Tricky real integral: $int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi$











up vote
8
down vote

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7












I'm trying to prove the following:
$$ int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi $$



Numerical analysis agrees with this to very high accuracy, so I'm almost sure it's true. Mathematica gives this answer after thinking for a long, but gives an insane antiderivative in terms of exponential integrals. I'd like to evaluate the integral with purely real methods (I've never done complex analysis), as elegantly as possible.



How can I tackle this integral?






calculus integration definite-integrals







share|cite|improve this question




















  • 2




    Have you heard of Cauchy integral theorem?
    – Frank W.
    Dec 1 at 2:35










  • It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
    – user170231
    Dec 1 at 2:41












  • $x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
    – herb steinberg
    Dec 1 at 2:48






  • 1




    You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
    – J.G.
    Dec 1 at 9:20















up vote
8
down vote

favorite
7












I'm trying to prove the following:
$$ int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi $$



Numerical analysis agrees with this to very high accuracy, so I'm almost sure it's true. Mathematica gives this answer after thinking for a long, but gives an insane antiderivative in terms of exponential integrals. I'd like to evaluate the integral with purely real methods (I've never done complex analysis), as elegantly as possible.



How can I tackle this integral?






calculus integration definite-integrals







share|cite|improve this question




















  • 2




    Have you heard of Cauchy integral theorem?
    – Frank W.
    Dec 1 at 2:35










  • It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
    – user170231
    Dec 1 at 2:41












  • $x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
    – herb steinberg
    Dec 1 at 2:48






  • 1




    You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
    – J.G.
    Dec 1 at 9:20













up vote
8
down vote

favorite
7









up vote
8
down vote

favorite
7






7





I'm trying to prove the following:
$$ int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi $$



Numerical analysis agrees with this to very high accuracy, so I'm almost sure it's true. Mathematica gives this answer after thinking for a long, but gives an insane antiderivative in terms of exponential integrals. I'd like to evaluate the integral with purely real methods (I've never done complex analysis), as elegantly as possible.



How can I tackle this integral?






calculus integration definite-integrals







share|cite|improve this question















I'm trying to prove the following:
$$ int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi $$



Numerical analysis agrees with this to very high accuracy, so I'm almost sure it's true. Mathematica gives this answer after thinking for a long, but gives an insane antiderivative in terms of exponential integrals. I'd like to evaluate the integral with purely real methods (I've never done complex analysis), as elegantly as possible.



How can I tackle this integral?






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 at 9:34









Martin Sleziak

44.6k7115269




44.6k7115269










asked Dec 1 at 2:15









NMister

376110




376110








  • 2




    Have you heard of Cauchy integral theorem?
    – Frank W.
    Dec 1 at 2:35










  • It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
    – user170231
    Dec 1 at 2:41












  • $x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
    – herb steinberg
    Dec 1 at 2:48






  • 1




    You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
    – J.G.
    Dec 1 at 9:20














  • 2




    Have you heard of Cauchy integral theorem?
    – Frank W.
    Dec 1 at 2:35










  • It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
    – user170231
    Dec 1 at 2:41












  • $x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
    – herb steinberg
    Dec 1 at 2:48






  • 1




    You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
    – J.G.
    Dec 1 at 9:20








2




2




Have you heard of Cauchy integral theorem?
– Frank W.
Dec 1 at 2:35




Have you heard of Cauchy integral theorem?
– Frank W.
Dec 1 at 2:35












It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
– user170231
Dec 1 at 2:41






It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
– user170231
Dec 1 at 2:41














$x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
– herb steinberg
Dec 1 at 2:48




$x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
– herb steinberg
Dec 1 at 2:48




1




1




You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
– J.G.
Dec 1 at 9:20




You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
– J.G.
Dec 1 at 9:20










3 Answers
3






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up vote
22
down vote













Write



$$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$



Then $I(0) = 2pi$, and for $alpha > 0$,



begin{align*}
I'(alpha)
&= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
&= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
&= 0.
end{align*}



So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.




A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by



$$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$



Then



$$ I'(r)
= int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
= left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
= 0 $$



and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.






share|cite|improve this answer






























    up vote
    5
    down vote













    Assuming that you could enjoy special functions.



    Consider
    $$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$



    $$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
    $$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
    $$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
    $$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
    t}right)right)$$
    $$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.






    share|cite|improve this answer






























      up vote
      0
      down vote













      Put $z=e^{it}$ and using the formulas:
      $$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$



      $$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$



      apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        22
        down vote













        Write



        $$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$



        Then $I(0) = 2pi$, and for $alpha > 0$,



        begin{align*}
        I'(alpha)
        &= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
        &= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
        &= 0.
        end{align*}



        So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.




        A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by



        $$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$



        Then



        $$ I'(r)
        = int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
        = left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
        = 0 $$



        and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.






        share|cite|improve this answer



























          up vote
          22
          down vote













          Write



          $$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$



          Then $I(0) = 2pi$, and for $alpha > 0$,



          begin{align*}
          I'(alpha)
          &= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
          &= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
          &= 0.
          end{align*}



          So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.




          A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by



          $$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$



          Then



          $$ I'(r)
          = int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
          = left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
          = 0 $$



          and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.






          share|cite|improve this answer

























            up vote
            22
            down vote










            up vote
            22
            down vote









            Write



            $$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$



            Then $I(0) = 2pi$, and for $alpha > 0$,



            begin{align*}
            I'(alpha)
            &= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
            &= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
            &= 0.
            end{align*}



            So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.




            A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by



            $$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$



            Then



            $$ I'(r)
            = int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
            = left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
            = 0 $$



            and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.






            share|cite|improve this answer














            Write



            $$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$



            Then $I(0) = 2pi$, and for $alpha > 0$,



            begin{align*}
            I'(alpha)
            &= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
            &= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
            &= 0.
            end{align*}



            So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.




            A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by



            $$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$



            Then



            $$ I'(r)
            = int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
            = left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
            = 0 $$



            and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 1 at 9:13









            Mutantoe

            558411




            558411










            answered Dec 1 at 2:42









            Sangchul Lee

            91k12163263




            91k12163263






















                up vote
                5
                down vote













                Assuming that you could enjoy special functions.



                Consider
                $$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$



                $$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
                $$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
                $$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
                $$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
                t}right)right)$$
                $$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.






                share|cite|improve this answer



























                  up vote
                  5
                  down vote













                  Assuming that you could enjoy special functions.



                  Consider
                  $$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$



                  $$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
                  $$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
                  $$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
                  $$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
                  t}right)right)$$
                  $$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.






                  share|cite|improve this answer

























                    up vote
                    5
                    down vote










                    up vote
                    5
                    down vote









                    Assuming that you could enjoy special functions.



                    Consider
                    $$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$



                    $$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
                    $$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
                    $$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
                    $$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
                    t}right)right)$$
                    $$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.






                    share|cite|improve this answer














                    Assuming that you could enjoy special functions.



                    Consider
                    $$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$



                    $$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
                    $$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
                    $$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
                    $$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
                    t}right)right)$$
                    $$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 1 at 15:21

























                    answered Dec 1 at 5:35









                    Claude Leibovici

                    118k1156131




                    118k1156131






















                        up vote
                        0
                        down vote













                        Put $z=e^{it}$ and using the formulas:
                        $$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$



                        $$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$



                        apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          Put $z=e^{it}$ and using the formulas:
                          $$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$



                          $$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$



                          apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Put $z=e^{it}$ and using the formulas:
                            $$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$



                            $$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$



                            apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$






                            share|cite|improve this answer












                            Put $z=e^{it}$ and using the formulas:
                            $$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$



                            $$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$



                            apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 1 at 15:38









                            Marios Gretsas

                            8,42711437




                            8,42711437






























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