Diverging Integral with Bessel Function
$begingroup$
I am looking for the solution to the integral:
$$int_{a}^{infty} x J_n(alpha x);dx$$
where $a<< alpha$ and $n$. I get something out of Mathematica for $a=1,2,0.1...$ in terms of the Hypergeometric function. It makes me think that some conditional expression for the solution exists, I just fail to find it.
The original integral of interest is actually this one
$$int_{0}^{infty} x J_n(alpha x);dx$$
However, I know, this one does not converge. Maybe someone knows how to estimate it at least in certain limits (n should be any)... I need to extract something sensible, not just "you failed, go on with your life" kind of a situation...
I have tried to truncate it on top and take a limit to infinity, but that also doesn't work.
integration definite-integrals bessel-functions parameter-estimation
edited Mar 10 at 20:10
Robert Howard
2,2112933
asked Jan 25 at 23:16
MsTaisMsTais
1808
$endgroup$
add a comment |
$begingroup$
I am looking for the solution to the integral:
$$int_{a}^{infty} x J_n(alpha x);dx$$
where $a<< alpha$ and $n$. I get something out of Mathematica for $a=1,2,0.1...$ in terms of the Hypergeometric function. It makes me think that some conditional expression for the solution exists, I just fail to find it.
The original integral of interest is actually this one
$$int_{0}^{infty} x J_n(alpha x);dx$$
However, I know, this one does not converge. Maybe someone knows how to estimate it at least in certain limits (n should be any)... I need to extract something sensible, not just "you failed, go on with your life" kind of a situation...
I have tried to truncate it on top and take a limit to infinity, but that also doesn't work.
integration definite-integrals bessel-functions parameter-estimation
edited Mar 10 at 20:10
Robert Howard
2,2112933
asked Jan 25 at 23:16
MsTaisMsTais
1808
$endgroup$
$begingroup$
You sure that's not a product of two Bessel functions? Because there's an orthogonality relation $int_0^{infty}xJ_n(alpha x)J_n(beta x),dx=0$ for $alpha neq beta$. The sensible thing to extract may well be that we should have been working with a different integral.
$endgroup$
– jmerry
Jan 25 at 23:47
add a comment |
$begingroup$
I am looking for the solution to the integral:
$$int_{a}^{infty} x J_n(alpha x);dx$$
where $a<< alpha$ and $n$. I get something out of Mathematica for $a=1,2,0.1...$ in terms of the Hypergeometric function. It makes me think that some conditional expression for the solution exists, I just fail to find it.
The original integral of interest is actually this one
$$int_{0}^{infty} x J_n(alpha x);dx$$
However, I know, this one does not converge. Maybe someone knows how to estimate it at least in certain limits (n should be any)... I need to extract something sensible, not just "you failed, go on with your life" kind of a situation...
I have tried to truncate it on top and take a limit to infinity, but that also doesn't work.
integration definite-integrals bessel-functions parameter-estimation
edited Mar 10 at 20:10
Robert Howard
2,2112933
asked Jan 25 at 23:16
MsTaisMsTais
1808
$endgroup$
I am looking for the solution to the integral:
$$int_{a}^{infty} x J_n(alpha x);dx$$
where $a<< alpha$ and $n$. I get something out of Mathematica for $a=1,2,0.1...$ in terms of the Hypergeometric function. It makes me think that some conditional expression for the solution exists, I just fail to find it.
The original integral of interest is actually this one
$$int_{0}^{infty} x J_n(alpha x);dx$$
However, I know, this one does not converge. Maybe someone knows how to estimate it at least in certain limits (n should be any)... I need to extract something sensible, not just "you failed, go on with your life" kind of a situation...
I have tried to truncate it on top and take a limit to infinity, but that also doesn't work.
integration definite-integrals bessel-functions parameter-estimation
integration definite-integrals bessel-functions parameter-estimation
edited Mar 10 at 20:10
Robert Howard
2,2112933
asked Jan 25 at 23:16
MsTaisMsTais
1808
edited Mar 10 at 20:10
Robert Howard
2,2112933
asked Jan 25 at 23:16
MsTaisMsTais
1808
edited Mar 10 at 20:10
Robert Howard
2,2112933
edited Mar 10 at 20:10
Robert Howard
2,2112933
edited Mar 10 at 20:10
Robert Howard
2,2112933
2,2112933
asked Jan 25 at 23:16
MsTaisMsTais
1808
asked Jan 25 at 23:16
MsTaisMsTais
1808
asked Jan 25 at 23:16
MsTaisMsTais
1808
1808
$begingroup$
You sure that's not a product of two Bessel functions? Because there's an orthogonality relation $int_0^{infty}xJ_n(alpha x)J_n(beta x),dx=0$ for $alpha neq beta$. The sensible thing to extract may well be that we should have been working with a different integral.
$endgroup$
– jmerry
Jan 25 at 23:47
add a comment |
$begingroup$
You sure that's not a product of two Bessel functions? Because there's an orthogonality relation $int_0^{infty}xJ_n(alpha x)J_n(beta x),dx=0$ for $alpha neq beta$. The sensible thing to extract may well be that we should have been working with a different integral.
$endgroup$
– jmerry
Jan 25 at 23:47
$begingroup$
You sure that's not a product of two Bessel functions? Because there's an orthogonality relation $int_0^{infty}xJ_n(alpha x)J_n(beta x),dx=0$ for $alpha neq beta$. The sensible thing to extract may well be that we should have been working with a different integral.
$endgroup$
– jmerry
Jan 25 at 23:47
$begingroup$
You sure that's not a product of two Bessel functions? Because there's an orthogonality relation $int_0^{infty}xJ_n(alpha x)J_n(beta x),dx=0$ for $alpha neq beta$. The sensible thing to extract may well be that we should have been working with a different integral.
$endgroup$
– jmerry
Jan 25 at 23:47
add a comment |
1 Answer
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$begingroup$
Your original integral is the Hankel Transform of order n of the function $f(x) = 1$.
The Hankel Transform of order n, and its inverse, can be defined as
$$mathscr{H}_nleft{f(r)right} = F(q) = int_0^infty f(r)J_n(qr)rspace dr$$
$$mathscr{H}^{-1}_nleft{F(q)right} = f(r) = int_0^infty F(q)J_n(qr)qspace dq$$
Note that the Hankel Transform is its own inverse.
For $n = 0$, and for an appropriate definition of the Dirac Delta distribution, it is easy to show the answer to your integral by using the inverse transform:
$$mathscr{H}^{-1}_0left{dfrac{delta(q)}{q}right} = int_0^infty dfrac{delta(q)}{q}J_0(qr)qspace dq = 1$$
$$mathscr{H}_0left{1right} = int_0^infty J_0(qr)rspace dr = dfrac{delta(q)}{q}$$
For higher orders of $n$, try to find a table of Hankel Transforms from a reliable source.
answered Jan 26 at 0:09
Andy WallsAndy Walls
1,754139
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
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votes
$begingroup$
Your original integral is the Hankel Transform of order n of the function $f(x) = 1$.
The Hankel Transform of order n, and its inverse, can be defined as
$$mathscr{H}_nleft{f(r)right} = F(q) = int_0^infty f(r)J_n(qr)rspace dr$$
$$mathscr{H}^{-1}_nleft{F(q)right} = f(r) = int_0^infty F(q)J_n(qr)qspace dq$$
Note that the Hankel Transform is its own inverse.
For $n = 0$, and for an appropriate definition of the Dirac Delta distribution, it is easy to show the answer to your integral by using the inverse transform:
$$mathscr{H}^{-1}_0left{dfrac{delta(q)}{q}right} = int_0^infty dfrac{delta(q)}{q}J_0(qr)qspace dq = 1$$
$$mathscr{H}_0left{1right} = int_0^infty J_0(qr)rspace dr = dfrac{delta(q)}{q}$$
For higher orders of $n$, try to find a table of Hankel Transforms from a reliable source.
answered Jan 26 at 0:09
Andy WallsAndy Walls
1,754139
$endgroup$
add a comment |
$begingroup$
Your original integral is the Hankel Transform of order n of the function $f(x) = 1$.
The Hankel Transform of order n, and its inverse, can be defined as
$$mathscr{H}_nleft{f(r)right} = F(q) = int_0^infty f(r)J_n(qr)rspace dr$$
$$mathscr{H}^{-1}_nleft{F(q)right} = f(r) = int_0^infty F(q)J_n(qr)qspace dq$$
Note that the Hankel Transform is its own inverse.
For $n = 0$, and for an appropriate definition of the Dirac Delta distribution, it is easy to show the answer to your integral by using the inverse transform:
$$mathscr{H}^{-1}_0left{dfrac{delta(q)}{q}right} = int_0^infty dfrac{delta(q)}{q}J_0(qr)qspace dq = 1$$
$$mathscr{H}_0left{1right} = int_0^infty J_0(qr)rspace dr = dfrac{delta(q)}{q}$$
For higher orders of $n$, try to find a table of Hankel Transforms from a reliable source.
answered Jan 26 at 0:09
Andy WallsAndy Walls
1,754139
$endgroup$
add a comment |
$begingroup$
Your original integral is the Hankel Transform of order n of the function $f(x) = 1$.
The Hankel Transform of order n, and its inverse, can be defined as
$$mathscr{H}_nleft{f(r)right} = F(q) = int_0^infty f(r)J_n(qr)rspace dr$$
$$mathscr{H}^{-1}_nleft{F(q)right} = f(r) = int_0^infty F(q)J_n(qr)qspace dq$$
Note that the Hankel Transform is its own inverse.
For $n = 0$, and for an appropriate definition of the Dirac Delta distribution, it is easy to show the answer to your integral by using the inverse transform:
$$mathscr{H}^{-1}_0left{dfrac{delta(q)}{q}right} = int_0^infty dfrac{delta(q)}{q}J_0(qr)qspace dq = 1$$
$$mathscr{H}_0left{1right} = int_0^infty J_0(qr)rspace dr = dfrac{delta(q)}{q}$$
For higher orders of $n$, try to find a table of Hankel Transforms from a reliable source.
answered Jan 26 at 0:09
Andy WallsAndy Walls
1,754139
$endgroup$
Your original integral is the Hankel Transform of order n of the function $f(x) = 1$.
The Hankel Transform of order n, and its inverse, can be defined as
$$mathscr{H}_nleft{f(r)right} = F(q) = int_0^infty f(r)J_n(qr)rspace dr$$
$$mathscr{H}^{-1}_nleft{F(q)right} = f(r) = int_0^infty F(q)J_n(qr)qspace dq$$
Note that the Hankel Transform is its own inverse.
For $n = 0$, and for an appropriate definition of the Dirac Delta distribution, it is easy to show the answer to your integral by using the inverse transform:
$$mathscr{H}^{-1}_0left{dfrac{delta(q)}{q}right} = int_0^infty dfrac{delta(q)}{q}J_0(qr)qspace dq = 1$$
$$mathscr{H}_0left{1right} = int_0^infty J_0(qr)rspace dr = dfrac{delta(q)}{q}$$
For higher orders of $n$, try to find a table of Hankel Transforms from a reliable source.
answered Jan 26 at 0:09
Andy WallsAndy Walls
1,754139
answered Jan 26 at 0:09
Andy WallsAndy Walls
1,754139
answered Jan 26 at 0:09
Andy WallsAndy Walls
1,754139
answered Jan 26 at 0:09
Andy WallsAndy Walls
1,754139
1,754139
add a comment |
add a comment |
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$begingroup$
You sure that's not a product of two Bessel functions? Because there's an orthogonality relation $int_0^{infty}xJ_n(alpha x)J_n(beta x),dx=0$ for $alpha neq beta$. The sensible thing to extract may well be that we should have been working with a different integral.
$endgroup$
– jmerry
Jan 25 at 23:47