Dlnfrac {D+t}{D-t}
-
2rholeft(1+frac {rho^2}{3D^2}right)
.
end{aligned}
$$
Computer check for $D=2$, $rho=1$ (pari/gp code):
? D=2; r=1;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%19 = 0.030557910669552716123823807178384744388
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%20 = 0.030557910669552716123823807178384742634
?
? D=223; r=101;
? intnum(t=0,r, intnum(s=0, Pi, t^2*sin(s)*t*cos(s) / (t^2-2*t*D*cos(s)+D^2)^(3/2) ) )
%22 = 1.9969022076015148346071622544965636670
? D*log( (D+r)/(D-r) ) - 2*r*(1+r^2/3/D^2)
%23 = 1.9969022076015148346071622544965636629
The other integral. I will integrate here first w.r.t. $t$.
$$
begin{aligned}
J_2
&=
-D
int_0^pi
dphi
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&qquadtext{ and we consider separately (without the factor $-D$)}
\
J_2(phi)
&=
int_0^rho
frac
{t^2}
{left[(t-Dcosphi)^2+Dsin^2phiright]^{3/2}}
; dt
\
&=
int_{0-Dcosphi}^{rho-Dcosphi}
frac
{(u+Dcosphi)^2}
{(u^2+a^2)^{3/2}}
; du ,qquad a:= Dsinphi
.
\
&qquad
text{ Now the integrals can be computed}
\
int frac{u^2}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac t{(u^2+a^2)^{1/2}}+operatorname{arcsinh} frac ta+C ,
\
int frac{u}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac 1{(u^2+a^2)^{1/2}}+C ,
\
int frac{1}
{(u^2+a^2)^{3/2}}
; dt
&=
-frac {a^2;u}{(u^2+a^2)^{1/2}}+C ,
end{aligned}
$$
and the computation goes on.
If my calculus is ok, then
$$
begin{aligned}
J_2(phi)
&=
int_0^pi
dphi;
Bigg[
operatorname{arcsinh} frac{t-Dcos phi}{Dsinphi}
\&qquadqquadqquad+
frac{t-Dcosphi}{(t^2-2Dtcosphi+D^2)^{1/2}sin^2phi}
\&qquadqquadqquadqquadqquadqquad
+frac2{(t^2-2Dtcosphi+D^2)^{1/2}}
Bigg]_0^rho .
end{aligned}
$$
I have to submit, hope this is helpful to check with the own computations.
I'll be back, but typing kills a lot of time.
edited 9 hours ago
clathratus
5,2391438
5,2391438
answered 18 hours ago
dan_fuleadan_fulea
6,9031312
6,9031312
add a comment |
add a comment |
$begingroup$
Hint:
With the change of variable $u=cosphi$, the integral on $phi$ becomes
$$int_{-1}^1frac{t^2(tu-d)}{sqrt{(u-dt)^2+d^2(1-t^2)}}du.$$
By decomposition of the numerator, you will get a term
$$c(t)log((u-dt)^2+d^2(1-t^2))$$
and another
$$c'(t)arctanfrac{u-dt}{dsqrt{1-t^2}}.$$
These terms do not simplify at the bounds of the integration interval.
The integral on $t$ (cubic in $t$ at the denominator) is worse. I am not optimisitc about existence of a closed-form.
$endgroup$
add a comment |
$begingroup$
Hint:
With the change of variable $u=cosphi$, the integral on $phi$ becomes
$$int_{-1}^1frac{t^2(tu-d)}{sqrt{(u-dt)^2+d^2(1-t^2)}}du.$$
By decomposition of the numerator, you will get a term
$$c(t)log((u-dt)^2+d^2(1-t^2))$$
and another
$$c'(t)arctanfrac{u-dt}{dsqrt{1-t^2}}.$$
These terms do not simplify at the bounds of the integration interval.
The integral on $t$ (cubic in $t$ at the denominator) is worse. I am not optimisitc about existence of a closed-form.
$endgroup$
add a comment |
$begingroup$
Hint:
With the change of variable $u=cosphi$, the integral on $phi$ becomes
$$int_{-1}^1frac{t^2(tu-d)}{sqrt{(u-dt)^2+d^2(1-t^2)}}du.$$
By decomposition of the numerator, you will get a term
$$c(t)log((u-dt)^2+d^2(1-t^2))$$
and another
$$c'(t)arctanfrac{u-dt}{dsqrt{1-t^2}}.$$
These terms do not simplify at the bounds of the integration interval.
The integral on $t$ (cubic in $t$ at the denominator) is worse. I am not optimisitc about existence of a closed-form.
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Hint:
With the change of variable $u=cosphi$, the integral on $phi$ becomes
$$int_{-1}^1frac{t^2(tu-d)}{sqrt{(u-dt)^2+d^2(1-t^2)}}du.$$
By decomposition of the numerator, you will get a term
$$c(t)log((u-dt)^2+d^2(1-t^2))$$
and another
$$c'(t)arctanfrac{u-dt}{dsqrt{1-t^2}}.$$
These terms do not simplify at the bounds of the integration interval.
The integral on $t$ (cubic in $t$ at the denominator) is worse. I am not optimisitc about existence of a closed-form.
answered 17 hours ago
Yves DaoustYves Daoust
131k676229
131k676229
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Edited: Typo in the original post
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– Lanier Freeman
yesterday
1
$begingroup$
So $phiin[0,pi]$, and $tin[0,rho]$ for some ?fixed? constant $rho$? (And the first solution seems to not depend on $rho$. Unexpected, since i can take $rho=0$.) Please fix some framework for all used constants. Things seem to be important, please just fix these details for the eye of a first reader... Help will come in some seconds... (At least numerically, this is the easiest (experimental) validation when explicit choices are given.)
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– dan_fulea
yesterday
1
$begingroup$
@LanierFreeman I'm not sure if this is too helpful, but I think the result must depend on rho. If you call the integral $I= I(rho,d)$, then differentiate with rho, I got $I'= -2(frac{rho}{d})^{2}$, again, differentiated in rho. But this means that the original integral can't depend only on $d$, right?
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– Ryan Goulden
yesterday
1
$begingroup$
Yes this is the type of integration found in some older textbooks on electromagnetism; the aim would be to eventually calculate the magnetic moment of a "classical" spinning electron, say if its charge is uniformly distributed over a spherical volume of radius $d$.
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– James Arathoon
yesterday
1
$begingroup$
If it still matters, I also found the answer to be $frac{-2rho^3}{3d^2}$ by converting to rectangular.
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– Tom Himler
yesterday