Is there a closed form for the integral $int_0^infty frac{e^{-x^2} I_0 left(beta x right) d x}{sqrt{...
I encountered this integral in my work, and it would be really convenient if it had a closed form in terms of any known special functions (which Mathematica could handle):
$$J(alpha,beta)=int_0^infty frac{e^{-x^2} I_0 left(beta x right) d x}{sqrt{ alpha^2+x^2}}$$
It's easy to see that:
$$J(alpha,0)=frac12 e^{alpha^2/2} K_0 left( frac{alpha^2}{2} right)$$
Using series expansion of the Bessel function and a little help from Mathematica, I was able to get a series expression:
$$J(alpha,beta)=frac12 e^{alpha^2/2} K_0 left( frac{alpha^2}{2} right)+frac{sqrt{pi}}{16} beta^2 sum_{k=0}^infty frac{(2k+1)!}{k!^3 (k+1)^2} U left(frac{1}{2},-k,alpha^2 right) frac{beta^{2k}}{4^{2k}}$$
Where $U$ is the confluent hypergeometric function, which in this case (according to Mathematica) is a linear combination of Bessel functions $K_0$ and $K_1$ though I wasn't able to get the general expression for the coefficients yet.
This series is good for small values of $beta$, but for large $beta$ I need too many terms, and the computation time is slower than I'd like.
I know that it's simple to get a quadrature formula for the integral, but still, a closed form would be better.
The integral can be thought as a Hankel transform with imaginary argument, but I wasn't able to find it in the tables of Hankel transforms either.
Clarification re: R. Burton's comment:
The final expression has the form:
$$alpha e^{-beta^2/2} J(alpha,beta)$$
Which should make the expression finite for every choice of variables.
Now that I think about it, I should probably use asymptotics for the Bessel function for large $beta$:
$$I_0 left(beta x right) asymp frac{1}{sqrt{2 pi beta x}}e^{beta x}$$
Then I get:
$$J(alpha,beta) asymp frac{1}{sqrt{2 pi beta}} int_0^infty frac{e^{-x^2+beta x} d x}{sqrt{x}sqrt{ alpha^2+x^2}}, quad beta to infty$$
Which I still don't know the closed form for.
A more simple integral which does have a closed form:
$$int_0^infty e^{-x^2} I_0 left(beta x right) d x= frac{sqrt{pi}}{2} e^{beta^2/8} I_0 left(frac{beta^2}{8} right)$$
definite-integrals closed-form bessel-functions
asked Dec 26 '18 at 11:41
Yuriy S
15.7k433117
|
show 4 more comments
I encountered this integral in my work, and it would be really convenient if it had a closed form in terms of any known special functions (which Mathematica could handle):
$$J(alpha,beta)=int_0^infty frac{e^{-x^2} I_0 left(beta x right) d x}{sqrt{ alpha^2+x^2}}$$
It's easy to see that:
$$J(alpha,0)=frac12 e^{alpha^2/2} K_0 left( frac{alpha^2}{2} right)$$
Using series expansion of the Bessel function and a little help from Mathematica, I was able to get a series expression:
$$J(alpha,beta)=frac12 e^{alpha^2/2} K_0 left( frac{alpha^2}{2} right)+frac{sqrt{pi}}{16} beta^2 sum_{k=0}^infty frac{(2k+1)!}{k!^3 (k+1)^2} U left(frac{1}{2},-k,alpha^2 right) frac{beta^{2k}}{4^{2k}}$$
Where $U$ is the confluent hypergeometric function, which in this case (according to Mathematica) is a linear combination of Bessel functions $K_0$ and $K_1$ though I wasn't able to get the general expression for the coefficients yet.
This series is good for small values of $beta$, but for large $beta$ I need too many terms, and the computation time is slower than I'd like.
I know that it's simple to get a quadrature formula for the integral, but still, a closed form would be better.
The integral can be thought as a Hankel transform with imaginary argument, but I wasn't able to find it in the tables of Hankel transforms either.
Clarification re: R. Burton's comment:
The final expression has the form:
$$alpha e^{-beta^2/2} J(alpha,beta)$$
Which should make the expression finite for every choice of variables.
Now that I think about it, I should probably use asymptotics for the Bessel function for large $beta$:
$$I_0 left(beta x right) asymp frac{1}{sqrt{2 pi beta x}}e^{beta x}$$
Then I get:
$$J(alpha,beta) asymp frac{1}{sqrt{2 pi beta}} int_0^infty frac{e^{-x^2+beta x} d x}{sqrt{x}sqrt{ alpha^2+x^2}}, quad beta to infty$$
Which I still don't know the closed form for.
A more simple integral which does have a closed form:
$$int_0^infty e^{-x^2} I_0 left(beta x right) d x= frac{sqrt{pi}}{2} e^{beta^2/8} I_0 left(frac{beta^2}{8} right)$$
definite-integrals closed-form bessel-functions
asked Dec 26 '18 at 11:41
Yuriy S
15.7k433117
I don't know about a closed form, but you can use the definition of $$I_0(x)=sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2}$$ to get $$J(alpha,beta,x)=int_0^infty frac{e^{-x^2}}{leftVert(alpha,x)rightVert}sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2} dx$$ If you truncate the sum it should at least reduce the computation time.
– R. Burton
Dec 26 '18 at 16:39
Also, I tried running this through Scilab, and for very large $beta$ (ex 130+), I ended up with things like $xtimes10^{103}$, %inf, and error messages. Suffice to say that for sufficiently large $beta$, the output of $J$ becomes impractically large.
– R. Burton
Dec 26 '18 at 16:46
@R.Burton, re: your first comment: this is what I did to get the series
– Yuriy S
Dec 26 '18 at 17:15
@R.Burton, I posted an edit to answer your second observation. The point is that for large $beta$ there's an exponential factor before the integral which should make it small. I guess instead of the series we could derive some kind of asymptotic expression for that case
– Yuriy S
Dec 26 '18 at 18:07
Is this somehow related to probability distributions or wavefunctions?
– R. Burton
Dec 27 '18 at 22:57
|
show 4 more comments
I encountered this integral in my work, and it would be really convenient if it had a closed form in terms of any known special functions (which Mathematica could handle):
$$J(alpha,beta)=int_0^infty frac{e^{-x^2} I_0 left(beta x right) d x}{sqrt{ alpha^2+x^2}}$$
It's easy to see that:
$$J(alpha,0)=frac12 e^{alpha^2/2} K_0 left( frac{alpha^2}{2} right)$$
Using series expansion of the Bessel function and a little help from Mathematica, I was able to get a series expression:
$$J(alpha,beta)=frac12 e^{alpha^2/2} K_0 left( frac{alpha^2}{2} right)+frac{sqrt{pi}}{16} beta^2 sum_{k=0}^infty frac{(2k+1)!}{k!^3 (k+1)^2} U left(frac{1}{2},-k,alpha^2 right) frac{beta^{2k}}{4^{2k}}$$
Where $U$ is the confluent hypergeometric function, which in this case (according to Mathematica) is a linear combination of Bessel functions $K_0$ and $K_1$ though I wasn't able to get the general expression for the coefficients yet.
This series is good for small values of $beta$, but for large $beta$ I need too many terms, and the computation time is slower than I'd like.
I know that it's simple to get a quadrature formula for the integral, but still, a closed form would be better.
The integral can be thought as a Hankel transform with imaginary argument, but I wasn't able to find it in the tables of Hankel transforms either.
Clarification re: R. Burton's comment:
The final expression has the form:
$$alpha e^{-beta^2/2} J(alpha,beta)$$
Which should make the expression finite for every choice of variables.
Now that I think about it, I should probably use asymptotics for the Bessel function for large $beta$:
$$I_0 left(beta x right) asymp frac{1}{sqrt{2 pi beta x}}e^{beta x}$$
Then I get:
$$J(alpha,beta) asymp frac{1}{sqrt{2 pi beta}} int_0^infty frac{e^{-x^2+beta x} d x}{sqrt{x}sqrt{ alpha^2+x^2}}, quad beta to infty$$
Which I still don't know the closed form for.
A more simple integral which does have a closed form:
$$int_0^infty e^{-x^2} I_0 left(beta x right) d x= frac{sqrt{pi}}{2} e^{beta^2/8} I_0 left(frac{beta^2}{8} right)$$
definite-integrals closed-form bessel-functions
asked Dec 26 '18 at 11:41
Yuriy S
15.7k433117
I encountered this integral in my work, and it would be really convenient if it had a closed form in terms of any known special functions (which Mathematica could handle):
$$J(alpha,beta)=int_0^infty frac{e^{-x^2} I_0 left(beta x right) d x}{sqrt{ alpha^2+x^2}}$$
It's easy to see that:
$$J(alpha,0)=frac12 e^{alpha^2/2} K_0 left( frac{alpha^2}{2} right)$$
Using series expansion of the Bessel function and a little help from Mathematica, I was able to get a series expression:
$$J(alpha,beta)=frac12 e^{alpha^2/2} K_0 left( frac{alpha^2}{2} right)+frac{sqrt{pi}}{16} beta^2 sum_{k=0}^infty frac{(2k+1)!}{k!^3 (k+1)^2} U left(frac{1}{2},-k,alpha^2 right) frac{beta^{2k}}{4^{2k}}$$
Where $U$ is the confluent hypergeometric function, which in this case (according to Mathematica) is a linear combination of Bessel functions $K_0$ and $K_1$ though I wasn't able to get the general expression for the coefficients yet.
This series is good for small values of $beta$, but for large $beta$ I need too many terms, and the computation time is slower than I'd like.
I know that it's simple to get a quadrature formula for the integral, but still, a closed form would be better.
The integral can be thought as a Hankel transform with imaginary argument, but I wasn't able to find it in the tables of Hankel transforms either.
Clarification re: R. Burton's comment:
The final expression has the form:
$$alpha e^{-beta^2/2} J(alpha,beta)$$
Which should make the expression finite for every choice of variables.
Now that I think about it, I should probably use asymptotics for the Bessel function for large $beta$:
$$I_0 left(beta x right) asymp frac{1}{sqrt{2 pi beta x}}e^{beta x}$$
Then I get:
$$J(alpha,beta) asymp frac{1}{sqrt{2 pi beta}} int_0^infty frac{e^{-x^2+beta x} d x}{sqrt{x}sqrt{ alpha^2+x^2}}, quad beta to infty$$
Which I still don't know the closed form for.
A more simple integral which does have a closed form:
$$int_0^infty e^{-x^2} I_0 left(beta x right) d x= frac{sqrt{pi}}{2} e^{beta^2/8} I_0 left(frac{beta^2}{8} right)$$
definite-integrals closed-form bessel-functions
definite-integrals closed-form bessel-functions
asked Dec 26 '18 at 11:41
Yuriy S
15.7k433117
asked Dec 26 '18 at 11:41
Yuriy S
15.7k433117
edited Dec 27 '18 at 11:08
asked Dec 26 '18 at 11:41
Yuriy S
15.7k433117
asked Dec 26 '18 at 11:41
Yuriy S
15.7k433117
asked Dec 26 '18 at 11:41
Yuriy S
15.7k433117
15.7k433117
I don't know about a closed form, but you can use the definition of $$I_0(x)=sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2}$$ to get $$J(alpha,beta,x)=int_0^infty frac{e^{-x^2}}{leftVert(alpha,x)rightVert}sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2} dx$$ If you truncate the sum it should at least reduce the computation time.
– R. Burton
Dec 26 '18 at 16:39
Also, I tried running this through Scilab, and for very large $beta$ (ex 130+), I ended up with things like $xtimes10^{103}$, %inf, and error messages. Suffice to say that for sufficiently large $beta$, the output of $J$ becomes impractically large.
– R. Burton
Dec 26 '18 at 16:46
@R.Burton, re: your first comment: this is what I did to get the series
– Yuriy S
Dec 26 '18 at 17:15
@R.Burton, I posted an edit to answer your second observation. The point is that for large $beta$ there's an exponential factor before the integral which should make it small. I guess instead of the series we could derive some kind of asymptotic expression for that case
– Yuriy S
Dec 26 '18 at 18:07
Is this somehow related to probability distributions or wavefunctions?
– R. Burton
Dec 27 '18 at 22:57
|
show 4 more comments
I don't know about a closed form, but you can use the definition of $$I_0(x)=sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2}$$ to get $$J(alpha,beta,x)=int_0^infty frac{e^{-x^2}}{leftVert(alpha,x)rightVert}sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2} dx$$ If you truncate the sum it should at least reduce the computation time.
– R. Burton
Dec 26 '18 at 16:39
Also, I tried running this through Scilab, and for very large $beta$ (ex 130+), I ended up with things like $xtimes10^{103}$, %inf, and error messages. Suffice to say that for sufficiently large $beta$, the output of $J$ becomes impractically large.
– R. Burton
Dec 26 '18 at 16:46
@R.Burton, re: your first comment: this is what I did to get the series
– Yuriy S
Dec 26 '18 at 17:15
@R.Burton, I posted an edit to answer your second observation. The point is that for large $beta$ there's an exponential factor before the integral which should make it small. I guess instead of the series we could derive some kind of asymptotic expression for that case
– Yuriy S
Dec 26 '18 at 18:07
Is this somehow related to probability distributions or wavefunctions?
– R. Burton
Dec 27 '18 at 22:57
I don't know about a closed form, but you can use the definition of $$I_0(x)=sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2}$$ to get $$J(alpha,beta,x)=int_0^infty frac{e^{-x^2}}{leftVert(alpha,x)rightVert}sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2} dx$$ If you truncate the sum it should at least reduce the computation time.
– R. Burton
Dec 26 '18 at 16:39
I don't know about a closed form, but you can use the definition of $$I_0(x)=sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2}$$ to get $$J(alpha,beta,x)=int_0^infty frac{e^{-x^2}}{leftVert(alpha,x)rightVert}sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2} dx$$ If you truncate the sum it should at least reduce the computation time.
– R. Burton
Dec 26 '18 at 16:39
Also, I tried running this through Scilab, and for very large $beta$ (ex 130+), I ended up with things like $xtimes10^{103}$, %inf, and error messages. Suffice to say that for sufficiently large $beta$, the output of $J$ becomes impractically large.
– R. Burton
Dec 26 '18 at 16:46
Also, I tried running this through Scilab, and for very large $beta$ (ex 130+), I ended up with things like $xtimes10^{103}$, %inf, and error messages. Suffice to say that for sufficiently large $beta$, the output of $J$ becomes impractically large.
– R. Burton
Dec 26 '18 at 16:46
@R.Burton, re: your first comment: this is what I did to get the series
– Yuriy S
Dec 26 '18 at 17:15
@R.Burton, re: your first comment: this is what I did to get the series
– Yuriy S
Dec 26 '18 at 17:15
@R.Burton, I posted an edit to answer your second observation. The point is that for large $beta$ there's an exponential factor before the integral which should make it small. I guess instead of the series we could derive some kind of asymptotic expression for that case
– Yuriy S
Dec 26 '18 at 18:07
@R.Burton, I posted an edit to answer your second observation. The point is that for large $beta$ there's an exponential factor before the integral which should make it small. I guess instead of the series we could derive some kind of asymptotic expression for that case
– Yuriy S
Dec 26 '18 at 18:07
Is this somehow related to probability distributions or wavefunctions?
– R. Burton
Dec 27 '18 at 22:57
Is this somehow related to probability distributions or wavefunctions?
– R. Burton
Dec 27 '18 at 22:57
|
show 4 more comments
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I don't know about a closed form, but you can use the definition of $$I_0(x)=sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2}$$ to get $$J(alpha,beta,x)=int_0^infty frac{e^{-x^2}}{leftVert(alpha,x)rightVert}sum_{n=0}^infty frac{(beta x)^{2n}}{4^nn!^2} dx$$ If you truncate the sum it should at least reduce the computation time.
– R. Burton
Dec 26 '18 at 16:39
Also, I tried running this through Scilab, and for very large $beta$ (ex 130+), I ended up with things like $xtimes10^{103}$, %inf, and error messages. Suffice to say that for sufficiently large $beta$, the output of $J$ becomes impractically large.
– R. Burton
Dec 26 '18 at 16:46
@R.Burton, re: your first comment: this is what I did to get the series
– Yuriy S
Dec 26 '18 at 17:15
@R.Burton, I posted an edit to answer your second observation. The point is that for large $beta$ there's an exponential factor before the integral which should make it small. I guess instead of the series we could derive some kind of asymptotic expression for that case
– Yuriy S
Dec 26 '18 at 18:07
Is this somehow related to probability distributions or wavefunctions?
– R. Burton
Dec 27 '18 at 22:57