Mixing
Computation of a double exponential integral
$begingroup$
I want to understand the behavior of this integral
$$ int_0^x e^{-frac{c}{t^2}} frac{1}{t^5} e^{-frac{c_2}{(x-t)^2}} frac{1}{(x-t)^5} dt. $$
The ideal answer would be a way to explicitly compute this, but I have no clue. This integral is clearly definite, but my problem is to understand quantitatively its behavior for small and large values of $c$ and $c_2$. Any idea on how to proceed?
analysis definite-integrals
asked Jan 19 at 12:30
Gâteau-GalloisGâteau-Gallois
362113
$endgroup$
add a comment |
$begingroup$
I want to understand the behavior of this integral
$$ int_0^x e^{-frac{c}{t^2}} frac{1}{t^5} e^{-frac{c_2}{(x-t)^2}} frac{1}{(x-t)^5} dt. $$
The ideal answer would be a way to explicitly compute this, but I have no clue. This integral is clearly definite, but my problem is to understand quantitatively its behavior for small and large values of $c$ and $c_2$. Any idea on how to proceed?
analysis definite-integrals
asked Jan 19 at 12:30
Gâteau-GalloisGâteau-Gallois
362113
$endgroup$
$begingroup$
At first i would simplify the integrand!
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 12:32
$begingroup$
What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
$endgroup$
– Gâteau-Gallois
Jan 19 at 12:44
$begingroup$
Why did you delete the second part of the question ?
$endgroup$
– Claude Leibovici
Jan 19 at 13:14
$begingroup$
I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
$endgroup$
– Gâteau-Gallois
Jan 19 at 13:38
add a comment |
$begingroup$
I want to understand the behavior of this integral
$$ int_0^x e^{-frac{c}{t^2}} frac{1}{t^5} e^{-frac{c_2}{(x-t)^2}} frac{1}{(x-t)^5} dt. $$
The ideal answer would be a way to explicitly compute this, but I have no clue. This integral is clearly definite, but my problem is to understand quantitatively its behavior for small and large values of $c$ and $c_2$. Any idea on how to proceed?
analysis definite-integrals
asked Jan 19 at 12:30
Gâteau-GalloisGâteau-Gallois
362113
$endgroup$
I want to understand the behavior of this integral
$$ int_0^x e^{-frac{c}{t^2}} frac{1}{t^5} e^{-frac{c_2}{(x-t)^2}} frac{1}{(x-t)^5} dt. $$
The ideal answer would be a way to explicitly compute this, but I have no clue. This integral is clearly definite, but my problem is to understand quantitatively its behavior for small and large values of $c$ and $c_2$. Any idea on how to proceed?
analysis definite-integrals
analysis definite-integrals
asked Jan 19 at 12:30
Gâteau-GalloisGâteau-Gallois
362113
asked Jan 19 at 12:30
Gâteau-GalloisGâteau-Gallois
362113
edited Jan 19 at 13:00
Gâteau-Gallois
asked Jan 19 at 12:30
Gâteau-GalloisGâteau-Gallois
362113
asked Jan 19 at 12:30
Gâteau-GalloisGâteau-Gallois
362113
asked Jan 19 at 12:30
Gâteau-GalloisGâteau-Gallois
362113
362113
$begingroup$
At first i would simplify the integrand!
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 12:32
$begingroup$
What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
$endgroup$
– Gâteau-Gallois
Jan 19 at 12:44
$begingroup$
Why did you delete the second part of the question ?
$endgroup$
– Claude Leibovici
Jan 19 at 13:14
$begingroup$
I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
$endgroup$
– Gâteau-Gallois
Jan 19 at 13:38
add a comment |
$begingroup$
At first i would simplify the integrand!
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 12:32
$begingroup$
What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
$endgroup$
– Gâteau-Gallois
Jan 19 at 12:44
$begingroup$
Why did you delete the second part of the question ?
$endgroup$
– Claude Leibovici
Jan 19 at 13:14
$begingroup$
I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
$endgroup$
– Gâteau-Gallois
Jan 19 at 13:38
$begingroup$
At first i would simplify the integrand!
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 12:32
$begingroup$
At first i would simplify the integrand!
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 12:32
$begingroup$
What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
$endgroup$
– Gâteau-Gallois
Jan 19 at 12:44
$begingroup$
What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
$endgroup$
– Gâteau-Gallois
Jan 19 at 12:44
$begingroup$
Why did you delete the second part of the question ?
$endgroup$
– Claude Leibovici
Jan 19 at 13:14
$begingroup$
Why did you delete the second part of the question ?
$endgroup$
– Claude Leibovici
Jan 19 at 13:14
$begingroup$
I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
$endgroup$
– Gâteau-Gallois
Jan 19 at 13:38
$begingroup$
I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
$endgroup$
– Gâteau-Gallois
Jan 19 at 13:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$
answered Jan 19 at 13:09
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$
answered Jan 19 at 12:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
add a comment |
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%2f3079311%2fcomputation-of-a-double-exponential-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$
answered Jan 19 at 13:09
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
Hint
Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$
answered Jan 19 at 13:09
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
add a comment |
$begingroup$
Hint
Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$
answered Jan 19 at 13:09
Claude LeiboviciClaude Leibovici
123k1157134
$endgroup$
Hint
Assuming $t<x$, $x>0$,$$intfrac{1}{t^k} frac{1}{(x-t)^l},dt=x^{-k-l+1} B_{frac{t}{x}}(1-k,1-l)$$
answered Jan 19 at 13:09
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 13:09
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 13:09
Claude LeiboviciClaude Leibovici
123k1157134
answered Jan 19 at 13:09
Claude LeiboviciClaude Leibovici
123k1157134
123k1157134
add a comment |
add a comment |
$begingroup$
I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$
answered Jan 19 at 12:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
add a comment |
$begingroup$
I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$
answered Jan 19 at 12:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
add a comment |
$begingroup$
I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$
answered Jan 19 at 12:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
$endgroup$
I mean this here $$e^{-frac{c}{t^2}-frac{c_2}{(x-t)^2}}cdot frac{1}{(t(x-t))^5}$$
answered Jan 19 at 12:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 19 at 12:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 19 at 12:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
answered Jan 19 at 12:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.7k42866
76.7k42866
add a comment |
add a comment |
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%2f3079311%2fcomputation-of-a-double-exponential-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
$begingroup$
At first i would simplify the integrand!
$endgroup$
– Dr. Sonnhard Graubner
Jan 19 at 12:32
$begingroup$
What would be the best way to do so? The two obvious changes of variable ($t to frac{1}{t}$ or $t to frac{1}{x-t}$) seem to lead to something more complicated.
$endgroup$
– Gâteau-Gallois
Jan 19 at 12:44
$begingroup$
Why did you delete the second part of the question ?
$endgroup$
– Claude Leibovici
Jan 19 at 13:14
$begingroup$
I think the integral I had was not convergent, near 0 it is homogeneous to $1/t^k$ for $k geq 2$... I need to check again the entire series computation
$endgroup$
– Gâteau-Gallois
Jan 19 at 13:38