definite integration - solution breakdown












0












$begingroup$


$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$



Unfortunately i cannot follow the steps. I guess integration by parts has been used?



How can Gamma density approach be applied here?






integration probability-distributions definite-integrals distribution-theory gamma-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
    $endgroup$
    – Luke Mathwalker
    Dec 12 '18 at 18:35


















0












$begingroup$


$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$



Unfortunately i cannot follow the steps. I guess integration by parts has been used?



How can Gamma density approach be applied here?






integration probability-distributions definite-integrals distribution-theory gamma-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
    $endgroup$
    – Luke Mathwalker
    Dec 12 '18 at 18:35
















0












0








0


1



$begingroup$


$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$



Unfortunately i cannot follow the steps. I guess integration by parts has been used?



How can Gamma density approach be applied here?






integration probability-distributions definite-integrals distribution-theory gamma-function







share|cite|improve this question











$endgroup$




$$ int_0^{infty} xe^{-x(y+1)}dx$$ $$=-frac x {y+1}e^{-x(y+1)}|_0^{infty} +frac 1 {y+1} int_0^{infty}e^{-x(y+1)}dx$$ $$=frac 1{(y+1)^{2}}.$$



Unfortunately i cannot follow the steps. I guess integration by parts has been used?



How can Gamma density approach be applied here?






integration probability-distributions definite-integrals distribution-theory gamma-function




integration probability-distributions definite-integrals distribution-theory gamma-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 18:48







user1607

















asked Dec 12 '18 at 18:25









user1607user1607

1718




1718












  • $begingroup$
    Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
    $endgroup$
    – Luke Mathwalker
    Dec 12 '18 at 18:35




















  • $begingroup$
    Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
    $endgroup$
    – Luke Mathwalker
    Dec 12 '18 at 18:35


















$begingroup$
Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
$endgroup$
– Luke Mathwalker
Dec 12 '18 at 18:35






$begingroup$
Yes. Integration by parts with $f(x):=x$ and $g'(x)=e^{-x(y+1)}$ works. ($g(x):=-frac{1}{y+1}e^{-x(y+1)}$
$endgroup$
– Luke Mathwalker
Dec 12 '18 at 18:35












1 Answer
1






active

oldest

votes


















0












$begingroup$

by parts, with $a=y+1$



$u=x$ and $v'=e^{-ax}$.



$$int_0^Xxe^{-ax}dx=$$



$$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



$$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



$$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



now take the limit when $Xto+infty$.






share|cite|improve this answer









$endgroup$













    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%2f3037054%2fdefinite-integration-solution-breakdown%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









    0












    $begingroup$

    by parts, with $a=y+1$



    $u=x$ and $v'=e^{-ax}$.



    $$int_0^Xxe^{-ax}dx=$$



    $$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



    $$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



    $$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



    now take the limit when $Xto+infty$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      by parts, with $a=y+1$



      $u=x$ and $v'=e^{-ax}$.



      $$int_0^Xxe^{-ax}dx=$$



      $$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



      $$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



      $$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



      now take the limit when $Xto+infty$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        by parts, with $a=y+1$



        $u=x$ and $v'=e^{-ax}$.



        $$int_0^Xxe^{-ax}dx=$$



        $$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



        $$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



        $$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



        now take the limit when $Xto+infty$.






        share|cite|improve this answer









        $endgroup$



        by parts, with $a=y+1$



        $u=x$ and $v'=e^{-ax}$.



        $$int_0^Xxe^{-ax}dx=$$



        $$Bigl[xfrac{-1}{a}e^{-ax}Bigr]_0^X-int_0^Xfrac{-1}{a}e^{-ax}dx$$



        $$frac{Xe^{-aX}}{a}+frac{1}{a}Bigl[frac{-1}{a}e^{-ax}Bigr]_0^X=$$



        $$frac{Xe^{-aX}}{a}+frac{1}{a^2}-frac{e^{-aX}}{a^2}$$



        now take the limit when $Xto+infty$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 18:44









        hamam_Abdallahhamam_Abdallah

        38.1k21634




        38.1k21634






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037054%2fdefinite-integration-solution-breakdown%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

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
            Read more

            “minimization” problem in Euclidean space related to orthonormal basis

            Image
            3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
            Read more