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












8












$begingroup$


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











$endgroup$








  • 2




    $begingroup$
    Have you heard of Cauchy integral theorem?
    $endgroup$
    – Frank W.
    Dec 1 '18 at 2:35










  • $begingroup$
    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 $a$.
    $endgroup$
    – user170231
    Dec 1 '18 at 2:41












  • $begingroup$
    $x=cos(2t)$ 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 $frac{pi}{2}$.
    $endgroup$
    – herb steinberg
    Dec 1 '18 at 2:48






  • 1




    $begingroup$
    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.
    $endgroup$
    – J.G.
    Dec 1 '18 at 9:20
















8












$begingroup$


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











$endgroup$








  • 2




    $begingroup$
    Have you heard of Cauchy integral theorem?
    $endgroup$
    – Frank W.
    Dec 1 '18 at 2:35










  • $begingroup$
    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 $a$.
    $endgroup$
    – user170231
    Dec 1 '18 at 2:41












  • $begingroup$
    $x=cos(2t)$ 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 $frac{pi}{2}$.
    $endgroup$
    – herb steinberg
    Dec 1 '18 at 2:48






  • 1




    $begingroup$
    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.
    $endgroup$
    – J.G.
    Dec 1 '18 at 9:20














8












8








8


6



$begingroup$


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











$endgroup$




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 '18 at 9:34









Martin Sleziak

44.7k8117272




44.7k8117272










asked Dec 1 '18 at 2:15









NMisterNMister

376110




376110








  • 2




    $begingroup$
    Have you heard of Cauchy integral theorem?
    $endgroup$
    – Frank W.
    Dec 1 '18 at 2:35










  • $begingroup$
    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 $a$.
    $endgroup$
    – user170231
    Dec 1 '18 at 2:41












  • $begingroup$
    $x=cos(2t)$ 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 $frac{pi}{2}$.
    $endgroup$
    – herb steinberg
    Dec 1 '18 at 2:48






  • 1




    $begingroup$
    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.
    $endgroup$
    – J.G.
    Dec 1 '18 at 9:20














  • 2




    $begingroup$
    Have you heard of Cauchy integral theorem?
    $endgroup$
    – Frank W.
    Dec 1 '18 at 2:35










  • $begingroup$
    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 $a$.
    $endgroup$
    – user170231
    Dec 1 '18 at 2:41












  • $begingroup$
    $x=cos(2t)$ 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 $frac{pi}{2}$.
    $endgroup$
    – herb steinberg
    Dec 1 '18 at 2:48






  • 1




    $begingroup$
    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.
    $endgroup$
    – J.G.
    Dec 1 '18 at 9:20








2




2




$begingroup$
Have you heard of Cauchy integral theorem?
$endgroup$
– Frank W.
Dec 1 '18 at 2:35




$begingroup$
Have you heard of Cauchy integral theorem?
$endgroup$
– Frank W.
Dec 1 '18 at 2:35












$begingroup$
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 $a$.
$endgroup$
– user170231
Dec 1 '18 at 2:41






$begingroup$
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 $a$.
$endgroup$
– user170231
Dec 1 '18 at 2:41














$begingroup$
$x=cos(2t)$ 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 $frac{pi}{2}$.
$endgroup$
– herb steinberg
Dec 1 '18 at 2:48




$begingroup$
$x=cos(2t)$ 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 $frac{pi}{2}$.
$endgroup$
– herb steinberg
Dec 1 '18 at 2:48




1




1




$begingroup$
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.
$endgroup$
– J.G.
Dec 1 '18 at 9:20




$begingroup$
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.
$endgroup$
– J.G.
Dec 1 '18 at 9:20










3 Answers
3






active

oldest

votes


















23












$begingroup$

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











$endgroup$





















    5












    $begingroup$

    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











    $endgroup$





















      0












      $begingroup$

      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









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        23












        $begingroup$

        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











        $endgroup$


















          23












          $begingroup$

          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











          $endgroup$
















            23












            23








            23





            $begingroup$

            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











            $endgroup$



            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 '18 at 9:13









            Mutantoe

            582412




            582412










            answered Dec 1 '18 at 2:42









            Sangchul LeeSangchul Lee

            91.8k12165266




            91.8k12165266























                5












                $begingroup$

                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











                $endgroup$


















                  5












                  $begingroup$

                  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











                  $endgroup$
















                    5












                    5








                    5





                    $begingroup$

                    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











                    $endgroup$



                    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 $a$.







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                    share|cite|improve this answer








                    edited Dec 1 '18 at 15:21

























                    answered Dec 1 '18 at 5:35









                    Claude LeiboviciClaude Leibovici

                    120k1157132




                    120k1157132























                        0












                        $begingroup$

                        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









                        $endgroup$


















                          0












                          $begingroup$

                          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









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            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









                            $endgroup$



                            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 '18 at 15:38









                            Marios GretsasMarios Gretsas

                            8,45111437




                            8,45111437






























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