Bear of an integral
$begingroup$
I have a pretty ferocious integral to solve, and would be over the moon if I were able to get some sort of analytic expression / insight for it.
$$ I = int_{r}^{infty} r_0^{-5/2} W_{-ialpha'/2, nu}left(frac{-ir_0^2omega'}{L}right) W_{-ialpha/2, nu}left(frac{-ir_0^2omega}{L}right) (i^beta J_{beta}left(-ihat{k}r_0right)) dr_0$$
All I've really done so far is look at the integrand in Mathematica (plotting Abs(integrand)):
Where $W$ is the $W$ Whittaker function (http://mathworld.wolfram.com/WhittakerFunction.html) and $J$ is the Bessel function of the first kind (http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html )
And also expand the integrand for small $r_0$ to gain some insight as to what happens at small $r$, but this is far too imprecise for the application
(particular solution of a differential equation using Sturm-Liouville theory). For the application, the best thing I could do is write $I = I_0 + f(r)$ where $I_0$ is some analytically determined constant. r-dependence is the first datum of importance, and then $hat{k},omega',alpha, beta, omega, alpha'$ dependence. Note that these constants should be positive.
If it helps, note that $omega' = omega + hat{omega}$ and $alpha = k^2L/2omega, alpha' = (k+hat{k})^2 L /(2 omega')$
calculus asymptotics special-functions
edited Dec 28 '18 at 20:34
Antonio Vargas
20.8k245111
asked Aug 17 '15 at 19:03
SchwingerSchwinger
1048
$endgroup$
add a comment |
$begingroup$
I have a pretty ferocious integral to solve, and would be over the moon if I were able to get some sort of analytic expression / insight for it.
$$ I = int_{r}^{infty} r_0^{-5/2} W_{-ialpha'/2, nu}left(frac{-ir_0^2omega'}{L}right) W_{-ialpha/2, nu}left(frac{-ir_0^2omega}{L}right) (i^beta J_{beta}left(-ihat{k}r_0right)) dr_0$$
All I've really done so far is look at the integrand in Mathematica (plotting Abs(integrand)):
Where $W$ is the $W$ Whittaker function (http://mathworld.wolfram.com/WhittakerFunction.html) and $J$ is the Bessel function of the first kind (http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html )
And also expand the integrand for small $r_0$ to gain some insight as to what happens at small $r$, but this is far too imprecise for the application
(particular solution of a differential equation using Sturm-Liouville theory). For the application, the best thing I could do is write $I = I_0 + f(r)$ where $I_0$ is some analytically determined constant. r-dependence is the first datum of importance, and then $hat{k},omega',alpha, beta, omega, alpha'$ dependence. Note that these constants should be positive.
If it helps, note that $omega' = omega + hat{omega}$ and $alpha = k^2L/2omega, alpha' = (k+hat{k})^2 L /(2 omega')$
calculus asymptotics special-functions
edited Dec 28 '18 at 20:34
Antonio Vargas
20.8k245111
asked Aug 17 '15 at 19:03
SchwingerSchwinger
1048
$endgroup$
$begingroup$
This doesn't help for $rrightarrowinfty$ ? Together with an asymptotic expansion of $J_0$ dlmf.nist.gov/13.19 (13.19.3)
$endgroup$
– tired
Aug 17 '15 at 19:32
$begingroup$
So I'm interested in $r rightarrow 0$ (should've specified). In this case the asymptotic forms hold for the piece of the integral that is very large, when the functions become oscillatory and those pieces of the integral are zero. Really I'm interested in what happens in between very small r and large r (the middle portion)
$endgroup$
– Schwinger
Aug 17 '15 at 19:43
6
$begingroup$
When I first read the title, I thought you were developing a new concept, as in defining the "bear" of the integral.
$endgroup$
– Hasan Saad
Aug 17 '15 at 19:43
$begingroup$
If you're interested in an analytical solution, you may want to take a look at the method of bracket, which appears to be quite powerful at computing these types of integral that can be expressed in series form. Here are a few of papers that outline and give examples of the approach: arxiv.org/abs/0812.3356, arxiv.org/abs/1004.2062, and arxiv.org/abs/1707.08942.
$endgroup$
– Zachary
Dec 28 '18 at 21:34
add a comment |
$begingroup$
I have a pretty ferocious integral to solve, and would be over the moon if I were able to get some sort of analytic expression / insight for it.
$$ I = int_{r}^{infty} r_0^{-5/2} W_{-ialpha'/2, nu}left(frac{-ir_0^2omega'}{L}right) W_{-ialpha/2, nu}left(frac{-ir_0^2omega}{L}right) (i^beta J_{beta}left(-ihat{k}r_0right)) dr_0$$
All I've really done so far is look at the integrand in Mathematica (plotting Abs(integrand)):
Where $W$ is the $W$ Whittaker function (http://mathworld.wolfram.com/WhittakerFunction.html) and $J$ is the Bessel function of the first kind (http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html )
And also expand the integrand for small $r_0$ to gain some insight as to what happens at small $r$, but this is far too imprecise for the application
(particular solution of a differential equation using Sturm-Liouville theory). For the application, the best thing I could do is write $I = I_0 + f(r)$ where $I_0$ is some analytically determined constant. r-dependence is the first datum of importance, and then $hat{k},omega',alpha, beta, omega, alpha'$ dependence. Note that these constants should be positive.
If it helps, note that $omega' = omega + hat{omega}$ and $alpha = k^2L/2omega, alpha' = (k+hat{k})^2 L /(2 omega')$
calculus asymptotics special-functions
edited Dec 28 '18 at 20:34
Antonio Vargas
20.8k245111
asked Aug 17 '15 at 19:03
SchwingerSchwinger
1048
$endgroup$
I have a pretty ferocious integral to solve, and would be over the moon if I were able to get some sort of analytic expression / insight for it.
$$ I = int_{r}^{infty} r_0^{-5/2} W_{-ialpha'/2, nu}left(frac{-ir_0^2omega'}{L}right) W_{-ialpha/2, nu}left(frac{-ir_0^2omega}{L}right) (i^beta J_{beta}left(-ihat{k}r_0right)) dr_0$$
All I've really done so far is look at the integrand in Mathematica (plotting Abs(integrand)):
Where $W$ is the $W$ Whittaker function (http://mathworld.wolfram.com/WhittakerFunction.html) and $J$ is the Bessel function of the first kind (http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html )
And also expand the integrand for small $r_0$ to gain some insight as to what happens at small $r$, but this is far too imprecise for the application
(particular solution of a differential equation using Sturm-Liouville theory). For the application, the best thing I could do is write $I = I_0 + f(r)$ where $I_0$ is some analytically determined constant. r-dependence is the first datum of importance, and then $hat{k},omega',alpha, beta, omega, alpha'$ dependence. Note that these constants should be positive.
If it helps, note that $omega' = omega + hat{omega}$ and $alpha = k^2L/2omega, alpha' = (k+hat{k})^2 L /(2 omega')$
calculus asymptotics special-functions
calculus asymptotics special-functions
edited Dec 28 '18 at 20:34
Antonio Vargas
20.8k245111
asked Aug 17 '15 at 19:03
SchwingerSchwinger
1048
edited Dec 28 '18 at 20:34
Antonio Vargas
20.8k245111
asked Aug 17 '15 at 19:03
SchwingerSchwinger
1048
edited Dec 28 '18 at 20:34
Antonio Vargas
20.8k245111
edited Dec 28 '18 at 20:34
Antonio Vargas
20.8k245111
edited Dec 28 '18 at 20:34
Antonio Vargas
20.8k245111
20.8k245111
asked Aug 17 '15 at 19:03
SchwingerSchwinger
1048
asked Aug 17 '15 at 19:03
SchwingerSchwinger
1048
asked Aug 17 '15 at 19:03
SchwingerSchwinger
1048
1048
$begingroup$
This doesn't help for $rrightarrowinfty$ ? Together with an asymptotic expansion of $J_0$ dlmf.nist.gov/13.19 (13.19.3)
$endgroup$
– tired
Aug 17 '15 at 19:32
$begingroup$
So I'm interested in $r rightarrow 0$ (should've specified). In this case the asymptotic forms hold for the piece of the integral that is very large, when the functions become oscillatory and those pieces of the integral are zero. Really I'm interested in what happens in between very small r and large r (the middle portion)
$endgroup$
– Schwinger
Aug 17 '15 at 19:43
6
$begingroup$
When I first read the title, I thought you were developing a new concept, as in defining the "bear" of the integral.
$endgroup$
– Hasan Saad
Aug 17 '15 at 19:43
$begingroup$
If you're interested in an analytical solution, you may want to take a look at the method of bracket, which appears to be quite powerful at computing these types of integral that can be expressed in series form. Here are a few of papers that outline and give examples of the approach: arxiv.org/abs/0812.3356, arxiv.org/abs/1004.2062, and arxiv.org/abs/1707.08942.
$endgroup$
– Zachary
Dec 28 '18 at 21:34
add a comment |
$begingroup$
This doesn't help for $rrightarrowinfty$ ? Together with an asymptotic expansion of $J_0$ dlmf.nist.gov/13.19 (13.19.3)
$endgroup$
– tired
Aug 17 '15 at 19:32
$begingroup$
So I'm interested in $r rightarrow 0$ (should've specified). In this case the asymptotic forms hold for the piece of the integral that is very large, when the functions become oscillatory and those pieces of the integral are zero. Really I'm interested in what happens in between very small r and large r (the middle portion)
$endgroup$
– Schwinger
Aug 17 '15 at 19:43
6
$begingroup$
When I first read the title, I thought you were developing a new concept, as in defining the "bear" of the integral.
$endgroup$
– Hasan Saad
Aug 17 '15 at 19:43
$begingroup$
If you're interested in an analytical solution, you may want to take a look at the method of bracket, which appears to be quite powerful at computing these types of integral that can be expressed in series form. Here are a few of papers that outline and give examples of the approach: arxiv.org/abs/0812.3356, arxiv.org/abs/1004.2062, and arxiv.org/abs/1707.08942.
$endgroup$
– Zachary
Dec 28 '18 at 21:34
$begingroup$
This doesn't help for $rrightarrowinfty$ ? Together with an asymptotic expansion of $J_0$ dlmf.nist.gov/13.19 (13.19.3)
$endgroup$
– tired
Aug 17 '15 at 19:32
$begingroup$
This doesn't help for $rrightarrowinfty$ ? Together with an asymptotic expansion of $J_0$ dlmf.nist.gov/13.19 (13.19.3)
$endgroup$
– tired
Aug 17 '15 at 19:32
$begingroup$
So I'm interested in $r rightarrow 0$ (should've specified). In this case the asymptotic forms hold for the piece of the integral that is very large, when the functions become oscillatory and those pieces of the integral are zero. Really I'm interested in what happens in between very small r and large r (the middle portion)
$endgroup$
– Schwinger
Aug 17 '15 at 19:43
$begingroup$
So I'm interested in $r rightarrow 0$ (should've specified). In this case the asymptotic forms hold for the piece of the integral that is very large, when the functions become oscillatory and those pieces of the integral are zero. Really I'm interested in what happens in between very small r and large r (the middle portion)
$endgroup$
– Schwinger
Aug 17 '15 at 19:43
6
6
$begingroup$
When I first read the title, I thought you were developing a new concept, as in defining the "bear" of the integral.
$endgroup$
– Hasan Saad
Aug 17 '15 at 19:43
$begingroup$
When I first read the title, I thought you were developing a new concept, as in defining the "bear" of the integral.
$endgroup$
– Hasan Saad
Aug 17 '15 at 19:43
$begingroup$
If you're interested in an analytical solution, you may want to take a look at the method of bracket, which appears to be quite powerful at computing these types of integral that can be expressed in series form. Here are a few of papers that outline and give examples of the approach: arxiv.org/abs/0812.3356, arxiv.org/abs/1004.2062, and arxiv.org/abs/1707.08942.
$endgroup$
– Zachary
Dec 28 '18 at 21:34
$begingroup$
If you're interested in an analytical solution, you may want to take a look at the method of bracket, which appears to be quite powerful at computing these types of integral that can be expressed in series form. Here are a few of papers that outline and give examples of the approach: arxiv.org/abs/0812.3356, arxiv.org/abs/1004.2062, and arxiv.org/abs/1707.08942.
$endgroup$
– Zachary
Dec 28 '18 at 21:34
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1400596%2fbear-of-an-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1400596%2fbear-of-an-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
This doesn't help for $rrightarrowinfty$ ? Together with an asymptotic expansion of $J_0$ dlmf.nist.gov/13.19 (13.19.3)
$endgroup$
– tired
Aug 17 '15 at 19:32
$begingroup$
So I'm interested in $r rightarrow 0$ (should've specified). In this case the asymptotic forms hold for the piece of the integral that is very large, when the functions become oscillatory and those pieces of the integral are zero. Really I'm interested in what happens in between very small r and large r (the middle portion)
$endgroup$
– Schwinger
Aug 17 '15 at 19:43
6
$begingroup$
When I first read the title, I thought you were developing a new concept, as in defining the "bear" of the integral.
$endgroup$
– Hasan Saad
Aug 17 '15 at 19:43
$begingroup$
If you're interested in an analytical solution, you may want to take a look at the method of bracket, which appears to be quite powerful at computing these types of integral that can be expressed in series form. Here are a few of papers that outline and give examples of the approach: arxiv.org/abs/0812.3356, arxiv.org/abs/1004.2062, and arxiv.org/abs/1707.08942.
$endgroup$
– Zachary
Dec 28 '18 at 21:34