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How can prove this integral

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How can prove this integral


How to calculate the derivative of this integral?how prove this integral inequality?Could anybody check this integral?Help! How to solve this integral?Can anyone help me with this improper integral?Any idea how to solve this integral?How can I integrate this? Improper Integral.Integral smaller than $frac{1}{2}epsilon$Stochastic Geometry : Obtaining an IntegralHow would one prove the existence of the following indefinite 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 6 hours ago









      Alan Muniz

      2,2711829




      2,2711829






      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 6 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
            6 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
            4 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





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            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
              6 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
              4 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
              6 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
              4 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 6 hours ago









            FredFred

            47k1848




            47k1848








            • 2




              $begingroup$
              I can't figure it out how you derive from (*) to answer
              $endgroup$
              – Reza
              6 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
              4 hours ago
















            • 2




              $begingroup$
              I can't figure it out how you derive from (*) to answer
              $endgroup$
              – Reza
              6 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
              4 hours ago










            2




            2




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




            $begingroup$
            I can't figure it out how you derive from (*) to answer
            $endgroup$
            – Reza
            6 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
            4 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
            4 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 5 hours ago

























                answered 6 hours ago









                Paras KhoslaParas Khosla

                1,385219




                1,385219






















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