Integration trick $int^{2pi}_{0} f(a+ r cos theta, b+rsin theta)dtheta=2pi f(a,b)$












10












$begingroup$


While looking for some nice integrals that are not taught in school, I found this theorem:




Suppose $f$ is a bivariate harmonic function; $(a,b)$ is a point in the plane; and $r$ is a positive real number. Then,
$$
int^{2pi}_{0} f(a+ r cos theta, b+rsin theta)dtheta=2pi f(a,b)
$$




Is there a nice way to prove the theorem above?



Unfortunately I have no idea how to start proving this, but I suspect it's related to Euler's formula since there is an example there that can be solved that way.




Source $longrightarrow$ Integration Tricks | Brilliant Math & Science Wiki






integration proof-writing harmonic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I wonder why the Frullani integral was chosen for demonstration of differential "trick".
    $endgroup$
    – user
    Mar 17 '18 at 13:09












  • $begingroup$
    where can you see that proof?
    $endgroup$
    – Zacky
    Mar 17 '18 at 17:07
















10












$begingroup$


While looking for some nice integrals that are not taught in school, I found this theorem:




Suppose $f$ is a bivariate harmonic function; $(a,b)$ is a point in the plane; and $r$ is a positive real number. Then,
$$
int^{2pi}_{0} f(a+ r cos theta, b+rsin theta)dtheta=2pi f(a,b)
$$




Is there a nice way to prove the theorem above?



Unfortunately I have no idea how to start proving this, but I suspect it's related to Euler's formula since there is an example there that can be solved that way.




Source $longrightarrow$ Integration Tricks | Brilliant Math & Science Wiki






integration proof-writing harmonic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I wonder why the Frullani integral was chosen for demonstration of differential "trick".
    $endgroup$
    – user
    Mar 17 '18 at 13:09












  • $begingroup$
    where can you see that proof?
    $endgroup$
    – Zacky
    Mar 17 '18 at 17:07














10












10








10


3



$begingroup$


While looking for some nice integrals that are not taught in school, I found this theorem:




Suppose $f$ is a bivariate harmonic function; $(a,b)$ is a point in the plane; and $r$ is a positive real number. Then,
$$
int^{2pi}_{0} f(a+ r cos theta, b+rsin theta)dtheta=2pi f(a,b)
$$




Is there a nice way to prove the theorem above?



Unfortunately I have no idea how to start proving this, but I suspect it's related to Euler's formula since there is an example there that can be solved that way.




Source $longrightarrow$ Integration Tricks | Brilliant Math & Science Wiki






integration proof-writing harmonic-functions







share|cite|improve this question











$endgroup$




While looking for some nice integrals that are not taught in school, I found this theorem:




Suppose $f$ is a bivariate harmonic function; $(a,b)$ is a point in the plane; and $r$ is a positive real number. Then,
$$
int^{2pi}_{0} f(a+ r cos theta, b+rsin theta)dtheta=2pi f(a,b)
$$




Is there a nice way to prove the theorem above?



Unfortunately I have no idea how to start proving this, but I suspect it's related to Euler's formula since there is an example there that can be solved that way.




Source $longrightarrow$ Integration Tricks | Brilliant Math & Science Wiki






integration proof-writing harmonic-functions




integration proof-writing harmonic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 19:34







Zacky

















asked Mar 17 '18 at 12:52









ZackyZacky

6,6851958




6,6851958








  • 1




    $begingroup$
    I wonder why the Frullani integral was chosen for demonstration of differential "trick".
    $endgroup$
    – user
    Mar 17 '18 at 13:09












  • $begingroup$
    where can you see that proof?
    $endgroup$
    – Zacky
    Mar 17 '18 at 17:07














  • 1




    $begingroup$
    I wonder why the Frullani integral was chosen for demonstration of differential "trick".
    $endgroup$
    – user
    Mar 17 '18 at 13:09












  • $begingroup$
    where can you see that proof?
    $endgroup$
    – Zacky
    Mar 17 '18 at 17:07








1




1




$begingroup$
I wonder why the Frullani integral was chosen for demonstration of differential "trick".
$endgroup$
– user
Mar 17 '18 at 13:09






$begingroup$
I wonder why the Frullani integral was chosen for demonstration of differential "trick".
$endgroup$
– user
Mar 17 '18 at 13:09














$begingroup$
where can you see that proof?
$endgroup$
– Zacky
Mar 17 '18 at 17:07




$begingroup$
where can you see that proof?
$endgroup$
– Zacky
Mar 17 '18 at 17:07










2 Answers
2






active

oldest

votes


















8












$begingroup$

Start by writing $f$ as a function of a complex variable in the usual way, so the claimed result is



$$int_0^{2pi} dtheta ,fbigl(z_0+re^{itheta}bigr)=2pi,f(z_0)$$



with $z_0:=a+ib$. Expand the integrand as a Taylor series, viz.



$$sum_{nge 0}frac{f^{(n)}(z_0)}{n!}r^nint_0^{2pi} dtheta ,e^{intheta}.$$



The required result follows from $int_0^{2pi} dtheta, e^{intheta}=2pidelta_{n0}$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I am sorry if it's obvious but could you explain how expanding $$f(z_0+re^{itheta})$$ gives $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!} e^{intheta}$$?
    $endgroup$
    – Zacky
    Mar 17 '18 at 13:19






  • 3




    $begingroup$
    @Zacky $f(u+v)=f(u)+f'(u)v+cdots$
    $endgroup$
    – J.G.
    Mar 17 '18 at 13:32










  • $begingroup$
    What is $delta_{n0}$?
    $endgroup$
    – Chase Ryan Taylor
    Mar 17 '18 at 19:35










  • $begingroup$
    @ChaseRyanTaylor en.wikipedia.org/wiki/Kronecker_delta
    $endgroup$
    – J.G.
    Mar 17 '18 at 19:37



















3












$begingroup$

This follows from the Mean Value Theorem for harmonic functions. Let $fin C^2(Omega)$ be a harmonic function on some open domain $Omegasubseteqmathbb{R}^n$ then
$$f(x_0)=frac{1}{text{vol}(B(x_0,r))}int_{B(x_0,r)}f(x),dx=frac{1}{text{vol}(partial B(x_0,r))}int_{partial B(x_0,r)}f(x),dS$$
for every ball $B(x_0,r)subseteq Omega$. Here $partial B(x_0,r)$ is the sphere centered at $x_0$ with radius $r>0$ and $,dS$ is the element of integration on $partial B(x_0,r)$. Using the second formula for $n=2$ we obtain
$$f(x_0)=frac{1}{2pi r}int_{||x-x_0||leqslant r}f(x),dS=frac{1}{2pi }int_0^{2pi}f(x_0+rw(theta)),dtheta$$
where $w(theta):=(cos(theta),sin(theta))$. In particular for $x_0:=(a,b)$ you get the desired result.






share|cite|improve this answer











$endgroup$













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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

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    8












    $begingroup$

    Start by writing $f$ as a function of a complex variable in the usual way, so the claimed result is



    $$int_0^{2pi} dtheta ,fbigl(z_0+re^{itheta}bigr)=2pi,f(z_0)$$



    with $z_0:=a+ib$. Expand the integrand as a Taylor series, viz.



    $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!}r^nint_0^{2pi} dtheta ,e^{intheta}.$$



    The required result follows from $int_0^{2pi} dtheta, e^{intheta}=2pidelta_{n0}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I am sorry if it's obvious but could you explain how expanding $$f(z_0+re^{itheta})$$ gives $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!} e^{intheta}$$?
      $endgroup$
      – Zacky
      Mar 17 '18 at 13:19






    • 3




      $begingroup$
      @Zacky $f(u+v)=f(u)+f'(u)v+cdots$
      $endgroup$
      – J.G.
      Mar 17 '18 at 13:32










    • $begingroup$
      What is $delta_{n0}$?
      $endgroup$
      – Chase Ryan Taylor
      Mar 17 '18 at 19:35










    • $begingroup$
      @ChaseRyanTaylor en.wikipedia.org/wiki/Kronecker_delta
      $endgroup$
      – J.G.
      Mar 17 '18 at 19:37
















    8












    $begingroup$

    Start by writing $f$ as a function of a complex variable in the usual way, so the claimed result is



    $$int_0^{2pi} dtheta ,fbigl(z_0+re^{itheta}bigr)=2pi,f(z_0)$$



    with $z_0:=a+ib$. Expand the integrand as a Taylor series, viz.



    $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!}r^nint_0^{2pi} dtheta ,e^{intheta}.$$



    The required result follows from $int_0^{2pi} dtheta, e^{intheta}=2pidelta_{n0}$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I am sorry if it's obvious but could you explain how expanding $$f(z_0+re^{itheta})$$ gives $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!} e^{intheta}$$?
      $endgroup$
      – Zacky
      Mar 17 '18 at 13:19






    • 3




      $begingroup$
      @Zacky $f(u+v)=f(u)+f'(u)v+cdots$
      $endgroup$
      – J.G.
      Mar 17 '18 at 13:32










    • $begingroup$
      What is $delta_{n0}$?
      $endgroup$
      – Chase Ryan Taylor
      Mar 17 '18 at 19:35










    • $begingroup$
      @ChaseRyanTaylor en.wikipedia.org/wiki/Kronecker_delta
      $endgroup$
      – J.G.
      Mar 17 '18 at 19:37














    8












    8








    8





    $begingroup$

    Start by writing $f$ as a function of a complex variable in the usual way, so the claimed result is



    $$int_0^{2pi} dtheta ,fbigl(z_0+re^{itheta}bigr)=2pi,f(z_0)$$



    with $z_0:=a+ib$. Expand the integrand as a Taylor series, viz.



    $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!}r^nint_0^{2pi} dtheta ,e^{intheta}.$$



    The required result follows from $int_0^{2pi} dtheta, e^{intheta}=2pidelta_{n0}$.






    share|cite|improve this answer











    $endgroup$



    Start by writing $f$ as a function of a complex variable in the usual way, so the claimed result is



    $$int_0^{2pi} dtheta ,fbigl(z_0+re^{itheta}bigr)=2pi,f(z_0)$$



    with $z_0:=a+ib$. Expand the integrand as a Taylor series, viz.



    $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!}r^nint_0^{2pi} dtheta ,e^{intheta}.$$



    The required result follows from $int_0^{2pi} dtheta, e^{intheta}=2pidelta_{n0}$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 23 '18 at 19:49

























    answered Mar 17 '18 at 13:02









    J.G.J.G.

    27.5k22843




    27.5k22843












    • $begingroup$
      I am sorry if it's obvious but could you explain how expanding $$f(z_0+re^{itheta})$$ gives $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!} e^{intheta}$$?
      $endgroup$
      – Zacky
      Mar 17 '18 at 13:19






    • 3




      $begingroup$
      @Zacky $f(u+v)=f(u)+f'(u)v+cdots$
      $endgroup$
      – J.G.
      Mar 17 '18 at 13:32










    • $begingroup$
      What is $delta_{n0}$?
      $endgroup$
      – Chase Ryan Taylor
      Mar 17 '18 at 19:35










    • $begingroup$
      @ChaseRyanTaylor en.wikipedia.org/wiki/Kronecker_delta
      $endgroup$
      – J.G.
      Mar 17 '18 at 19:37


















    • $begingroup$
      I am sorry if it's obvious but could you explain how expanding $$f(z_0+re^{itheta})$$ gives $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!} e^{intheta}$$?
      $endgroup$
      – Zacky
      Mar 17 '18 at 13:19






    • 3




      $begingroup$
      @Zacky $f(u+v)=f(u)+f'(u)v+cdots$
      $endgroup$
      – J.G.
      Mar 17 '18 at 13:32










    • $begingroup$
      What is $delta_{n0}$?
      $endgroup$
      – Chase Ryan Taylor
      Mar 17 '18 at 19:35










    • $begingroup$
      @ChaseRyanTaylor en.wikipedia.org/wiki/Kronecker_delta
      $endgroup$
      – J.G.
      Mar 17 '18 at 19:37
















    $begingroup$
    I am sorry if it's obvious but could you explain how expanding $$f(z_0+re^{itheta})$$ gives $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!} e^{intheta}$$?
    $endgroup$
    – Zacky
    Mar 17 '18 at 13:19




    $begingroup$
    I am sorry if it's obvious but could you explain how expanding $$f(z_0+re^{itheta})$$ gives $$sum_{nge 0}frac{f^{(n)}(z_0)}{n!} e^{intheta}$$?
    $endgroup$
    – Zacky
    Mar 17 '18 at 13:19




    3




    3




    $begingroup$
    @Zacky $f(u+v)=f(u)+f'(u)v+cdots$
    $endgroup$
    – J.G.
    Mar 17 '18 at 13:32




    $begingroup$
    @Zacky $f(u+v)=f(u)+f'(u)v+cdots$
    $endgroup$
    – J.G.
    Mar 17 '18 at 13:32












    $begingroup$
    What is $delta_{n0}$?
    $endgroup$
    – Chase Ryan Taylor
    Mar 17 '18 at 19:35




    $begingroup$
    What is $delta_{n0}$?
    $endgroup$
    – Chase Ryan Taylor
    Mar 17 '18 at 19:35












    $begingroup$
    @ChaseRyanTaylor en.wikipedia.org/wiki/Kronecker_delta
    $endgroup$
    – J.G.
    Mar 17 '18 at 19:37




    $begingroup$
    @ChaseRyanTaylor en.wikipedia.org/wiki/Kronecker_delta
    $endgroup$
    – J.G.
    Mar 17 '18 at 19:37











    3












    $begingroup$

    This follows from the Mean Value Theorem for harmonic functions. Let $fin C^2(Omega)$ be a harmonic function on some open domain $Omegasubseteqmathbb{R}^n$ then
    $$f(x_0)=frac{1}{text{vol}(B(x_0,r))}int_{B(x_0,r)}f(x),dx=frac{1}{text{vol}(partial B(x_0,r))}int_{partial B(x_0,r)}f(x),dS$$
    for every ball $B(x_0,r)subseteq Omega$. Here $partial B(x_0,r)$ is the sphere centered at $x_0$ with radius $r>0$ and $,dS$ is the element of integration on $partial B(x_0,r)$. Using the second formula for $n=2$ we obtain
    $$f(x_0)=frac{1}{2pi r}int_{||x-x_0||leqslant r}f(x),dS=frac{1}{2pi }int_0^{2pi}f(x_0+rw(theta)),dtheta$$
    where $w(theta):=(cos(theta),sin(theta))$. In particular for $x_0:=(a,b)$ you get the desired result.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      This follows from the Mean Value Theorem for harmonic functions. Let $fin C^2(Omega)$ be a harmonic function on some open domain $Omegasubseteqmathbb{R}^n$ then
      $$f(x_0)=frac{1}{text{vol}(B(x_0,r))}int_{B(x_0,r)}f(x),dx=frac{1}{text{vol}(partial B(x_0,r))}int_{partial B(x_0,r)}f(x),dS$$
      for every ball $B(x_0,r)subseteq Omega$. Here $partial B(x_0,r)$ is the sphere centered at $x_0$ with radius $r>0$ and $,dS$ is the element of integration on $partial B(x_0,r)$. Using the second formula for $n=2$ we obtain
      $$f(x_0)=frac{1}{2pi r}int_{||x-x_0||leqslant r}f(x),dS=frac{1}{2pi }int_0^{2pi}f(x_0+rw(theta)),dtheta$$
      where $w(theta):=(cos(theta),sin(theta))$. In particular for $x_0:=(a,b)$ you get the desired result.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        This follows from the Mean Value Theorem for harmonic functions. Let $fin C^2(Omega)$ be a harmonic function on some open domain $Omegasubseteqmathbb{R}^n$ then
        $$f(x_0)=frac{1}{text{vol}(B(x_0,r))}int_{B(x_0,r)}f(x),dx=frac{1}{text{vol}(partial B(x_0,r))}int_{partial B(x_0,r)}f(x),dS$$
        for every ball $B(x_0,r)subseteq Omega$. Here $partial B(x_0,r)$ is the sphere centered at $x_0$ with radius $r>0$ and $,dS$ is the element of integration on $partial B(x_0,r)$. Using the second formula for $n=2$ we obtain
        $$f(x_0)=frac{1}{2pi r}int_{||x-x_0||leqslant r}f(x),dS=frac{1}{2pi }int_0^{2pi}f(x_0+rw(theta)),dtheta$$
        where $w(theta):=(cos(theta),sin(theta))$. In particular for $x_0:=(a,b)$ you get the desired result.






        share|cite|improve this answer











        $endgroup$



        This follows from the Mean Value Theorem for harmonic functions. Let $fin C^2(Omega)$ be a harmonic function on some open domain $Omegasubseteqmathbb{R}^n$ then
        $$f(x_0)=frac{1}{text{vol}(B(x_0,r))}int_{B(x_0,r)}f(x),dx=frac{1}{text{vol}(partial B(x_0,r))}int_{partial B(x_0,r)}f(x),dS$$
        for every ball $B(x_0,r)subseteq Omega$. Here $partial B(x_0,r)$ is the sphere centered at $x_0$ with radius $r>0$ and $,dS$ is the element of integration on $partial B(x_0,r)$. Using the second formula for $n=2$ we obtain
        $$f(x_0)=frac{1}{2pi r}int_{||x-x_0||leqslant r}f(x),dS=frac{1}{2pi }int_0^{2pi}f(x_0+rw(theta)),dtheta$$
        where $w(theta):=(cos(theta),sin(theta))$. In particular for $x_0:=(a,b)$ you get the desired result.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 17 '18 at 16:01

























        answered Mar 17 '18 at 13:21









        ArianArian

        5,310917




        5,310917






























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