Integral with two different answers using real and complex analysis












3














The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
$$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






complex-analysis definite-integrals cauchy-integral-formula







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    The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



    Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
    where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
    $$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



    Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






    complex-analysis definite-integrals cauchy-integral-formula







    share|cite|improve this question



























      3












      3








      3


      1





      The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



      Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
      where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
      $$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



      Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






      complex-analysis definite-integrals cauchy-integral-formula







      share|cite|improve this question















      The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



      Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
      where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
      $$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



      Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






      complex-analysis definite-integrals cauchy-integral-formula




      complex-analysis definite-integrals cauchy-integral-formula






      share|cite|improve this question















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      edited 19 mins ago









      Saad

      19.7k92252




      19.7k92252










      asked 1 hour ago









      khaled014z

      787




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          2 Answers
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          The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



          This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



          Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



          $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
          = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






          share|cite|improve this answer























          • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
            – khaled014z
            1 hour ago










          • @khaled014z See the edit. Let me know if you want more details.
            – N. S.
            1 hour ago












          • Brilliant, that was kind of a tricky substitution, thank you
            – khaled014z
            1 hour ago



















          1














          Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



          Instead, we have



          $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






          share|cite|improve this answer





















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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            5














            The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



            This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



            Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



            $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
            = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






            share|cite|improve this answer























            • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
              – khaled014z
              1 hour ago










            • @khaled014z See the edit. Let me know if you want more details.
              – N. S.
              1 hour ago












            • Brilliant, that was kind of a tricky substitution, thank you
              – khaled014z
              1 hour ago
















            5














            The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



            This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



            Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



            $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
            = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






            share|cite|improve this answer























            • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
              – khaled014z
              1 hour ago










            • @khaled014z See the edit. Let me know if you want more details.
              – N. S.
              1 hour ago












            • Brilliant, that was kind of a tricky substitution, thank you
              – khaled014z
              1 hour ago














            5












            5








            5






            The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



            This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



            Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



            $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
            = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






            share|cite|improve this answer














            The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



            This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



            Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



            $$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
            = lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 46 mins ago









            Saad

            19.7k92252




            19.7k92252










            answered 1 hour ago









            N. S.

            102k5109204




            102k5109204












            • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
              – khaled014z
              1 hour ago










            • @khaled014z See the edit. Let me know if you want more details.
              – N. S.
              1 hour ago












            • Brilliant, that was kind of a tricky substitution, thank you
              – khaled014z
              1 hour ago


















            • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
              – khaled014z
              1 hour ago










            • @khaled014z See the edit. Let me know if you want more details.
              – N. S.
              1 hour ago












            • Brilliant, that was kind of a tricky substitution, thank you
              – khaled014z
              1 hour ago
















            Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
            – khaled014z
            1 hour ago




            Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
            – khaled014z
            1 hour ago












            @khaled014z See the edit. Let me know if you want more details.
            – N. S.
            1 hour ago






            @khaled014z See the edit. Let me know if you want more details.
            – N. S.
            1 hour ago














            Brilliant, that was kind of a tricky substitution, thank you
            – khaled014z
            1 hour ago




            Brilliant, that was kind of a tricky substitution, thank you
            – khaled014z
            1 hour ago











            1














            Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



            Instead, we have



            $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






            share|cite|improve this answer


























              1














              Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



              Instead, we have



              $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






              share|cite|improve this answer
























                1












                1








                1






                Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



                Instead, we have



                $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






                share|cite|improve this answer












                Note that that tangent function, $tan(x)$, is discontinuous when $x=npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



                Instead, we have



                $$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Mark Viola

                130k1274170




                130k1274170






























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