How can prove this integral












4












$begingroup$


I was reading a textbook which these two equations posed . The second one was the result of the first one .
How can we say that ?



If we know :
$$int_{0^+}^{+infty} frac{sin(x)}{x} = frac{pi}{2}$$



How can we prove :
$$int_{0^+}^{+infty} left(frac{sin(x)}{x}right)^2 = frac{pi}{2}$$



Thanks in advance






real-analysis calculus integration







share|cite|improve this question









New contributor




Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    4












    $begingroup$


    I was reading a textbook which these two equations posed . The second one was the result of the first one .
    How can we say that ?



    If we know :
    $$int_{0^+}^{+infty} frac{sin(x)}{x} = frac{pi}{2}$$



    How can we prove :
    $$int_{0^+}^{+infty} left(frac{sin(x)}{x}right)^2 = frac{pi}{2}$$



    Thanks in advance






    real-analysis calculus integration







    share|cite|improve this question









    New contributor




    Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4


      1



      $begingroup$


      I was reading a textbook which these two equations posed . The second one was the result of the first one .
      How can we say that ?



      If we know :
      $$int_{0^+}^{+infty} frac{sin(x)}{x} = frac{pi}{2}$$



      How can we prove :
      $$int_{0^+}^{+infty} left(frac{sin(x)}{x}right)^2 = frac{pi}{2}$$



      Thanks in advance






      real-analysis calculus integration







      share|cite|improve this question









      New contributor




      Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I was reading a textbook which these two equations posed . The second one was the result of the first one .
      How can we say that ?



      If we know :
      $$int_{0^+}^{+infty} frac{sin(x)}{x} = frac{pi}{2}$$



      How can we prove :
      $$int_{0^+}^{+infty} left(frac{sin(x)}{x}right)^2 = frac{pi}{2}$$



      Thanks in advance






      real-analysis calculus integration




      real-analysis calculus integration






      share|cite|improve this question









      New contributor




      Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 16 hours ago









      Alan Muniz

      2,5711830




      2,5711830






      New contributor




      Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 16 hours ago









      RezaReza

      233




      233




      New contributor




      Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Reza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          6












          $begingroup$

          Use $int_{0}^{infty} frac{sin(x)}{x} = frac{pi}{2}$ and $sin (2x)= 2 sin(x) cos(x)$ to get



          $$ (*) quadint_{0}^{infty} frac{sin(x) cos (x)}{x} =frac{pi}{4}.$$



          Then use integration by parts in $(*)$ to derive



          $$int_{0}^{infty} frac{sin^2(x)}{x^2} = frac{pi}{2}.$$






          share|cite|improve this answer









          $endgroup$









          • 2




            $begingroup$
            I can't figure it out how you derive from (*) to answer
            $endgroup$
            – Reza
            16 hours ago










          • $begingroup$
            Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
            $endgroup$
            – FDP
            14 hours ago





















          2












          $begingroup$

          $$I(a)=int_{-infty}^{+infty}dfrac{sin^2ax}{x^2}mathrm dx\ dfrac{mathrm dI}{mathrm da}=int_{-infty}^{+infty}partial_a dfrac{sin^2ax}{x^2}mathrm dx=2int_{0}^{infty}dfrac{sin 2ax}{x}mathrm dx=pi\ I(a)=pi a implies int_{0}^{infty}dfrac{sin^2x}{x^2}mathrm dx =dfrac{1}{2}I(1)=dfrac{pi}{2}$$






          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
            });


            }
            });






            Reza is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3128751%2fhow-can-prove-this-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









            6












            $begingroup$

            Use $int_{0}^{infty} frac{sin(x)}{x} = frac{pi}{2}$ and $sin (2x)= 2 sin(x) cos(x)$ to get



            $$ (*) quadint_{0}^{infty} frac{sin(x) cos (x)}{x} =frac{pi}{4}.$$



            Then use integration by parts in $(*)$ to derive



            $$int_{0}^{infty} frac{sin^2(x)}{x^2} = frac{pi}{2}.$$






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              I can't figure it out how you derive from (*) to answer
              $endgroup$
              – Reza
              16 hours ago










            • $begingroup$
              Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
              $endgroup$
              – FDP
              14 hours ago


















            6












            $begingroup$

            Use $int_{0}^{infty} frac{sin(x)}{x} = frac{pi}{2}$ and $sin (2x)= 2 sin(x) cos(x)$ to get



            $$ (*) quadint_{0}^{infty} frac{sin(x) cos (x)}{x} =frac{pi}{4}.$$



            Then use integration by parts in $(*)$ to derive



            $$int_{0}^{infty} frac{sin^2(x)}{x^2} = frac{pi}{2}.$$






            share|cite|improve this answer









            $endgroup$









            • 2




              $begingroup$
              I can't figure it out how you derive from (*) to answer
              $endgroup$
              – Reza
              16 hours ago










            • $begingroup$
              Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
              $endgroup$
              – FDP
              14 hours ago
















            6












            6








            6





            $begingroup$

            Use $int_{0}^{infty} frac{sin(x)}{x} = frac{pi}{2}$ and $sin (2x)= 2 sin(x) cos(x)$ to get



            $$ (*) quadint_{0}^{infty} frac{sin(x) cos (x)}{x} =frac{pi}{4}.$$



            Then use integration by parts in $(*)$ to derive



            $$int_{0}^{infty} frac{sin^2(x)}{x^2} = frac{pi}{2}.$$






            share|cite|improve this answer









            $endgroup$



            Use $int_{0}^{infty} frac{sin(x)}{x} = frac{pi}{2}$ and $sin (2x)= 2 sin(x) cos(x)$ to get



            $$ (*) quadint_{0}^{infty} frac{sin(x) cos (x)}{x} =frac{pi}{4}.$$



            Then use integration by parts in $(*)$ to derive



            $$int_{0}^{infty} frac{sin^2(x)}{x^2} = frac{pi}{2}.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 16 hours ago









            FredFred

            47k1848




            47k1848








            • 2




              $begingroup$
              I can't figure it out how you derive from (*) to answer
              $endgroup$
              – Reza
              16 hours ago










            • $begingroup$
              Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
              $endgroup$
              – FDP
              14 hours ago
















            • 2




              $begingroup$
              I can't figure it out how you derive from (*) to answer
              $endgroup$
              – Reza
              16 hours ago










            • $begingroup$
              Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
              $endgroup$
              – FDP
              14 hours ago










            2




            2




            $begingroup$
            I can't figure it out how you derive from (*) to answer
            $endgroup$
            – Reza
            16 hours ago




            $begingroup$
            I can't figure it out how you derive from (*) to answer
            $endgroup$
            – Reza
            16 hours ago












            $begingroup$
            Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
            $endgroup$
            – FDP
            14 hours ago






            $begingroup$
            Reza: What is an antiderivative of the function cos? Can you derive the function defined for $x> 0$ by $f(x)=frac{sin x}{x}$ (product of functions)?
            $endgroup$
            – FDP
            14 hours ago













            2












            $begingroup$

            $$I(a)=int_{-infty}^{+infty}dfrac{sin^2ax}{x^2}mathrm dx\ dfrac{mathrm dI}{mathrm da}=int_{-infty}^{+infty}partial_a dfrac{sin^2ax}{x^2}mathrm dx=2int_{0}^{infty}dfrac{sin 2ax}{x}mathrm dx=pi\ I(a)=pi a implies int_{0}^{infty}dfrac{sin^2x}{x^2}mathrm dx =dfrac{1}{2}I(1)=dfrac{pi}{2}$$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              $$I(a)=int_{-infty}^{+infty}dfrac{sin^2ax}{x^2}mathrm dx\ dfrac{mathrm dI}{mathrm da}=int_{-infty}^{+infty}partial_a dfrac{sin^2ax}{x^2}mathrm dx=2int_{0}^{infty}dfrac{sin 2ax}{x}mathrm dx=pi\ I(a)=pi a implies int_{0}^{infty}dfrac{sin^2x}{x^2}mathrm dx =dfrac{1}{2}I(1)=dfrac{pi}{2}$$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                $$I(a)=int_{-infty}^{+infty}dfrac{sin^2ax}{x^2}mathrm dx\ dfrac{mathrm dI}{mathrm da}=int_{-infty}^{+infty}partial_a dfrac{sin^2ax}{x^2}mathrm dx=2int_{0}^{infty}dfrac{sin 2ax}{x}mathrm dx=pi\ I(a)=pi a implies int_{0}^{infty}dfrac{sin^2x}{x^2}mathrm dx =dfrac{1}{2}I(1)=dfrac{pi}{2}$$






                share|cite|improve this answer











                $endgroup$



                $$I(a)=int_{-infty}^{+infty}dfrac{sin^2ax}{x^2}mathrm dx\ dfrac{mathrm dI}{mathrm da}=int_{-infty}^{+infty}partial_a dfrac{sin^2ax}{x^2}mathrm dx=2int_{0}^{infty}dfrac{sin 2ax}{x}mathrm dx=pi\ I(a)=pi a implies int_{0}^{infty}dfrac{sin^2x}{x^2}mathrm dx =dfrac{1}{2}I(1)=dfrac{pi}{2}$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 15 hours ago

























                answered 16 hours ago









                Paras KhoslaParas Khosla

                1,410219




                1,410219






















                    Reza is a new contributor. Be nice, and check out our Code of Conduct.










                    draft saved

                    draft discarded


















                    Reza is a new contributor. Be nice, and check out our Code of Conduct.













                    Reza is a new contributor. Be nice, and check out our Code of Conduct.












                    Reza is a new contributor. Be nice, and check out our Code of Conduct.
















                    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%2f3128751%2fhow-can-prove-this-integral%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

                    "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness. ...
                    Read more

                    Alcedinidae

                    Image
                    Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout ...
                    Read more

                    Origin of the phrase “under your belt”?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 17 1 Today, I found myself discussing what students should have "under their belts" during a lecture, and I wondered to myself if there was some inappropriate undertone here that I might be unaware of. Question : What is the origin of the phrase "under your belt"? A google search revealed some webpages (e.g. [1]) that assert its origin is related to consumption of food -- once a meal has been eaten, it's under your belt (which is a relief). However, with such websites, I have no way to determine fact from "random guy on the internet making stuff up". etymology idioms ...
                    Read more