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.
calculus real-analysis fourier-analysis
asked Aug 24 at 3:47
Math
67
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up vote
2
down vote
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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.
calculus real-analysis fourier-analysis
asked Aug 24 at 3:47
Math
67
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
add a comment |Â
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.
calculus real-analysis fourier-analysis
asked Aug 24 at 3:47
Math
67
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.
calculus real-analysis fourier-analysis
asked Aug 24 at 3:47
Math
67
asked Aug 24 at 3:47
Math
67
asked Aug 24 at 3:47
Math
67
asked Aug 24 at 3:47
Math
67
67
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
add a comment |Â
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
add a comment |Â
1 Answer
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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 $yinmathbbR$.
answered Aug 26 at 21:42
Sangchul Lee
86.1k12155253
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
(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 $yinmathbbR$.
answered Aug 26 at 21:42
Sangchul Lee
86.1k12155253
add a comment |Â
up vote
0
down vote
accepted
(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 $yinmathbbR$.
answered Aug 26 at 21:42
Sangchul Lee
86.1k12155253
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
(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 $yinmathbbR$.
answered Aug 26 at 21:42
Sangchul Lee
86.1k12155253
(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 $yinmathbbR$.
answered Aug 26 at 21:42
Sangchul Lee
86.1k12155253
answered Aug 26 at 21:42
Sangchul Lee
86.1k12155253
answered Aug 26 at 21:42
Sangchul Lee
86.1k12155253
answered Aug 26 at 21:42
Sangchul Lee
86.1k12155253
86.1k12155253
add a comment |Â
add a comment |Â
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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