Can $int_{-pi}^pi e^{cos(theta)}sin(sin(theta))sin(nx)dx=frac{pi}{n!}$ be extened to non-integers?












2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














2












2








2


1



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 $nin[1,20]$:
enter image description here



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
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018807%2fcan-int-pi-pi-e-cos-theta-sin-sin-theta-sinnxdx-frac-pin%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $nin[1,20]$:
enter image description here



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
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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 $nin[1,20]$:
enter image description here



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
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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 $nin[1,20]$:
enter image description here



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
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$






share|cite|improve this answer









$endgroup$



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 $nin[1,20]$:
enter image description here



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
$$
s(n) simeq frac{pi}{Gamma(n+1)} + frac{2sinbigl((n-1)pibigr)}{bigl(2+sqrt e(n-1)bigr)^2} .
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018807%2fcan-int-pi-pi-e-cos-theta-sin-sin-theta-sinnxdx-frac-pin%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more