Mixing
Verification of Integral identity
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The purpose of this thread is to
show that
beginequation
int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'=\=(n-1)frac2sqrt2pisqrt(2l+1)int_0^infty dr'int_-1^1 dxfracP_l^0(x)r'^2f(r')left(rr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)
endequation
Consider the Mellin transform
$$(Uf)(s) = frac1sqrt2pi int_0^infty f(r) r^-is+1/2 dr$$
and inverse Mellin transform $$(U^-1g)(s)= frac1sqrt2pi int_-infty^infty g(r) r^is-3/2 dr.$$
Consider the multiplication operator (2.31b) (I chose $l=0$, $n=2$ in the formula stated in the paper)
$$(mathcal M f)(r)=2pi int_0^inftyint_-1^1 fracr'^2 f(r') (r r')^1/2(r^2+r'^2+rr'x) dx dr' $$
then it is claimed that $U mathcal M U^-1$
real-analysis integration complex-analysis functional-analysis analysis
asked Jul 26 at 19:53
Sascha
60317
add a comment |Â
up vote
5
down vote
favorite
The purpose of this thread is to
show that
beginequation
int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'=\=(n-1)frac2sqrt2pisqrt(2l+1)int_0^infty dr'int_-1^1 dxfracP_l^0(x)r'^2f(r')left(rr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)
endequation
Consider the Mellin transform
$$(Uf)(s) = frac1sqrt2pi int_0^infty f(r) r^-is+1/2 dr$$
and inverse Mellin transform $$(U^-1g)(s)= frac1sqrt2pi int_-infty^infty g(r) r^is-3/2 dr.$$
Consider the multiplication operator (2.31b) (I chose $l=0$, $n=2$ in the formula stated in the paper)
$$(mathcal M f)(r)=2pi int_0^inftyint_-1^1 fracr'^2 f(r') (r r')^1/2(r^2+r'^2+rr'x) dx dr' $$
then it is claimed that $U mathcal M U^-1$
real-analysis integration complex-analysis functional-analysis analysis
asked Jul 26 at 19:53
Sascha
60317
What is the expression of $P_l^0$ ?
– Yves Daoust
Jul 27 at 8:40
Given the other expressions, one may suspect that the $+1$ from $2l+1$ was dropped.
– Yves Daoust
Jul 27 at 8:52
3
Sorry, too much work.
– Yves Daoust
Jul 27 at 9:16
I think it’s meant $,displaystyle frac2sqrt2pisqrt2l+1cdotfrac1sqrt2pi,$ (see $text2.31b,c$) and for $,textA.1,$ follows $,displaystyle frac1sqrt(2l+1)pi2pienspace$ (the paper is of course wrong). In case of doubt do simply a numerical calculation for $,l = 0,$ and compare $,text2.32,$ with $,textA.1,$ .
– user90369
Aug 1 at 15:59
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
The purpose of this thread is to
show that
beginequation
int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'=\=(n-1)frac2sqrt2pisqrt(2l+1)int_0^infty dr'int_-1^1 dxfracP_l^0(x)r'^2f(r')left(rr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)
endequation
Consider the Mellin transform
$$(Uf)(s) = frac1sqrt2pi int_0^infty f(r) r^-is+1/2 dr$$
and inverse Mellin transform $$(U^-1g)(s)= frac1sqrt2pi int_-infty^infty g(r) r^is-3/2 dr.$$
Consider the multiplication operator (2.31b) (I chose $l=0$, $n=2$ in the formula stated in the paper)
$$(mathcal M f)(r)=2pi int_0^inftyint_-1^1 fracr'^2 f(r') (r r')^1/2(r^2+r'^2+rr'x) dx dr' $$
then it is claimed that $U mathcal M U^-1$
real-analysis integration complex-analysis functional-analysis analysis
asked Jul 26 at 19:53
Sascha
60317
The purpose of this thread is to
show that
beginequation
int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'=\=(n-1)frac2sqrt2pisqrt(2l+1)int_0^infty dr'int_-1^1 dxfracP_l^0(x)r'^2f(r')left(rr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)
endequation
Consider the Mellin transform
$$(Uf)(s) = frac1sqrt2pi int_0^infty f(r) r^-is+1/2 dr$$
and inverse Mellin transform $$(U^-1g)(s)= frac1sqrt2pi int_-infty^infty g(r) r^is-3/2 dr.$$
Consider the multiplication operator (2.31b) (I chose $l=0$, $n=2$ in the formula stated in the paper)
$$(mathcal M f)(r)=2pi int_0^inftyint_-1^1 fracr'^2 f(r') (r r')^1/2(r^2+r'^2+rr'x) dx dr' $$
then it is claimed that $U mathcal M U^-1$
real-analysis integration complex-analysis functional-analysis analysis
asked Jul 26 at 19:53
Sascha
60317
edited 15 hours ago
asked Jul 26 at 19:53
Sascha
60317
asked Jul 26 at 19:53
Sascha
60317
asked Jul 26 at 19:53
Sascha
60317
60317
What is the expression of $P_l^0$ ?
– Yves Daoust
Jul 27 at 8:40
Given the other expressions, one may suspect that the $+1$ from $2l+1$ was dropped.
– Yves Daoust
Jul 27 at 8:52
3
Sorry, too much work.
– Yves Daoust
Jul 27 at 9:16
I think it’s meant $,displaystyle frac2sqrt2pisqrt2l+1cdotfrac1sqrt2pi,$ (see $text2.31b,c$) and for $,textA.1,$ follows $,displaystyle frac1sqrt(2l+1)pi2pienspace$ (the paper is of course wrong). In case of doubt do simply a numerical calculation for $,l = 0,$ and compare $,text2.32,$ with $,textA.1,$ .
– user90369
Aug 1 at 15:59
add a comment |Â
What is the expression of $P_l^0$ ?
– Yves Daoust
Jul 27 at 8:40
Given the other expressions, one may suspect that the $+1$ from $2l+1$ was dropped.
– Yves Daoust
Jul 27 at 8:52
3
Sorry, too much work.
– Yves Daoust
Jul 27 at 9:16
I think it’s meant $,displaystyle frac2sqrt2pisqrt2l+1cdotfrac1sqrt2pi,$ (see $text2.31b,c$) and for $,textA.1,$ follows $,displaystyle frac1sqrt(2l+1)pi2pienspace$ (the paper is of course wrong). In case of doubt do simply a numerical calculation for $,l = 0,$ and compare $,text2.32,$ with $,textA.1,$ .
– user90369
Aug 1 at 15:59
What is the expression of $P_l^0$ ?
– Yves Daoust
Jul 27 at 8:40
What is the expression of $P_l^0$ ?
– Yves Daoust
Jul 27 at 8:40
Given the other expressions, one may suspect that the $+1$ from $2l+1$ was dropped.
– Yves Daoust
Jul 27 at 8:52
Given the other expressions, one may suspect that the $+1$ from $2l+1$ was dropped.
– Yves Daoust
Jul 27 at 8:52
3
3
Sorry, too much work.
– Yves Daoust
Jul 27 at 9:16
Sorry, too much work.
– Yves Daoust
Jul 27 at 9:16
I think it’s meant $,displaystyle frac2sqrt2pisqrt2l+1cdotfrac1sqrt2pi,$ (see $text2.31b,c$) and for $,textA.1,$ follows $,displaystyle frac1sqrt(2l+1)pi2pienspace$ (the paper is of course wrong). In case of doubt do simply a numerical calculation for $,l = 0,$ and compare $,text2.32,$ with $,textA.1,$ .
– user90369
Aug 1 at 15:59
I think it’s meant $,displaystyle frac2sqrt2pisqrt2l+1cdotfrac1sqrt2pi,$ (see $text2.31b,c$) and for $,textA.1,$ follows $,displaystyle frac1sqrt(2l+1)pi2pienspace$ (the paper is of course wrong). In case of doubt do simply a numerical calculation for $,l = 0,$ and compare $,text2.32,$ with $,textA.1,$ .
– user90369
Aug 1 at 15:59
add a comment |Â
1 Answer
1
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2
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The Author says that it must be:
$$U^-1Lambda_lcdot(Uf)(r)=cal(hat M_lf)(r)$$
So let's verify it.
By the convolution theorem for the Mellin transformation it follows that:
$$U^-1Lambda_lcdot(Uf)(r)=int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'$$
The Mellin inverse transformation of $(textA.1)$ (with the misprint corrected)
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-isxi^2+frac2m+1xi x+1P_l^0(x)$$
that can be rewritten as
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-is+1/2xi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
is simply given by:
$$(U^-1Lambda_l)(r)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracxi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
from which:
$$left(U^-1Lambda_lmiddle)middle(fracrr'right)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracr'^2left(fracrr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)P_l^0(x)$$
and then, by substitution:
beginequation
int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'=\=(n-1)frac2sqrt2pisqrt(2l+1)int_0^infty dr'int_-1^1 dxfracP_l^0(x)r'^2f(r')left(rr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)
endequation
that is just $(2.31textb)$.
That proves the assertion.
Now go from the bottom up in this sequence of steps and you will get the answer.
answered Aug 4 at 16:43
trying
4,0461722
Thanks to you. You can find it on any text of physical mathematics. For instance, Integral Transforms for Engineers by Andrews and Shivamoggi 1988 Macmillan Publishing Company, republished in 1999 by The Society of Photo-Optical Instrumentation Engineers.
– trying
Aug 4 at 19:22
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The Author says that it must be:
$$U^-1Lambda_lcdot(Uf)(r)=cal(hat M_lf)(r)$$
So let's verify it.
By the convolution theorem for the Mellin transformation it follows that:
$$U^-1Lambda_lcdot(Uf)(r)=int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'$$
The Mellin inverse transformation of $(textA.1)$ (with the misprint corrected)
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-isxi^2+frac2m+1xi x+1P_l^0(x)$$
that can be rewritten as
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-is+1/2xi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
is simply given by:
$$(U^-1Lambda_l)(r)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracxi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
from which:
$$left(U^-1Lambda_lmiddle)middle(fracrr'right)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracr'^2left(fracrr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)P_l^0(x)$$
and then, by substitution:
beginequation
int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'=\=(n-1)frac2sqrt2pisqrt(2l+1)int_0^infty dr'int_-1^1 dxfracP_l^0(x)r'^2f(r')left(rr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)
endequation
that is just $(2.31textb)$.
That proves the assertion.
Now go from the bottom up in this sequence of steps and you will get the answer.
answered Aug 4 at 16:43
trying
4,0461722
Thanks to you. You can find it on any text of physical mathematics. For instance, Integral Transforms for Engineers by Andrews and Shivamoggi 1988 Macmillan Publishing Company, republished in 1999 by The Society of Photo-Optical Instrumentation Engineers.
– trying
Aug 4 at 19:22
add a comment |Â
up vote
2
down vote
accepted
The Author says that it must be:
$$U^-1Lambda_lcdot(Uf)(r)=cal(hat M_lf)(r)$$
So let's verify it.
By the convolution theorem for the Mellin transformation it follows that:
$$U^-1Lambda_lcdot(Uf)(r)=int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'$$
The Mellin inverse transformation of $(textA.1)$ (with the misprint corrected)
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-isxi^2+frac2m+1xi x+1P_l^0(x)$$
that can be rewritten as
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-is+1/2xi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
is simply given by:
$$(U^-1Lambda_l)(r)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracxi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
from which:
$$left(U^-1Lambda_lmiddle)middle(fracrr'right)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracr'^2left(fracrr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)P_l^0(x)$$
and then, by substitution:
beginequation
int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'=\=(n-1)frac2sqrt2pisqrt(2l+1)int_0^infty dr'int_-1^1 dxfracP_l^0(x)r'^2f(r')left(rr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)
endequation
that is just $(2.31textb)$.
That proves the assertion.
Now go from the bottom up in this sequence of steps and you will get the answer.
answered Aug 4 at 16:43
trying
4,0461722
Thanks to you. You can find it on any text of physical mathematics. For instance, Integral Transforms for Engineers by Andrews and Shivamoggi 1988 Macmillan Publishing Company, republished in 1999 by The Society of Photo-Optical Instrumentation Engineers.
– trying
Aug 4 at 19:22
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The Author says that it must be:
$$U^-1Lambda_lcdot(Uf)(r)=cal(hat M_lf)(r)$$
So let's verify it.
By the convolution theorem for the Mellin transformation it follows that:
$$U^-1Lambda_lcdot(Uf)(r)=int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'$$
The Mellin inverse transformation of $(textA.1)$ (with the misprint corrected)
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-isxi^2+frac2m+1xi x+1P_l^0(x)$$
that can be rewritten as
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-is+1/2xi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
is simply given by:
$$(U^-1Lambda_l)(r)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracxi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
from which:
$$left(U^-1Lambda_lmiddle)middle(fracrr'right)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracr'^2left(fracrr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)P_l^0(x)$$
and then, by substitution:
beginequation
int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'=\=(n-1)frac2sqrt2pisqrt(2l+1)int_0^infty dr'int_-1^1 dxfracP_l^0(x)r'^2f(r')left(rr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)
endequation
that is just $(2.31textb)$.
That proves the assertion.
Now go from the bottom up in this sequence of steps and you will get the answer.
answered Aug 4 at 16:43
trying
4,0461722
The Author says that it must be:
$$U^-1Lambda_lcdot(Uf)(r)=cal(hat M_lf)(r)$$
So let's verify it.
By the convolution theorem for the Mellin transformation it follows that:
$$U^-1Lambda_lcdot(Uf)(r)=int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'$$
The Mellin inverse transformation of $(textA.1)$ (with the misprint corrected)
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-isxi^2+frac2m+1xi x+1P_l^0(x)$$
that can be rewritten as
$$Lambda_l(s)=(n-1)sqrtfrac1(2l+1)pi2piint_0^infty dxiint_-1^1 dxfracxi^-is+1/2xi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
is simply given by:
$$(U^-1Lambda_l)(r)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracxi^-1/2xi^2+frac2m+1xi x+1P_l^0(x)$$
from which:
$$left(U^-1Lambda_lmiddle)middle(fracrr'right)=(n-1)sqrtfrac1(2l+1)pi2pisqrt2piint_-1^1 dxfracr'^2left(fracrr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)P_l^0(x)$$
and then, by substitution:
beginequation
int_0^infty left(U^-1Lambda_lmiddle)middle(fracrr'right)cdot f(r')fracdr'r'=\=(n-1)frac2sqrt2pisqrt(2l+1)int_0^infty dr'int_-1^1 dxfracP_l^0(x)r'^2f(r')left(rr'right)^1/2left(r^2+frac2m+1rr' x+r'^2right)
endequation
that is just $(2.31textb)$.
That proves the assertion.
Now go from the bottom up in this sequence of steps and you will get the answer.
answered Aug 4 at 16:43
trying
4,0461722
answered Aug 4 at 16:43
trying
4,0461722
answered Aug 4 at 16:43
trying
4,0461722
answered Aug 4 at 16:43
trying
4,0461722
4,0461722
Thanks to you. You can find it on any text of physical mathematics. For instance, Integral Transforms for Engineers by Andrews and Shivamoggi 1988 Macmillan Publishing Company, republished in 1999 by The Society of Photo-Optical Instrumentation Engineers.
– trying
Aug 4 at 19:22
add a comment |Â
Thanks to you. You can find it on any text of physical mathematics. For instance, Integral Transforms for Engineers by Andrews and Shivamoggi 1988 Macmillan Publishing Company, republished in 1999 by The Society of Photo-Optical Instrumentation Engineers.
– trying
Aug 4 at 19:22
Thanks to you. You can find it on any text of physical mathematics. For instance, Integral Transforms for Engineers by Andrews and Shivamoggi 1988 Macmillan Publishing Company, republished in 1999 by The Society of Photo-Optical Instrumentation Engineers.
– trying
Aug 4 at 19:22
Thanks to you. You can find it on any text of physical mathematics. For instance, Integral Transforms for Engineers by Andrews and Shivamoggi 1988 Macmillan Publishing Company, republished in 1999 by The Society of Photo-Optical Instrumentation Engineers.
– trying
Aug 4 at 19:22
add a comment |Â
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What is the expression of $P_l^0$ ?
– Yves Daoust
Jul 27 at 8:40
Given the other expressions, one may suspect that the $+1$ from $2l+1$ was dropped.
– Yves Daoust
Jul 27 at 8:52
3
Sorry, too much work.
– Yves Daoust
Jul 27 at 9:16
I think it’s meant $,displaystyle frac2sqrt2pisqrt2l+1cdotfrac1sqrt2pi,$ (see $text2.31b,c$) and for $,textA.1,$ follows $,displaystyle frac1sqrt(2l+1)pi2pienspace$ (the paper is of course wrong). In case of doubt do simply a numerical calculation for $,l = 0,$ and compare $,text2.32,$ with $,textA.1,$ .
– user90369
Aug 1 at 15:59