Definite Integral in functional relation format












7














If $$2f(x) + f(-x) = frac{1}{x}sinBiggl(x-frac{1}{x}Biggl)$$ then find the value of
$$int_{frac{1}{e}}^ef(x)dx$$(its not given that the function is odd or even)
I don't know where to start on this one I'm not much experienced in solving functional relations like these and I'm not even able to come up with an approach to solve it can someone help






calculus definite-integrals functional-equations







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    7














    If $$2f(x) + f(-x) = frac{1}{x}sinBiggl(x-frac{1}{x}Biggl)$$ then find the value of
    $$int_{frac{1}{e}}^ef(x)dx$$(its not given that the function is odd or even)
    I don't know where to start on this one I'm not much experienced in solving functional relations like these and I'm not even able to come up with an approach to solve it can someone help






    calculus definite-integrals functional-equations







    share|cite|improve this question



























      7












      7








      7


      1





      If $$2f(x) + f(-x) = frac{1}{x}sinBiggl(x-frac{1}{x}Biggl)$$ then find the value of
      $$int_{frac{1}{e}}^ef(x)dx$$(its not given that the function is odd or even)
      I don't know where to start on this one I'm not much experienced in solving functional relations like these and I'm not even able to come up with an approach to solve it can someone help






      calculus definite-integrals functional-equations







      share|cite|improve this question















      If $$2f(x) + f(-x) = frac{1}{x}sinBiggl(x-frac{1}{x}Biggl)$$ then find the value of
      $$int_{frac{1}{e}}^ef(x)dx$$(its not given that the function is odd or even)
      I don't know where to start on this one I'm not much experienced in solving functional relations like these and I'm not even able to come up with an approach to solve it can someone help






      calculus definite-integrals functional-equations




      calculus definite-integrals functional-equations






      share|cite|improve this question















      share|cite|improve this question













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      edited Dec 17 at 18:44









      Rebellos

      14.4k31245




      14.4k31245










      asked Dec 17 at 18:43









      aditya prakash

      423




      423






















          2 Answers
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          8














          $textbf{Hint:}$ $2f(x) + f(-x)=2f(-x) + f(x)implies f(-x)=f(x)$. So $$f(x)=dfrac{1}{3x}sinleft(x-dfrac{1}{x}right)$$
          $$fleft(dfrac1xright)=-x^2f(x) label{a}tag{1}$$



          $$displaystyleint_frac1e^ef(x)dx=int_frac1e^1f(x)dx:+:int_1^ef(x)dx$$



          For $1^{st}$ integral, $y=dfrac1x implies dfrac{-dy}{y^2}=dx$ with limits $dfrac1e=x=dfrac1yimplies y=e$



          to $1=x=dfrac1yimplies y=1$.



          $$implies int_frac1e^1f(x)dx=displaystyleint_e^1 fleft(dfrac1yright)dfrac{-dy}{y^2}=int_1^e fleft(dfrac1yright)dfrac{dy}{y^2}overset{(1)}{=}int_1^e -y^2f(y)dfrac{dy}{y^2}=-int_1^e f(y)dy$$



          Hence $$displaystyleint_frac1e^ef(x)dx=0$$






          share|cite|improve this answer























          • Substitute $x$ as $-x$ in your original equation. Assuming you know $sin(-x)=-sin x$.
            – Yadati Kiran
            Dec 17 at 18:58












          • Yeah I got it thanks attempting it again
            – aditya prakash
            Dec 17 at 19:00










          • Stuck with the integration now 😭
            – aditya prakash
            Dec 17 at 19:12



















          5














          The integral is zero.



          Proof:



          With $g(x) = frac{1}{x}sin(x-frac{1}{x})$ we have $2f(x)+f(-x) = g(x) = g(-x) = 2f(-x) + f(x)implies f(-x)=f(x)$ which gives



          $$f(x) = frac{1}{3 x}sin(x-frac{1}{x})$$



          Now the integral can be split up as



          $$i = int_{frac{1}{e}}^ef(x),dx =int_{frac{1}{e}}^1f(x),dx+int_1^ef(x),dx=i_1 + i_2$$



          Substituting $xto frac{1}{y}$ i.e. $dx to - frac{1}{y^2}dy$ in the second integral gives



          $$i_2=int_1^frac{1}{e}f(x),dx = int_{frac{1}{e}}^1 frac{1}{y^2}f(frac{1}{y}),dy=int_{frac{1}{e}}^1 frac{1}{y^2}frac{1}{3frac{1}{y}}sin(frac{1}{y}-y),dy\=int_{frac{1}{e}}^1 frac{1}{3 y}sin(frac{1}{y}-y),dy=-int_{frac{1}{e}}^1 frac{1}{3 y}sin(y-frac{1}{y}),dy=- int_{frac{1}{e}}^1f(y),dy= - i_1 $$



          Hence the integral in question is $i = i_1 +(- i_1) = 0$

          Q.E.D.



          Remark: from the proof it is clear that "$e$" can be any positive real number.






          share|cite|improve this answer























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            2 Answers
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            2 Answers
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            8














            $textbf{Hint:}$ $2f(x) + f(-x)=2f(-x) + f(x)implies f(-x)=f(x)$. So $$f(x)=dfrac{1}{3x}sinleft(x-dfrac{1}{x}right)$$
            $$fleft(dfrac1xright)=-x^2f(x) label{a}tag{1}$$



            $$displaystyleint_frac1e^ef(x)dx=int_frac1e^1f(x)dx:+:int_1^ef(x)dx$$



            For $1^{st}$ integral, $y=dfrac1x implies dfrac{-dy}{y^2}=dx$ with limits $dfrac1e=x=dfrac1yimplies y=e$



            to $1=x=dfrac1yimplies y=1$.



            $$implies int_frac1e^1f(x)dx=displaystyleint_e^1 fleft(dfrac1yright)dfrac{-dy}{y^2}=int_1^e fleft(dfrac1yright)dfrac{dy}{y^2}overset{(1)}{=}int_1^e -y^2f(y)dfrac{dy}{y^2}=-int_1^e f(y)dy$$



            Hence $$displaystyleint_frac1e^ef(x)dx=0$$






            share|cite|improve this answer























            • Substitute $x$ as $-x$ in your original equation. Assuming you know $sin(-x)=-sin x$.
              – Yadati Kiran
              Dec 17 at 18:58












            • Yeah I got it thanks attempting it again
              – aditya prakash
              Dec 17 at 19:00










            • Stuck with the integration now 😭
              – aditya prakash
              Dec 17 at 19:12
















            8














            $textbf{Hint:}$ $2f(x) + f(-x)=2f(-x) + f(x)implies f(-x)=f(x)$. So $$f(x)=dfrac{1}{3x}sinleft(x-dfrac{1}{x}right)$$
            $$fleft(dfrac1xright)=-x^2f(x) label{a}tag{1}$$



            $$displaystyleint_frac1e^ef(x)dx=int_frac1e^1f(x)dx:+:int_1^ef(x)dx$$



            For $1^{st}$ integral, $y=dfrac1x implies dfrac{-dy}{y^2}=dx$ with limits $dfrac1e=x=dfrac1yimplies y=e$



            to $1=x=dfrac1yimplies y=1$.



            $$implies int_frac1e^1f(x)dx=displaystyleint_e^1 fleft(dfrac1yright)dfrac{-dy}{y^2}=int_1^e fleft(dfrac1yright)dfrac{dy}{y^2}overset{(1)}{=}int_1^e -y^2f(y)dfrac{dy}{y^2}=-int_1^e f(y)dy$$



            Hence $$displaystyleint_frac1e^ef(x)dx=0$$






            share|cite|improve this answer























            • Substitute $x$ as $-x$ in your original equation. Assuming you know $sin(-x)=-sin x$.
              – Yadati Kiran
              Dec 17 at 18:58












            • Yeah I got it thanks attempting it again
              – aditya prakash
              Dec 17 at 19:00










            • Stuck with the integration now 😭
              – aditya prakash
              Dec 17 at 19:12














            8












            8








            8






            $textbf{Hint:}$ $2f(x) + f(-x)=2f(-x) + f(x)implies f(-x)=f(x)$. So $$f(x)=dfrac{1}{3x}sinleft(x-dfrac{1}{x}right)$$
            $$fleft(dfrac1xright)=-x^2f(x) label{a}tag{1}$$



            $$displaystyleint_frac1e^ef(x)dx=int_frac1e^1f(x)dx:+:int_1^ef(x)dx$$



            For $1^{st}$ integral, $y=dfrac1x implies dfrac{-dy}{y^2}=dx$ with limits $dfrac1e=x=dfrac1yimplies y=e$



            to $1=x=dfrac1yimplies y=1$.



            $$implies int_frac1e^1f(x)dx=displaystyleint_e^1 fleft(dfrac1yright)dfrac{-dy}{y^2}=int_1^e fleft(dfrac1yright)dfrac{dy}{y^2}overset{(1)}{=}int_1^e -y^2f(y)dfrac{dy}{y^2}=-int_1^e f(y)dy$$



            Hence $$displaystyleint_frac1e^ef(x)dx=0$$






            share|cite|improve this answer














            $textbf{Hint:}$ $2f(x) + f(-x)=2f(-x) + f(x)implies f(-x)=f(x)$. So $$f(x)=dfrac{1}{3x}sinleft(x-dfrac{1}{x}right)$$
            $$fleft(dfrac1xright)=-x^2f(x) label{a}tag{1}$$



            $$displaystyleint_frac1e^ef(x)dx=int_frac1e^1f(x)dx:+:int_1^ef(x)dx$$



            For $1^{st}$ integral, $y=dfrac1x implies dfrac{-dy}{y^2}=dx$ with limits $dfrac1e=x=dfrac1yimplies y=e$



            to $1=x=dfrac1yimplies y=1$.



            $$implies int_frac1e^1f(x)dx=displaystyleint_e^1 fleft(dfrac1yright)dfrac{-dy}{y^2}=int_1^e fleft(dfrac1yright)dfrac{dy}{y^2}overset{(1)}{=}int_1^e -y^2f(y)dfrac{dy}{y^2}=-int_1^e f(y)dy$$



            Hence $$displaystyleint_frac1e^ef(x)dx=0$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 17 at 20:17

























            answered Dec 17 at 18:46









            Yadati Kiran

            1,690619




            1,690619












            • Substitute $x$ as $-x$ in your original equation. Assuming you know $sin(-x)=-sin x$.
              – Yadati Kiran
              Dec 17 at 18:58












            • Yeah I got it thanks attempting it again
              – aditya prakash
              Dec 17 at 19:00










            • Stuck with the integration now 😭
              – aditya prakash
              Dec 17 at 19:12


















            • Substitute $x$ as $-x$ in your original equation. Assuming you know $sin(-x)=-sin x$.
              – Yadati Kiran
              Dec 17 at 18:58












            • Yeah I got it thanks attempting it again
              – aditya prakash
              Dec 17 at 19:00










            • Stuck with the integration now 😭
              – aditya prakash
              Dec 17 at 19:12
















            Substitute $x$ as $-x$ in your original equation. Assuming you know $sin(-x)=-sin x$.
            – Yadati Kiran
            Dec 17 at 18:58






            Substitute $x$ as $-x$ in your original equation. Assuming you know $sin(-x)=-sin x$.
            – Yadati Kiran
            Dec 17 at 18:58














            Yeah I got it thanks attempting it again
            – aditya prakash
            Dec 17 at 19:00




            Yeah I got it thanks attempting it again
            – aditya prakash
            Dec 17 at 19:00












            Stuck with the integration now 😭
            – aditya prakash
            Dec 17 at 19:12




            Stuck with the integration now 😭
            – aditya prakash
            Dec 17 at 19:12











            5














            The integral is zero.



            Proof:



            With $g(x) = frac{1}{x}sin(x-frac{1}{x})$ we have $2f(x)+f(-x) = g(x) = g(-x) = 2f(-x) + f(x)implies f(-x)=f(x)$ which gives



            $$f(x) = frac{1}{3 x}sin(x-frac{1}{x})$$



            Now the integral can be split up as



            $$i = int_{frac{1}{e}}^ef(x),dx =int_{frac{1}{e}}^1f(x),dx+int_1^ef(x),dx=i_1 + i_2$$



            Substituting $xto frac{1}{y}$ i.e. $dx to - frac{1}{y^2}dy$ in the second integral gives



            $$i_2=int_1^frac{1}{e}f(x),dx = int_{frac{1}{e}}^1 frac{1}{y^2}f(frac{1}{y}),dy=int_{frac{1}{e}}^1 frac{1}{y^2}frac{1}{3frac{1}{y}}sin(frac{1}{y}-y),dy\=int_{frac{1}{e}}^1 frac{1}{3 y}sin(frac{1}{y}-y),dy=-int_{frac{1}{e}}^1 frac{1}{3 y}sin(y-frac{1}{y}),dy=- int_{frac{1}{e}}^1f(y),dy= - i_1 $$



            Hence the integral in question is $i = i_1 +(- i_1) = 0$

            Q.E.D.



            Remark: from the proof it is clear that "$e$" can be any positive real number.






            share|cite|improve this answer




























              5














              The integral is zero.



              Proof:



              With $g(x) = frac{1}{x}sin(x-frac{1}{x})$ we have $2f(x)+f(-x) = g(x) = g(-x) = 2f(-x) + f(x)implies f(-x)=f(x)$ which gives



              $$f(x) = frac{1}{3 x}sin(x-frac{1}{x})$$



              Now the integral can be split up as



              $$i = int_{frac{1}{e}}^ef(x),dx =int_{frac{1}{e}}^1f(x),dx+int_1^ef(x),dx=i_1 + i_2$$



              Substituting $xto frac{1}{y}$ i.e. $dx to - frac{1}{y^2}dy$ in the second integral gives



              $$i_2=int_1^frac{1}{e}f(x),dx = int_{frac{1}{e}}^1 frac{1}{y^2}f(frac{1}{y}),dy=int_{frac{1}{e}}^1 frac{1}{y^2}frac{1}{3frac{1}{y}}sin(frac{1}{y}-y),dy\=int_{frac{1}{e}}^1 frac{1}{3 y}sin(frac{1}{y}-y),dy=-int_{frac{1}{e}}^1 frac{1}{3 y}sin(y-frac{1}{y}),dy=- int_{frac{1}{e}}^1f(y),dy= - i_1 $$



              Hence the integral in question is $i = i_1 +(- i_1) = 0$

              Q.E.D.



              Remark: from the proof it is clear that "$e$" can be any positive real number.






              share|cite|improve this answer


























                5












                5








                5






                The integral is zero.



                Proof:



                With $g(x) = frac{1}{x}sin(x-frac{1}{x})$ we have $2f(x)+f(-x) = g(x) = g(-x) = 2f(-x) + f(x)implies f(-x)=f(x)$ which gives



                $$f(x) = frac{1}{3 x}sin(x-frac{1}{x})$$



                Now the integral can be split up as



                $$i = int_{frac{1}{e}}^ef(x),dx =int_{frac{1}{e}}^1f(x),dx+int_1^ef(x),dx=i_1 + i_2$$



                Substituting $xto frac{1}{y}$ i.e. $dx to - frac{1}{y^2}dy$ in the second integral gives



                $$i_2=int_1^frac{1}{e}f(x),dx = int_{frac{1}{e}}^1 frac{1}{y^2}f(frac{1}{y}),dy=int_{frac{1}{e}}^1 frac{1}{y^2}frac{1}{3frac{1}{y}}sin(frac{1}{y}-y),dy\=int_{frac{1}{e}}^1 frac{1}{3 y}sin(frac{1}{y}-y),dy=-int_{frac{1}{e}}^1 frac{1}{3 y}sin(y-frac{1}{y}),dy=- int_{frac{1}{e}}^1f(y),dy= - i_1 $$



                Hence the integral in question is $i = i_1 +(- i_1) = 0$

                Q.E.D.



                Remark: from the proof it is clear that "$e$" can be any positive real number.






                share|cite|improve this answer














                The integral is zero.



                Proof:



                With $g(x) = frac{1}{x}sin(x-frac{1}{x})$ we have $2f(x)+f(-x) = g(x) = g(-x) = 2f(-x) + f(x)implies f(-x)=f(x)$ which gives



                $$f(x) = frac{1}{3 x}sin(x-frac{1}{x})$$



                Now the integral can be split up as



                $$i = int_{frac{1}{e}}^ef(x),dx =int_{frac{1}{e}}^1f(x),dx+int_1^ef(x),dx=i_1 + i_2$$



                Substituting $xto frac{1}{y}$ i.e. $dx to - frac{1}{y^2}dy$ in the second integral gives



                $$i_2=int_1^frac{1}{e}f(x),dx = int_{frac{1}{e}}^1 frac{1}{y^2}f(frac{1}{y}),dy=int_{frac{1}{e}}^1 frac{1}{y^2}frac{1}{3frac{1}{y}}sin(frac{1}{y}-y),dy\=int_{frac{1}{e}}^1 frac{1}{3 y}sin(frac{1}{y}-y),dy=-int_{frac{1}{e}}^1 frac{1}{3 y}sin(y-frac{1}{y}),dy=- int_{frac{1}{e}}^1f(y),dy= - i_1 $$



                Hence the integral in question is $i = i_1 +(- i_1) = 0$

                Q.E.D.



                Remark: from the proof it is clear that "$e$" can be any positive real number.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 18 at 13:05









                amWhy

                191k28224439




                191k28224439










                answered Dec 17 at 19:47









                Dr. Wolfgang Hintze

                3,150617




                3,150617






























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