Can $int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(nx)dx=frac{pi}{n!}$ be extened to non-integers?
$begingroup$
It is well known that the fourier series for $e^{cos(theta)}sin(sin(theta))$ is $sum_{n=1}^infty frac{sin(ntheta)}{n!}$ which implies that
$$int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(ntheta)dx=frac{pi}{n!}$$ for $n>0$. My question is if this identity can be extended to fractional values such as $$int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(frac{theta}{2})dtheta=2sqrt{pi} $$
definite-integrals fourier-series gamma-function
asked Nov 29 '18 at 15:50
aledenaleden
2,017511
$endgroup$
add a comment |
$begingroup$
It is well known that the fourier series for $e^{cos(theta)}sin(sin(theta))$ is $sum_{n=1}^infty frac{sin(ntheta)}{n!}$ which implies that
$$int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(ntheta)dx=frac{pi}{n!}$$ for $n>0$. My question is if this identity can be extended to fractional values such as $$int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(frac{theta}{2})dtheta=2sqrt{pi} $$
definite-integrals fourier-series gamma-function
asked Nov 29 '18 at 15:50
aledenaleden
2,017511
$endgroup$
$begingroup$
No! Try expanding with the half-angle formulas
$endgroup$
– TheOscillator
Nov 29 '18 at 15:59
$begingroup$
Numerical evidence suggests that $$ int_{-pi}^pi e^{costheta} sin(sintheta)sin(ntheta),dtheta simeq fracpi{n!} - frac34frac{sin((n-1)pi)}{n^2+1}, $$ where of course $n!=Gamma(n+1)$
$endgroup$
– Federico
Nov 29 '18 at 16:33
$begingroup$
Sorry, it should be $+tfrac34dots$, not $-$.
$endgroup$
– Federico
Nov 29 '18 at 16:43
add a comment |
$begingroup$
It is well known that the fourier series for $e^{cos(theta)}sin(sin(theta))$ is $sum_{n=1}^infty frac{sin(ntheta)}{n!}$ which implies that
$$int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(ntheta)dx=frac{pi}{n!}$$ for $n>0$. My question is if this identity can be extended to fractional values such as $$int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(frac{theta}{2})dtheta=2sqrt{pi} $$
definite-integrals fourier-series gamma-function
asked Nov 29 '18 at 15:50
aledenaleden
2,017511
$endgroup$
It is well known that the fourier series for $e^{cos(theta)}sin(sin(theta))$ is $sum_{n=1}^infty frac{sin(ntheta)}{n!}$ which implies that
$$int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(ntheta)dx=frac{pi}{n!}$$ for $n>0$. My question is if this identity can be extended to fractional values such as $$int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(frac{theta}{2})dtheta=2sqrt{pi} $$
definite-integrals fourier-series gamma-function
definite-integrals fourier-series gamma-function
asked Nov 29 '18 at 15:50
aledenaleden
2,017511
asked Nov 29 '18 at 15:50
aledenaleden
2,017511
asked Nov 29 '18 at 15:50
aledenaleden
2,017511
asked Nov 29 '18 at 15:50
aledenaleden
2,017511
asked Nov 29 '18 at 15:50
aledenaleden
2,017511
2,017511
$begingroup$
No! Try expanding with the half-angle formulas
$endgroup$
– TheOscillator
Nov 29 '18 at 15:59
$begingroup$
Numerical evidence suggests that $$ int_{-pi}^pi e^{costheta} sin(sintheta)sin(ntheta),dtheta simeq fracpi{n!} - frac34frac{sin((n-1)pi)}{n^2+1}, $$ where of course $n!=Gamma(n+1)$
$endgroup$
– Federico
Nov 29 '18 at 16:33
$begingroup$
Sorry, it should be $+tfrac34dots$, not $-$.
$endgroup$
– Federico
Nov 29 '18 at 16:43
add a comment |
$begingroup$
No! Try expanding with the half-angle formulas
$endgroup$
– TheOscillator
Nov 29 '18 at 15:59
$begingroup$
Numerical evidence suggests that $$ int_{-pi}^pi e^{costheta} sin(sintheta)sin(ntheta),dtheta simeq fracpi{n!} - frac34frac{sin((n-1)pi)}{n^2+1}, $$ where of course $n!=Gamma(n+1)$
$endgroup$
– Federico
Nov 29 '18 at 16:33
$begingroup$
Sorry, it should be $+tfrac34dots$, not $-$.
$endgroup$
– Federico
Nov 29 '18 at 16:43
$begingroup$
No! Try expanding with the half-angle formulas
$endgroup$
– TheOscillator
Nov 29 '18 at 15:59
$begingroup$
No! Try expanding with the half-angle formulas
$endgroup$
– TheOscillator
Nov 29 '18 at 15:59
$begingroup$
Numerical evidence suggests that $$ int_{-pi}^pi e^{costheta} sin(sintheta)sin(ntheta),dtheta simeq fracpi{n!} - frac34frac{sin((n-1)pi)}{n^2+1}, $$ where of course $n!=Gamma(n+1)$
$endgroup$
– Federico
Nov 29 '18 at 16:33
$begingroup$
Numerical evidence suggests that $$ int_{-pi}^pi e^{costheta} sin(sintheta)sin(ntheta),dtheta simeq fracpi{n!} - frac34frac{sin((n-1)pi)}{n^2+1}, $$ where of course $n!=Gamma(n+1)$
$endgroup$
– Federico
Nov 29 '18 at 16:33
$begingroup$
Sorry, it should be $+tfrac34dots$, not $-$.
$endgroup$
– Federico
Nov 29 '18 at 16:43
$begingroup$
Sorry, it should be $+tfrac34dots$, not $-$.
$endgroup$
– Federico
Nov 29 '18 at 16:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Incomplete answer: only numerical experiments.
Let me define
$$
s(n) = int_{-pi}^pi e^{costheta}sin(sintheta)sin(ntheta),dtheta
$$
and
$$
g(n) = frac{s(n)-frac{pi}{Gamma(n+1)}}{sinbigl((n-1)pibigr)}.
$$
Numerical evidence suggests that $g(n)to0$; but how fast? Let's try to extract some more information.
Here I plot $-logbigl(g(n)bigr)$ for $nin[1,20]$:
It looks like a logarithm, so it seems natural to try to fit it with
$$
-logbigl(g(n)bigr) = a+blog(n+c).
$$
By fiddling a bit (and looking at larger ranges of $n$) one can see that $b$ has to be $2$, so we have $a+2log(n+c)$. The value $g(1)$ seems to be around $0.5$, a bit less, so we impose $a+2log(1+b)=log(2)$. Moreover, by looking at large values, we deduce that $a=1-log(2)$ seems a good value. Therefore our approximation is
$$
g(n) simeq frac{2}{bigl(2+sqrt e(n-1)bigr)^2},
$$
which leads to
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$
answered Nov 29 '18 at 17:35
FedericoFederico
4,899514
$endgroup$
$begingroup$
I see, so at integer value it is equal to $frac{pi}{n!}$ but otherwise it is not.
$endgroup$
– aleden
Nov 29 '18 at 17:39
$begingroup$
Yes. Even just at $n=1/2$, we get $s(1/2)=tfrac2e + sqrtpi mathrm{erf}(1)$ instead of $2sqrtpi$.
$endgroup$
– Federico
Nov 29 '18 at 17:42
$begingroup$
I'm still intrigued to find a closed expression for $s(n)$
$endgroup$
– Federico
Nov 29 '18 at 17:43
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Incomplete answer: only numerical experiments.
Let me define
$$
s(n) = int_{-pi}^pi e^{costheta}sin(sintheta)sin(ntheta),dtheta
$$
and
$$
g(n) = frac{s(n)-frac{pi}{Gamma(n+1)}}{sinbigl((n-1)pibigr)}.
$$
Numerical evidence suggests that $g(n)to0$; but how fast? Let's try to extract some more information.
Here I plot $-logbigl(g(n)bigr)$ for $nin[1,20]$:
It looks like a logarithm, so it seems natural to try to fit it with
$$
-logbigl(g(n)bigr) = a+blog(n+c).
$$
By fiddling a bit (and looking at larger ranges of $n$) one can see that $b$ has to be $2$, so we have $a+2log(n+c)$. The value $g(1)$ seems to be around $0.5$, a bit less, so we impose $a+2log(1+b)=log(2)$. Moreover, by looking at large values, we deduce that $a=1-log(2)$ seems a good value. Therefore our approximation is
$$
g(n) simeq frac{2}{bigl(2+sqrt e(n-1)bigr)^2},
$$
which leads to
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$
answered Nov 29 '18 at 17:35
FedericoFederico
4,899514
$endgroup$
$begingroup$
I see, so at integer value it is equal to $frac{pi}{n!}$ but otherwise it is not.
$endgroup$
– aleden
Nov 29 '18 at 17:39
$begingroup$
Yes. Even just at $n=1/2$, we get $s(1/2)=tfrac2e + sqrtpi mathrm{erf}(1)$ instead of $2sqrtpi$.
$endgroup$
– Federico
Nov 29 '18 at 17:42
$begingroup$
I'm still intrigued to find a closed expression for $s(n)$
$endgroup$
– Federico
Nov 29 '18 at 17:43
add a comment |
$begingroup$
Incomplete answer: only numerical experiments.
Let me define
$$
s(n) = int_{-pi}^pi e^{costheta}sin(sintheta)sin(ntheta),dtheta
$$
and
$$
g(n) = frac{s(n)-frac{pi}{Gamma(n+1)}}{sinbigl((n-1)pibigr)}.
$$
Numerical evidence suggests that $g(n)to0$; but how fast? Let's try to extract some more information.
Here I plot $-logbigl(g(n)bigr)$ for $nin[1,20]$:
It looks like a logarithm, so it seems natural to try to fit it with
$$
-logbigl(g(n)bigr) = a+blog(n+c).
$$
By fiddling a bit (and looking at larger ranges of $n$) one can see that $b$ has to be $2$, so we have $a+2log(n+c)$. The value $g(1)$ seems to be around $0.5$, a bit less, so we impose $a+2log(1+b)=log(2)$. Moreover, by looking at large values, we deduce that $a=1-log(2)$ seems a good value. Therefore our approximation is
$$
g(n) simeq frac{2}{bigl(2+sqrt e(n-1)bigr)^2},
$$
which leads to
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$
answered Nov 29 '18 at 17:35
FedericoFederico
4,899514
$endgroup$
$begingroup$
I see, so at integer value it is equal to $frac{pi}{n!}$ but otherwise it is not.
$endgroup$
– aleden
Nov 29 '18 at 17:39
$begingroup$
Yes. Even just at $n=1/2$, we get $s(1/2)=tfrac2e + sqrtpi mathrm{erf}(1)$ instead of $2sqrtpi$.
$endgroup$
– Federico
Nov 29 '18 at 17:42
$begingroup$
I'm still intrigued to find a closed expression for $s(n)$
$endgroup$
– Federico
Nov 29 '18 at 17:43
add a comment |
$begingroup$
Incomplete answer: only numerical experiments.
Let me define
$$
s(n) = int_{-pi}^pi e^{costheta}sin(sintheta)sin(ntheta),dtheta
$$
and
$$
g(n) = frac{s(n)-frac{pi}{Gamma(n+1)}}{sinbigl((n-1)pibigr)}.
$$
Numerical evidence suggests that $g(n)to0$; but how fast? Let's try to extract some more information.
Here I plot $-logbigl(g(n)bigr)$ for $nin[1,20]$:
It looks like a logarithm, so it seems natural to try to fit it with
$$
-logbigl(g(n)bigr) = a+blog(n+c).
$$
By fiddling a bit (and looking at larger ranges of $n$) one can see that $b$ has to be $2$, so we have $a+2log(n+c)$. The value $g(1)$ seems to be around $0.5$, a bit less, so we impose $a+2log(1+b)=log(2)$. Moreover, by looking at large values, we deduce that $a=1-log(2)$ seems a good value. Therefore our approximation is
$$
g(n) simeq frac{2}{bigl(2+sqrt e(n-1)bigr)^2},
$$
which leads to
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$
answered Nov 29 '18 at 17:35
FedericoFederico
4,899514
$endgroup$
Incomplete answer: only numerical experiments.
Let me define
$$
s(n) = int_{-pi}^pi e^{costheta}sin(sintheta)sin(ntheta),dtheta
$$
and
$$
g(n) = frac{s(n)-frac{pi}{Gamma(n+1)}}{sinbigl((n-1)pibigr)}.
$$
Numerical evidence suggests that $g(n)to0$; but how fast? Let's try to extract some more information.
Here I plot $-logbigl(g(n)bigr)$ for $nin[1,20]$:
It looks like a logarithm, so it seems natural to try to fit it with
$$
-logbigl(g(n)bigr) = a+blog(n+c).
$$
By fiddling a bit (and looking at larger ranges of $n$) one can see that $b$ has to be $2$, so we have $a+2log(n+c)$. The value $g(1)$ seems to be around $0.5$, a bit less, so we impose $a+2log(1+b)=log(2)$. Moreover, by looking at large values, we deduce that $a=1-log(2)$ seems a good value. Therefore our approximation is
$$
g(n) simeq frac{2}{bigl(2+sqrt e(n-1)bigr)^2},
$$
which leads to
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$
answered Nov 29 '18 at 17:35
FedericoFederico
4,899514
answered Nov 29 '18 at 17:35
FedericoFederico
4,899514
answered Nov 29 '18 at 17:35
FedericoFederico
4,899514
answered Nov 29 '18 at 17:35
FedericoFederico
4,899514
4,899514
$begingroup$
I see, so at integer value it is equal to $frac{pi}{n!}$ but otherwise it is not.
$endgroup$
– aleden
Nov 29 '18 at 17:39
$begingroup$
Yes. Even just at $n=1/2$, we get $s(1/2)=tfrac2e + sqrtpi mathrm{erf}(1)$ instead of $2sqrtpi$.
$endgroup$
– Federico
Nov 29 '18 at 17:42
$begingroup$
I'm still intrigued to find a closed expression for $s(n)$
$endgroup$
– Federico
Nov 29 '18 at 17:43
add a comment |
$begingroup$
I see, so at integer value it is equal to $frac{pi}{n!}$ but otherwise it is not.
$endgroup$
– aleden
Nov 29 '18 at 17:39
$begingroup$
Yes. Even just at $n=1/2$, we get $s(1/2)=tfrac2e + sqrtpi mathrm{erf}(1)$ instead of $2sqrtpi$.
$endgroup$
– Federico
Nov 29 '18 at 17:42
$begingroup$
I'm still intrigued to find a closed expression for $s(n)$
$endgroup$
– Federico
Nov 29 '18 at 17:43
$begingroup$
I see, so at integer value it is equal to $frac{pi}{n!}$ but otherwise it is not.
$endgroup$
– aleden
Nov 29 '18 at 17:39
$begingroup$
I see, so at integer value it is equal to $frac{pi}{n!}$ but otherwise it is not.
$endgroup$
– aleden
Nov 29 '18 at 17:39
$begingroup$
Yes. Even just at $n=1/2$, we get $s(1/2)=tfrac2e + sqrtpi mathrm{erf}(1)$ instead of $2sqrtpi$.
$endgroup$
– Federico
Nov 29 '18 at 17:42
$begingroup$
Yes. Even just at $n=1/2$, we get $s(1/2)=tfrac2e + sqrtpi mathrm{erf}(1)$ instead of $2sqrtpi$.
$endgroup$
– Federico
Nov 29 '18 at 17:42
$begingroup$
I'm still intrigued to find a closed expression for $s(n)$
$endgroup$
– Federico
Nov 29 '18 at 17:43
$begingroup$
I'm still intrigued to find a closed expression for $s(n)$
$endgroup$
– Federico
Nov 29 '18 at 17:43
add a comment |
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$begingroup$
No! Try expanding with the half-angle formulas
$endgroup$
– TheOscillator
Nov 29 '18 at 15:59
$begingroup$
Numerical evidence suggests that $$ int_{-pi}^pi e^{costheta} sin(sintheta)sin(ntheta),dtheta simeq fracpi{n!} - frac34frac{sin((n-1)pi)}{n^2+1}, $$ where of course $n!=Gamma(n+1)$
$endgroup$
– Federico
Nov 29 '18 at 16:33
$begingroup$
Sorry, it should be $+tfrac34dots$, not $-$.
$endgroup$
– Federico
Nov 29 '18 at 16:43