Integral representation of a rational function
up vote
-3
down vote
favorite
I would like an integral representation for the function
$$ f(x)=frac{x(x^2-2)}{(x^2-1)^2},$$
that is I want to write it as $f(x)=int_a^b g(x,y)dy$ for some $g(x,y)$ which is not of the form $g(x,y)=f(x)g(y)$.
It must be a definite integral, with the range of integration independent of $x$.
My first guess were beta integrals, $int_0^1 y^a(1-y)^bdy$ with $a$ and $b$ functions of $x$, but I had no success with this.
real-analysis functions
asked Nov 21 at 20:21
thedude
670412
add a comment |
up vote
-3
down vote
favorite
I would like an integral representation for the function
$$ f(x)=frac{x(x^2-2)}{(x^2-1)^2},$$
that is I want to write it as $f(x)=int_a^b g(x,y)dy$ for some $g(x,y)$ which is not of the form $g(x,y)=f(x)g(y)$.
It must be a definite integral, with the range of integration independent of $x$.
My first guess were beta integrals, $int_0^1 y^a(1-y)^bdy$ with $a$ and $b$ functions of $x$, but I had no success with this.
real-analysis functions
asked Nov 21 at 20:21
thedude
670412
add a comment |
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
I would like an integral representation for the function
$$ f(x)=frac{x(x^2-2)}{(x^2-1)^2},$$
that is I want to write it as $f(x)=int_a^b g(x,y)dy$ for some $g(x,y)$ which is not of the form $g(x,y)=f(x)g(y)$.
It must be a definite integral, with the range of integration independent of $x$.
My first guess were beta integrals, $int_0^1 y^a(1-y)^bdy$ with $a$ and $b$ functions of $x$, but I had no success with this.
real-analysis functions
asked Nov 21 at 20:21
thedude
670412
I would like an integral representation for the function
$$ f(x)=frac{x(x^2-2)}{(x^2-1)^2},$$
that is I want to write it as $f(x)=int_a^b g(x,y)dy$ for some $g(x,y)$ which is not of the form $g(x,y)=f(x)g(y)$.
It must be a definite integral, with the range of integration independent of $x$.
My first guess were beta integrals, $int_0^1 y^a(1-y)^bdy$ with $a$ and $b$ functions of $x$, but I had no success with this.
real-analysis functions
real-analysis functions
asked Nov 21 at 20:21
thedude
670412
asked Nov 21 at 20:21
thedude
670412
edited Nov 22 at 17:13
asked Nov 21 at 20:21
thedude
670412
asked Nov 21 at 20:21
thedude
670412
asked Nov 21 at 20:21
thedude
670412
670412
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
As an indefinite integral,
$$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$
works. As a definite integral,
$$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$
works.
answered Nov 21 at 20:25
Ekesh
4855
An indefinite integral is not quite what I want, though.
– thedude
Nov 21 at 20:27
1
Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
– TrostAft
Nov 21 at 20:28
yes. @TrostAft is right. i updated my post with it for you.
– Ekesh
Nov 21 at 20:34
thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
– thedude
Nov 21 at 20:46
add a comment |
up vote
2
down vote
Of course, there is the trivial relation
$$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$
For example,
$$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$
for any $k > -1$, or
$$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$
As a potentially less trivial answer, you can perform any invertible integral transform
$$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$
Then, we can express the original function in terms of the inverse transform
$$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$
As an example, using the inverse Laplace transform,
$$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$
we can write the function
$$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$
answered Nov 21 at 20:46
AlexanderJ93
5,444622
Not helpful at all
– thedude
Nov 21 at 20:51
It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
– AlexanderJ93
Nov 21 at 20:52
@thedude I've updated the answer to include a less trivial solution
– AlexanderJ93
Nov 21 at 21:03
1
Yes, that is much better, thank you
– thedude
Nov 21 at 23:46
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',
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%2f3008303%2fintegral-representation-of-a-rational-function%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
up vote
2
down vote
As an indefinite integral,
$$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$
works. As a definite integral,
$$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$
works.
answered Nov 21 at 20:25
Ekesh
4855
An indefinite integral is not quite what I want, though.
– thedude
Nov 21 at 20:27
1
Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
– TrostAft
Nov 21 at 20:28
yes. @TrostAft is right. i updated my post with it for you.
– Ekesh
Nov 21 at 20:34
thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
– thedude
Nov 21 at 20:46
add a comment |
up vote
2
down vote
As an indefinite integral,
$$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$
works. As a definite integral,
$$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$
works.
answered Nov 21 at 20:25
Ekesh
4855
An indefinite integral is not quite what I want, though.
– thedude
Nov 21 at 20:27
1
Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
– TrostAft
Nov 21 at 20:28
yes. @TrostAft is right. i updated my post with it for you.
– Ekesh
Nov 21 at 20:34
thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
– thedude
Nov 21 at 20:46
add a comment |
up vote
2
down vote
up vote
2
down vote
As an indefinite integral,
$$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$
works. As a definite integral,
$$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$
works.
answered Nov 21 at 20:25
Ekesh
4855
As an indefinite integral,
$$int frac{-x^{4} +3x^{2} + 2}{(x^{2} - 1)^{3}} mathop{dx} $$
works. As a definite integral,
$$int_{sqrt{2}}^{x} frac{-y^{4} + 3y^{2} + 2}{(y^{2} - 1)^{3}} mathop{dy} $$
works.
answered Nov 21 at 20:25
Ekesh
4855
edited Nov 21 at 20:34
answered Nov 21 at 20:25
Ekesh
4855
answered Nov 21 at 20:25
Ekesh
4855
answered Nov 21 at 20:25
Ekesh
4855
4855
An indefinite integral is not quite what I want, though.
– thedude
Nov 21 at 20:27
1
Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
– TrostAft
Nov 21 at 20:28
yes. @TrostAft is right. i updated my post with it for you.
– Ekesh
Nov 21 at 20:34
thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
– thedude
Nov 21 at 20:46
add a comment |
An indefinite integral is not quite what I want, though.
– thedude
Nov 21 at 20:27
1
Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
– TrostAft
Nov 21 at 20:28
yes. @TrostAft is right. i updated my post with it for you.
– Ekesh
Nov 21 at 20:34
thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
– thedude
Nov 21 at 20:46
An indefinite integral is not quite what I want, though.
– thedude
Nov 21 at 20:27
An indefinite integral is not quite what I want, though.
– thedude
Nov 21 at 20:27
1
1
Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
– TrostAft
Nov 21 at 20:28
Yeah and if you want definite just choose $x$ as the upper bound of integration and some root of $x(x^2-2)$ as a lower bound of integration, i.e. $0, pm sqrt{2}$.
– TrostAft
Nov 21 at 20:28
yes. @TrostAft is right. i updated my post with it for you.
– Ekesh
Nov 21 at 20:34
yes. @TrostAft is right. i updated my post with it for you.
– Ekesh
Nov 21 at 20:34
thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
– thedude
Nov 21 at 20:46
thanks, I gave you an upvote, but this is not what I want. I updated the question to make it clear that the range must not depend on $x$ (otherwise the problem is very easy, of course)
– thedude
Nov 21 at 20:46
add a comment |
up vote
2
down vote
Of course, there is the trivial relation
$$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$
For example,
$$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$
for any $k > -1$, or
$$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$
As a potentially less trivial answer, you can perform any invertible integral transform
$$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$
Then, we can express the original function in terms of the inverse transform
$$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$
As an example, using the inverse Laplace transform,
$$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$
we can write the function
$$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$
answered Nov 21 at 20:46
AlexanderJ93
5,444622
Not helpful at all
– thedude
Nov 21 at 20:51
It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
– AlexanderJ93
Nov 21 at 20:52
@thedude I've updated the answer to include a less trivial solution
– AlexanderJ93
Nov 21 at 21:03
1
Yes, that is much better, thank you
– thedude
Nov 21 at 23:46
add a comment |
up vote
2
down vote
Of course, there is the trivial relation
$$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$
For example,
$$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$
for any $k > -1$, or
$$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$
As a potentially less trivial answer, you can perform any invertible integral transform
$$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$
Then, we can express the original function in terms of the inverse transform
$$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$
As an example, using the inverse Laplace transform,
$$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$
we can write the function
$$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$
answered Nov 21 at 20:46
AlexanderJ93
5,444622
Not helpful at all
– thedude
Nov 21 at 20:51
It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
– AlexanderJ93
Nov 21 at 20:52
@thedude I've updated the answer to include a less trivial solution
– AlexanderJ93
Nov 21 at 21:03
1
Yes, that is much better, thank you
– thedude
Nov 21 at 23:46
add a comment |
up vote
2
down vote
up vote
2
down vote
Of course, there is the trivial relation
$$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$
For example,
$$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$
for any $k > -1$, or
$$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$
As a potentially less trivial answer, you can perform any invertible integral transform
$$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$
Then, we can express the original function in terms of the inverse transform
$$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$
As an example, using the inverse Laplace transform,
$$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$
we can write the function
$$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$
answered Nov 21 at 20:46
AlexanderJ93
5,444622
Of course, there is the trivial relation
$$ frac{int_alpha^beta f(x)g(y)text dy}{int_alpha^beta g(y)text dy} = f(x) $$
For example,
$$ int_0^1 (k+1)y^kf(x)text dy = f(x) $$
for any $k > -1$, or
$$ int_{-infty}^infty frac{e^{-y^2/2}f(x)}{sqrt{2pi}}text dy = f(x) $$
As a potentially less trivial answer, you can perform any invertible integral transform
$$ F(y) = int_{x_1}^{x_2} K(x,y)f(x)text dx$$
Then, we can express the original function in terms of the inverse transform
$$ f(x) = int_{y_1}^{y_2} K^{-1}(y,x)F(y)text dy$$
As an example, using the inverse Laplace transform,
$$L_x^{-1}[f(x)](y) = frac{1}{2}((1-2y)sinh(y)+ycosh(y)) $$
we can write the function
$$ f(x) = L_y[F(y)](x) = int_0^infty frac{1}{2}((1-2y)sinh(y)+ycosh(y))exp(-xy)text dy $$
answered Nov 21 at 20:46
AlexanderJ93
5,444622
edited Nov 21 at 21:03
answered Nov 21 at 20:46
AlexanderJ93
5,444622
answered Nov 21 at 20:46
AlexanderJ93
5,444622
answered Nov 21 at 20:46
AlexanderJ93
5,444622
5,444622
Not helpful at all
– thedude
Nov 21 at 20:51
It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
– AlexanderJ93
Nov 21 at 20:52
@thedude I've updated the answer to include a less trivial solution
– AlexanderJ93
Nov 21 at 21:03
1
Yes, that is much better, thank you
– thedude
Nov 21 at 23:46
add a comment |
Not helpful at all
– thedude
Nov 21 at 20:51
It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
– AlexanderJ93
Nov 21 at 20:52
@thedude I've updated the answer to include a less trivial solution
– AlexanderJ93
Nov 21 at 21:03
1
Yes, that is much better, thank you
– thedude
Nov 21 at 23:46
Not helpful at all
– thedude
Nov 21 at 20:51
Not helpful at all
– thedude
Nov 21 at 20:51
It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
– AlexanderJ93
Nov 21 at 20:52
It's unclear what you're asking for. If this solution isn't valid, there should be some statement in your question that it is not.
– AlexanderJ93
Nov 21 at 20:52
@thedude I've updated the answer to include a less trivial solution
– AlexanderJ93
Nov 21 at 21:03
@thedude I've updated the answer to include a less trivial solution
– AlexanderJ93
Nov 21 at 21:03
1
1
Yes, that is much better, thank you
– thedude
Nov 21 at 23:46
Yes, that is much better, thank you
– thedude
Nov 21 at 23:46
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3008303%2fintegral-representation-of-a-rational-function%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
