Any ideas how to solve or perhaps simplify this integral? Wolfram is unable to.

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Define $f:mathbbR^2 to mathbbR$ by:



$$f(x,y) = int_0^2 pi fracexp(cos(x theta+y))-1cos(x theta+y)exp(cos(x theta+y))d theta $$



I was very much hoping to be able to write $f(x,y)$ without the integral there, any chance this is able to be simplified somehow? Wolfram has trouble doing definite integrals with symbolic manipulations.



I don't even need a completely closed form expression, I would be happy to know $f$ is a Bessel function or some such thing.







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  • 3




    Wolfram|Alpha can't even solve the simpler integral you get for $x=1$, $y=0$: wolframalpha.com/input/…, nor the one that results from that by substituting $u=costheta$: wolframalpha.com/input/…. So I think your chances are slim -- that's the sort of integral that Wolfram|Alpha is usually good at, though there are rare exceptions.
    – joriki
    Aug 24 at 4:47







  • 3




    You don't need a function of two variables. If $$ F(t) = int frac(cos t)-1cos t ;e^cos t dt, $$ which is presumably not an elementary function, then change variables to see $$ f(x,y) = fracF(2 x pi + y) - F(y)x $$
    – GEdgar
    Aug 24 at 10:54







  • 1




    At least for $ninmathbbZsetminus0$, $$f(n,y) = 2pi,_1F_2left(tfrac12;1,tfrac32;tfrac14right)approx6.82681cdots$$ is independent of $y$.
    – Sangchul Lee
    Aug 26 at 19:23















up vote
2
down vote

favorite












Define $f:mathbbR^2 to mathbbR$ by:



$$f(x,y) = int_0^2 pi fracexp(cos(x theta+y))-1cos(x theta+y)exp(cos(x theta+y))d theta $$



I was very much hoping to be able to write $f(x,y)$ without the integral there, any chance this is able to be simplified somehow? Wolfram has trouble doing definite integrals with symbolic manipulations.



I don't even need a completely closed form expression, I would be happy to know $f$ is a Bessel function or some such thing.







share|cite|improve this question
















  • 3




    Wolfram|Alpha can't even solve the simpler integral you get for $x=1$, $y=0$: wolframalpha.com/input/…, nor the one that results from that by substituting $u=costheta$: wolframalpha.com/input/…. So I think your chances are slim -- that's the sort of integral that Wolfram|Alpha is usually good at, though there are rare exceptions.
    – joriki
    Aug 24 at 4:47







  • 3




    You don't need a function of two variables. If $$ F(t) = int frac(cos t)-1cos t ;e^cos t dt, $$ which is presumably not an elementary function, then change variables to see $$ f(x,y) = fracF(2 x pi + y) - F(y)x $$
    – GEdgar
    Aug 24 at 10:54







  • 1




    At least for $ninmathbbZsetminus0$, $$f(n,y) = 2pi,_1F_2left(tfrac12;1,tfrac32;tfrac14right)approx6.82681cdots$$ is independent of $y$.
    – Sangchul Lee
    Aug 26 at 19:23













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Define $f:mathbbR^2 to mathbbR$ by:



$$f(x,y) = int_0^2 pi fracexp(cos(x theta+y))-1cos(x theta+y)exp(cos(x theta+y))d theta $$



I was very much hoping to be able to write $f(x,y)$ without the integral there, any chance this is able to be simplified somehow? Wolfram has trouble doing definite integrals with symbolic manipulations.



I don't even need a completely closed form expression, I would be happy to know $f$ is a Bessel function or some such thing.







share|cite|improve this question












Define $f:mathbbR^2 to mathbbR$ by:



$$f(x,y) = int_0^2 pi fracexp(cos(x theta+y))-1cos(x theta+y)exp(cos(x theta+y))d theta $$



I was very much hoping to be able to write $f(x,y)$ without the integral there, any chance this is able to be simplified somehow? Wolfram has trouble doing definite integrals with symbolic manipulations.



I don't even need a completely closed form expression, I would be happy to know $f$ is a Bessel function or some such thing.









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 24 at 3:47









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  • 3




    Wolfram|Alpha can't even solve the simpler integral you get for $x=1$, $y=0$: wolframalpha.com/input/…, nor the one that results from that by substituting $u=costheta$: wolframalpha.com/input/…. So I think your chances are slim -- that's the sort of integral that Wolfram|Alpha is usually good at, though there are rare exceptions.
    – joriki
    Aug 24 at 4:47







  • 3




    You don't need a function of two variables. If $$ F(t) = int frac(cos t)-1cos t ;e^cos t dt, $$ which is presumably not an elementary function, then change variables to see $$ f(x,y) = fracF(2 x pi + y) - F(y)x $$
    – GEdgar
    Aug 24 at 10:54







  • 1




    At least for $ninmathbbZsetminus0$, $$f(n,y) = 2pi,_1F_2left(tfrac12;1,tfrac32;tfrac14right)approx6.82681cdots$$ is independent of $y$.
    – Sangchul Lee
    Aug 26 at 19:23













  • 3




    Wolfram|Alpha can't even solve the simpler integral you get for $x=1$, $y=0$: wolframalpha.com/input/…, nor the one that results from that by substituting $u=costheta$: wolframalpha.com/input/…. So I think your chances are slim -- that's the sort of integral that Wolfram|Alpha is usually good at, though there are rare exceptions.
    – joriki
    Aug 24 at 4:47







  • 3




    You don't need a function of two variables. If $$ F(t) = int frac(cos t)-1cos t ;e^cos t dt, $$ which is presumably not an elementary function, then change variables to see $$ f(x,y) = fracF(2 x pi + y) - F(y)x $$
    – GEdgar
    Aug 24 at 10:54







  • 1




    At least for $ninmathbbZsetminus0$, $$f(n,y) = 2pi,_1F_2left(tfrac12;1,tfrac32;tfrac14right)approx6.82681cdots$$ is independent of $y$.
    – Sangchul Lee
    Aug 26 at 19:23








3




3




Wolfram|Alpha can't even solve the simpler integral you get for $x=1$, $y=0$: wolframalpha.com/input/…, nor the one that results from that by substituting $u=costheta$: wolframalpha.com/input/…. So I think your chances are slim -- that's the sort of integral that Wolfram|Alpha is usually good at, though there are rare exceptions.
– joriki
Aug 24 at 4:47





Wolfram|Alpha can't even solve the simpler integral you get for $x=1$, $y=0$: wolframalpha.com/input/…, nor the one that results from that by substituting $u=costheta$: wolframalpha.com/input/…. So I think your chances are slim -- that's the sort of integral that Wolfram|Alpha is usually good at, though there are rare exceptions.
– joriki
Aug 24 at 4:47





3




3




You don't need a function of two variables. If $$ F(t) = int frac(cos t)-1cos t ;e^cos t dt, $$ which is presumably not an elementary function, then change variables to see $$ f(x,y) = fracF(2 x pi + y) - F(y)x $$
– GEdgar
Aug 24 at 10:54





You don't need a function of two variables. If $$ F(t) = int frac(cos t)-1cos t ;e^cos t dt, $$ which is presumably not an elementary function, then change variables to see $$ f(x,y) = fracF(2 x pi + y) - F(y)x $$
– GEdgar
Aug 24 at 10:54





1




1




At least for $ninmathbbZsetminus0$, $$f(n,y) = 2pi,_1F_2left(tfrac12;1,tfrac32;tfrac14right)approx6.82681cdots$$ is independent of $y$.
– Sangchul Lee
Aug 26 at 19:23





At least for $ninmathbbZsetminus0$, $$f(n,y) = 2pi,_1F_2left(tfrac12;1,tfrac32;tfrac14right)approx6.82681cdots$$ is independent of $y$.
– Sangchul Lee
Aug 26 at 19:23











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(Too long for a comment)



A slightly more general observation is as follows: If we define



$$A(x,z):=int_0^xfrace^zcostheta-1zcostheta,dtheta,$$



then $A(x,z)=sum_a,b=0^inftyfrac(z/2)^a+ba!b!(a+b+1)xoperatornamesinc((a-b)x)$. From this, we can verify that $A$



$$A(x+pi,z)=A(pi,z)+A(x,-z).$$



In particular, using $A(pi,-z)=pisum_n=0^inftyfrac(z/2)^2nn!^2(2n+1)=A(pi,z)$, we obtain $A(npi,z)=nA(pi,z)$ for any integer $n$. This can be used to derive a special value of



$$f(x,y)=fracA(2pi x+y,-1)-A(y,-1)x, $$



such as $f(fracn2, 0) = 2A(pi,1) = f(n,y)$ for any non-zero integers $n$ and for any $yinmathbbR$.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted
    +100










    (Too long for a comment)



    A slightly more general observation is as follows: If we define



    $$A(x,z):=int_0^xfrace^zcostheta-1zcostheta,dtheta,$$



    then $A(x,z)=sum_a,b=0^inftyfrac(z/2)^a+ba!b!(a+b+1)xoperatornamesinc((a-b)x)$. From this, we can verify that $A$



    $$A(x+pi,z)=A(pi,z)+A(x,-z).$$



    In particular, using $A(pi,-z)=pisum_n=0^inftyfrac(z/2)^2nn!^2(2n+1)=A(pi,z)$, we obtain $A(npi,z)=nA(pi,z)$ for any integer $n$. This can be used to derive a special value of



    $$f(x,y)=fracA(2pi x+y,-1)-A(y,-1)x, $$



    such as $f(fracn2, 0) = 2A(pi,1) = f(n,y)$ for any non-zero integers $n$ and for any $yinmathbbR$.






    share|cite|improve this answer
























      up vote
      0
      down vote



      accepted
      +100










      (Too long for a comment)



      A slightly more general observation is as follows: If we define



      $$A(x,z):=int_0^xfrace^zcostheta-1zcostheta,dtheta,$$



      then $A(x,z)=sum_a,b=0^inftyfrac(z/2)^a+ba!b!(a+b+1)xoperatornamesinc((a-b)x)$. From this, we can verify that $A$



      $$A(x+pi,z)=A(pi,z)+A(x,-z).$$



      In particular, using $A(pi,-z)=pisum_n=0^inftyfrac(z/2)^2nn!^2(2n+1)=A(pi,z)$, we obtain $A(npi,z)=nA(pi,z)$ for any integer $n$. This can be used to derive a special value of



      $$f(x,y)=fracA(2pi x+y,-1)-A(y,-1)x, $$



      such as $f(fracn2, 0) = 2A(pi,1) = f(n,y)$ for any non-zero integers $n$ and for any $yinmathbbR$.






      share|cite|improve this answer






















        up vote
        0
        down vote



        accepted
        +100







        up vote
        0
        down vote



        accepted
        +100




        +100




        (Too long for a comment)



        A slightly more general observation is as follows: If we define



        $$A(x,z):=int_0^xfrace^zcostheta-1zcostheta,dtheta,$$



        then $A(x,z)=sum_a,b=0^inftyfrac(z/2)^a+ba!b!(a+b+1)xoperatornamesinc((a-b)x)$. From this, we can verify that $A$



        $$A(x+pi,z)=A(pi,z)+A(x,-z).$$



        In particular, using $A(pi,-z)=pisum_n=0^inftyfrac(z/2)^2nn!^2(2n+1)=A(pi,z)$, we obtain $A(npi,z)=nA(pi,z)$ for any integer $n$. This can be used to derive a special value of



        $$f(x,y)=fracA(2pi x+y,-1)-A(y,-1)x, $$



        such as $f(fracn2, 0) = 2A(pi,1) = f(n,y)$ for any non-zero integers $n$ and for any $yinmathbbR$.






        share|cite|improve this answer












        (Too long for a comment)



        A slightly more general observation is as follows: If we define



        $$A(x,z):=int_0^xfrace^zcostheta-1zcostheta,dtheta,$$



        then $A(x,z)=sum_a,b=0^inftyfrac(z/2)^a+ba!b!(a+b+1)xoperatornamesinc((a-b)x)$. From this, we can verify that $A$



        $$A(x+pi,z)=A(pi,z)+A(x,-z).$$



        In particular, using $A(pi,-z)=pisum_n=0^inftyfrac(z/2)^2nn!^2(2n+1)=A(pi,z)$, we obtain $A(npi,z)=nA(pi,z)$ for any integer $n$. This can be used to derive a special value of



        $$f(x,y)=fracA(2pi x+y,-1)-A(y,-1)x, $$



        such as $f(fracn2, 0) = 2A(pi,1) = f(n,y)$ for any non-zero integers $n$ and for any $yinmathbbR$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 26 at 21:42









        Sangchul Lee

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