Integrating a real nasty!Nasty Integral - Closed form solution?Integral evaluation $int_-infty^inftyfraccos (ax)pi (1+x^2)dx$Complex contour integralsMethod for integrating $int_-infty^infty x e^frac-x^22 dx$?Can this integral be evaluated/approximated?Integrating a Linear Operator $A:Hlongrightarrow H$ (Matrix)Is this way of integrating correct?Are there real definite integrals which can only be evaluated by contour integration?Real integrals in the complex planeHow can I use residues to solve this real integral?

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Integrating a real nasty!


Nasty Integral - Closed form solution?Integral evaluation $int_-infty^inftyfraccos (ax)pi (1+x^2)dx$Complex contour integralsMethod for integrating $int_-infty^infty x e^frac-x^22 dx$?Can this integral be evaluated/approximated?Integrating a Linear Operator $A:Hlongrightarrow H$ (Matrix)Is this way of integrating correct?Are there real definite integrals which can only be evaluated by contour integration?Real integrals in the complex planeHow can I use residues to solve this real integral?













1












$begingroup$


I came across this beast



$$int_0^pi ln(1-2a cosx +a^2)cosnx , dx $$



where $n=1,2,3,...$ and $a$ is an arbitrary real. Can this be down WITHOUT contour integration?










share|cite|improve this question









$endgroup$
















    1












    $begingroup$


    I came across this beast



    $$int_0^pi ln(1-2a cosx +a^2)cosnx , dx $$



    where $n=1,2,3,...$ and $a$ is an arbitrary real. Can this be down WITHOUT contour integration?










    share|cite|improve this question









    $endgroup$














      1












      1








      1


      1



      $begingroup$


      I came across this beast



      $$int_0^pi ln(1-2a cosx +a^2)cosnx , dx $$



      where $n=1,2,3,...$ and $a$ is an arbitrary real. Can this be down WITHOUT contour integration?










      share|cite|improve this question









      $endgroup$




      I came across this beast



      $$int_0^pi ln(1-2a cosx +a^2)cosnx , dx $$



      where $n=1,2,3,...$ and $a$ is an arbitrary real. Can this be down WITHOUT contour integration?







      integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 28 at 15:57









      keeran_q789keeran_q789

      184




      184




















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          For $|a|<1$, we have
          $$
          I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
          $$
          Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
          $$beginalign*
          I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
          endalign*$$
          which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
          $$
          I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
          $$
          to obtain
          $$
          I(a)=-fracpi a^-nn.
          $$






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Nice (+1). Where can one see a proof of that interesting Fourier series?
            $endgroup$
            – clathratus
            Mar 28 at 16:47










          • $begingroup$
            @clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
            $endgroup$
            – Nao
            Mar 28 at 16:51











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          For $|a|<1$, we have
          $$
          I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
          $$
          Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
          $$beginalign*
          I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
          endalign*$$
          which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
          $$
          I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
          $$
          to obtain
          $$
          I(a)=-fracpi a^-nn.
          $$






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Nice (+1). Where can one see a proof of that interesting Fourier series?
            $endgroup$
            – clathratus
            Mar 28 at 16:47










          • $begingroup$
            @clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
            $endgroup$
            – Nao
            Mar 28 at 16:51















          5












          $begingroup$

          For $|a|<1$, we have
          $$
          I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
          $$
          Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
          $$beginalign*
          I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
          endalign*$$
          which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
          $$
          I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
          $$
          to obtain
          $$
          I(a)=-fracpi a^-nn.
          $$






          share|cite|improve this answer









          $endgroup$








          • 1




            $begingroup$
            Nice (+1). Where can one see a proof of that interesting Fourier series?
            $endgroup$
            – clathratus
            Mar 28 at 16:47










          • $begingroup$
            @clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
            $endgroup$
            – Nao
            Mar 28 at 16:51













          5












          5








          5





          $begingroup$

          For $|a|<1$, we have
          $$
          I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
          $$
          Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
          $$beginalign*
          I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
          endalign*$$
          which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
          $$
          I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
          $$
          to obtain
          $$
          I(a)=-fracpi a^-nn.
          $$






          share|cite|improve this answer









          $endgroup$



          For $|a|<1$, we have
          $$
          I'(a)=2int_0^pi fraccos nx(a-cos x)1-2acos x+a^2dx=int_0^2pi fraccos nx(a-cos x)1-2acos x+a^2dx.
          $$
          Using the Fourier series $displaystylefrac11-a^2sum_k=-infty^infty a^e^ikx = frac11-2acos x+a^2$, we have for $nge 1$
          $$beginalign*
          I'(a)&=frac14(1-a^2)sum_kinBbb Z a^int_0^2pie^ikx(e^inx+e^-inx)(2a-e^ix-e^-ix) dx\&=fracpi2(1-a^2)sum_kinBbb Z a^left(2amathbf1_k-mathbf1_-mathbf1_kright)\&=fracpi2(1-a^2)left( 4a^n+1-2a^n+1-2a^n-1right)\&=-pi a^n-1,
          endalign*$$
          which implies that $displaystyle I(a)=-fracpi a^nn +I(0)=-fracpi a^nn$. For $|a|>1$, use
          $$
          I(a)=int_0^pi left(lnleft(1-2 fraccosxa +frac1a^2right)+2ln aright)cosnx dx
          $$
          to obtain
          $$
          I(a)=-fracpi a^-nn.
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 28 at 16:41









          NaoNao

          1836




          1836







          • 1




            $begingroup$
            Nice (+1). Where can one see a proof of that interesting Fourier series?
            $endgroup$
            – clathratus
            Mar 28 at 16:47










          • $begingroup$
            @clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
            $endgroup$
            – Nao
            Mar 28 at 16:51












          • 1




            $begingroup$
            Nice (+1). Where can one see a proof of that interesting Fourier series?
            $endgroup$
            – clathratus
            Mar 28 at 16:47










          • $begingroup$
            @clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
            $endgroup$
            – Nao
            Mar 28 at 16:51







          1




          1




          $begingroup$
          Nice (+1). Where can one see a proof of that interesting Fourier series?
          $endgroup$
          – clathratus
          Mar 28 at 16:47




          $begingroup$
          Nice (+1). Where can one see a proof of that interesting Fourier series?
          $endgroup$
          – clathratus
          Mar 28 at 16:47












          $begingroup$
          @clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
          $endgroup$
          – Nao
          Mar 28 at 16:51




          $begingroup$
          @clathratus Thank you :) In fact, it even has a famous name Poisson kernel in complex/harmonic analysis.
          $endgroup$
          – Nao
          Mar 28 at 16:51

















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