Double Integral unequal after switching $dy$ and $dx$. Exact reasoning required
$begingroup$
I have worked out $int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(x)) dlambda (y)=-frac{pi}{4}$ and $int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(y)) dlambda (x)=frac{pi}{4}$
What should my exact reasoning be as to why the double integral does not exist according to $lambda^2$?
Should I argue along the lines of $int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)$ not being well-defined as
$int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)=int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(x)) dlambda (y)=-frac{pi}{4}$
and
$int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)=int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(y)) dlambda (x)=frac{pi}{4}
$
But I am not sure on this argument.
real-analysis integration measure-theory multivariable-calculus
asked Dec 18 '18 at 11:23
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
I have worked out $int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(x)) dlambda (y)=-frac{pi}{4}$ and $int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(y)) dlambda (x)=frac{pi}{4}$
What should my exact reasoning be as to why the double integral does not exist according to $lambda^2$?
Should I argue along the lines of $int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)$ not being well-defined as
$int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)=int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(x)) dlambda (y)=-frac{pi}{4}$
and
$int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)=int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(y)) dlambda (x)=frac{pi}{4}
$
But I am not sure on this argument.
real-analysis integration measure-theory multivariable-calculus
asked Dec 18 '18 at 11:23
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
I have worked out $int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(x)) dlambda (y)=-frac{pi}{4}$ and $int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(y)) dlambda (x)=frac{pi}{4}$
What should my exact reasoning be as to why the double integral does not exist according to $lambda^2$?
Should I argue along the lines of $int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)$ not being well-defined as
$int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)=int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(x)) dlambda (y)=-frac{pi}{4}$
and
$int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)=int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(y)) dlambda (x)=frac{pi}{4}
$
But I am not sure on this argument.
real-analysis integration measure-theory multivariable-calculus
asked Dec 18 '18 at 11:23
SABOYSABOY
656311
$endgroup$
I have worked out $int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(x)) dlambda (y)=-frac{pi}{4}$ and $int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(y)) dlambda (x)=frac{pi}{4}$
What should my exact reasoning be as to why the double integral does not exist according to $lambda^2$?
Should I argue along the lines of $int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)$ not being well-defined as
$int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)=int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(x)) dlambda (y)=-frac{pi}{4}$
and
$int_{[0,1]times[0,1]}frac{x^2-y^2}{(x^2+y^2)^2}dlambda^2(x,y)=int_{[0,1]}(int_{[0,1]}frac{x^2-y^2}{(x^2+y^2)^2} dlambda(y)) dlambda (x)=frac{pi}{4}
$
But I am not sure on this argument.
real-analysis integration measure-theory multivariable-calculus
real-analysis integration measure-theory multivariable-calculus
asked Dec 18 '18 at 11:23
SABOYSABOY
656311
asked Dec 18 '18 at 11:23
SABOYSABOY
656311
asked Dec 18 '18 at 11:23
SABOYSABOY
656311
asked Dec 18 '18 at 11:23
SABOYSABOY
656311
asked Dec 18 '18 at 11:23
SABOYSABOY
656311
656311
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It's a direct application of Fubini: $fnotin mathcal L^1(Itimes I)$ because the iterated integrals are not equal. You can show this directly by using polar coordinates and noting that, if $E=left {(x,y):) 0le yle frac{1}{sqrt 3}xright },$ we have $ |f(x,y)|=frac{cos 2t}{r^{2}}ge frac{1}{2r^{2}}$ so $int|f|=int_E |f|+int_{E^c}|f|$ and since $int_E |f|=infty, fnotin L^1(Itimes I)$.
answered Dec 18 '18 at 16:31
MatematletaMatematleta
11.5k2920
$endgroup$
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%2f3045043%2fdouble-integral-unequal-after-switching-dy-and-dx-exact-reasoning-required%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's a direct application of Fubini: $fnotin mathcal L^1(Itimes I)$ because the iterated integrals are not equal. You can show this directly by using polar coordinates and noting that, if $E=left {(x,y):) 0le yle frac{1}{sqrt 3}xright },$ we have $ |f(x,y)|=frac{cos 2t}{r^{2}}ge frac{1}{2r^{2}}$ so $int|f|=int_E |f|+int_{E^c}|f|$ and since $int_E |f|=infty, fnotin L^1(Itimes I)$.
answered Dec 18 '18 at 16:31
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
It's a direct application of Fubini: $fnotin mathcal L^1(Itimes I)$ because the iterated integrals are not equal. You can show this directly by using polar coordinates and noting that, if $E=left {(x,y):) 0le yle frac{1}{sqrt 3}xright },$ we have $ |f(x,y)|=frac{cos 2t}{r^{2}}ge frac{1}{2r^{2}}$ so $int|f|=int_E |f|+int_{E^c}|f|$ and since $int_E |f|=infty, fnotin L^1(Itimes I)$.
answered Dec 18 '18 at 16:31
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
It's a direct application of Fubini: $fnotin mathcal L^1(Itimes I)$ because the iterated integrals are not equal. You can show this directly by using polar coordinates and noting that, if $E=left {(x,y):) 0le yle frac{1}{sqrt 3}xright },$ we have $ |f(x,y)|=frac{cos 2t}{r^{2}}ge frac{1}{2r^{2}}$ so $int|f|=int_E |f|+int_{E^c}|f|$ and since $int_E |f|=infty, fnotin L^1(Itimes I)$.
answered Dec 18 '18 at 16:31
MatematletaMatematleta
11.5k2920
$endgroup$
It's a direct application of Fubini: $fnotin mathcal L^1(Itimes I)$ because the iterated integrals are not equal. You can show this directly by using polar coordinates and noting that, if $E=left {(x,y):) 0le yle frac{1}{sqrt 3}xright },$ we have $ |f(x,y)|=frac{cos 2t}{r^{2}}ge frac{1}{2r^{2}}$ so $int|f|=int_E |f|+int_{E^c}|f|$ and since $int_E |f|=infty, fnotin L^1(Itimes I)$.
answered Dec 18 '18 at 16:31
MatematletaMatematleta
11.5k2920
answered Dec 18 '18 at 16:31
MatematletaMatematleta
11.5k2920
answered Dec 18 '18 at 16:31
MatematletaMatematleta
11.5k2920
answered Dec 18 '18 at 16:31
MatematletaMatematleta
11.5k2920
11.5k2920
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%2f3045043%2fdouble-integral-unequal-after-switching-dy-and-dx-exact-reasoning-required%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
