Help:For what $ alpha $ do the integrals exist?












2














Let
$ K_R (y) subset mathbb{R}^n $
be an n-dimensional Ball with radius $R>0$ and midpoint $y in mathbb{R}^n $



Far what $ alpha in mathbb{R} $ and $ n in mathbb{N}{0} $ do following integrals exist:



$ int_{K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n setminus K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n } |x-y|^{alpha}dx $



with what criteria can i solve this task?
I don't know how to begin.
Thanks for any help!!



Let's take
$ int_{K{_R(y)}} |x-y|^{alpha}dx $ ,$y=0, n=2 $



If I replace by polarcoordinates $ x= rsin phi $ :
$ int_0^{2 pi} int_0^R | r|^{alpha} r dr dphi $..is this the right way?






real-analysis integration multivariable-calculus







share|cite|improve this question
























  • Take $y=0$ and $n=2$ to start. You can use polar coordinates.
    – zhw.
    Nov 28 at 18:18
















2














Let
$ K_R (y) subset mathbb{R}^n $
be an n-dimensional Ball with radius $R>0$ and midpoint $y in mathbb{R}^n $



Far what $ alpha in mathbb{R} $ and $ n in mathbb{N}{0} $ do following integrals exist:



$ int_{K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n setminus K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n } |x-y|^{alpha}dx $



with what criteria can i solve this task?
I don't know how to begin.
Thanks for any help!!



Let's take
$ int_{K{_R(y)}} |x-y|^{alpha}dx $ ,$y=0, n=2 $



If I replace by polarcoordinates $ x= rsin phi $ :
$ int_0^{2 pi} int_0^R | r|^{alpha} r dr dphi $..is this the right way?






real-analysis integration multivariable-calculus







share|cite|improve this question
























  • Take $y=0$ and $n=2$ to start. You can use polar coordinates.
    – zhw.
    Nov 28 at 18:18














2












2








2







Let
$ K_R (y) subset mathbb{R}^n $
be an n-dimensional Ball with radius $R>0$ and midpoint $y in mathbb{R}^n $



Far what $ alpha in mathbb{R} $ and $ n in mathbb{N}{0} $ do following integrals exist:



$ int_{K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n setminus K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n } |x-y|^{alpha}dx $



with what criteria can i solve this task?
I don't know how to begin.
Thanks for any help!!



Let's take
$ int_{K{_R(y)}} |x-y|^{alpha}dx $ ,$y=0, n=2 $



If I replace by polarcoordinates $ x= rsin phi $ :
$ int_0^{2 pi} int_0^R | r|^{alpha} r dr dphi $..is this the right way?






real-analysis integration multivariable-calculus







share|cite|improve this question















Let
$ K_R (y) subset mathbb{R}^n $
be an n-dimensional Ball with radius $R>0$ and midpoint $y in mathbb{R}^n $



Far what $ alpha in mathbb{R} $ and $ n in mathbb{N}{0} $ do following integrals exist:



$ int_{K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n setminus K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n } |x-y|^{alpha}dx $



with what criteria can i solve this task?
I don't know how to begin.
Thanks for any help!!



Let's take
$ int_{K{_R(y)}} |x-y|^{alpha}dx $ ,$y=0, n=2 $



If I replace by polarcoordinates $ x= rsin phi $ :
$ int_0^{2 pi} int_0^R | r|^{alpha} r dr dphi $..is this the right way?






real-analysis integration multivariable-calculus




real-analysis integration multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 21:44

























asked Nov 28 at 18:16









constant94

1415




1415












  • Take $y=0$ and $n=2$ to start. You can use polar coordinates.
    – zhw.
    Nov 28 at 18:18


















  • Take $y=0$ and $n=2$ to start. You can use polar coordinates.
    – zhw.
    Nov 28 at 18:18
















Take $y=0$ and $n=2$ to start. You can use polar coordinates.
– zhw.
Nov 28 at 18:18




Take $y=0$ and $n=2$ to start. You can use polar coordinates.
– zhw.
Nov 28 at 18:18










1 Answer
1






active

oldest

votes


















1














Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



Your integrals, respectively, are
$$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
$$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
$$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



The remaining improper integrals are straight forward to work out.






share|cite|improve this answer





















    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017493%2fhelpfor-what-alpha-do-the-integrals-exist%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









    1














    Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



    Your integrals, respectively, are
    $$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
    $$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
    $$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
    where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



    The remaining improper integrals are straight forward to work out.






    share|cite|improve this answer


























      1














      Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



      Your integrals, respectively, are
      $$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
      $$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
      $$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
      where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



      The remaining improper integrals are straight forward to work out.






      share|cite|improve this answer
























        1












        1








        1






        Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



        Your integrals, respectively, are
        $$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
        $$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
        $$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
        where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



        The remaining improper integrals are straight forward to work out.






        share|cite|improve this answer












        Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



        Your integrals, respectively, are
        $$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
        $$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
        $$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
        where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



        The remaining improper integrals are straight forward to work out.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 at 21:52









        Umberto P.

        38.3k13063




        38.3k13063






























            draft saved

            draft discarded




















































            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3017493%2fhelpfor-what-alpha-do-the-integrals-exist%23new-answer', 'question_page');
            }
            );

            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







            -

            Popular posts from this blog

            Wiesbaden

            Image
            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
            Read more

            Marschland

            Image
            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more