An AMM-like integral $int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx$











up vote
7
down vote

favorite
7













How can we evaluate $$I=int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx=0?$$




I tried substitution $x=frac{1-t}{1+t}$ and got
$$I=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} arctan frac{t-1}{t+1}}{t^2-1}dt\
=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} (arctan t-fracpi4)}{t^2-1}dt$$
I'm able to evaluate $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.






calculus integration definite-integrals







share|cite|improve this question






















  • Just wondering: Why do you want to calculate this (by hand)
    – klirk
    Nov 20 at 20:38










  • @klirk Just an interest.
    – Kemono Chen
    Nov 21 at 0:18










  • $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
    – user178256
    Nov 21 at 11:00















up vote
7
down vote

favorite
7













How can we evaluate $$I=int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx=0?$$




I tried substitution $x=frac{1-t}{1+t}$ and got
$$I=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} arctan frac{t-1}{t+1}}{t^2-1}dt\
=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} (arctan t-fracpi4)}{t^2-1}dt$$
I'm able to evaluate $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.






calculus integration definite-integrals







share|cite|improve this question






















  • Just wondering: Why do you want to calculate this (by hand)
    – klirk
    Nov 20 at 20:38










  • @klirk Just an interest.
    – Kemono Chen
    Nov 21 at 0:18










  • $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
    – user178256
    Nov 21 at 11:00













up vote
7
down vote

favorite
7









up vote
7
down vote

favorite
7






7






How can we evaluate $$I=int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx=0?$$




I tried substitution $x=frac{1-t}{1+t}$ and got
$$I=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} arctan frac{t-1}{t+1}}{t^2-1}dt\
=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} (arctan t-fracpi4)}{t^2-1}dt$$
I'm able to evaluate $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.






calculus integration definite-integrals







share|cite|improve this question














How can we evaluate $$I=int_0^1frac{arctan x}xlnfrac{(1+x^2)^3}{(1+x)^2}dx=0?$$




I tried substitution $x=frac{1-t}{1+t}$ and got
$$I=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} arctan frac{t-1}{t+1}}{t^2-1}dt\
=int_0^1frac{2 ln frac{2 (t^2+1)^3}{(t+1)^4} (arctan t-fracpi4)}{t^2-1}dt$$
I'm able to evaluate $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.






calculus integration definite-integrals




calculus integration definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 9:20









Kemono Chen

2,050435




2,050435












  • Just wondering: Why do you want to calculate this (by hand)
    – klirk
    Nov 20 at 20:38










  • @klirk Just an interest.
    – Kemono Chen
    Nov 21 at 0:18










  • $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
    – user178256
    Nov 21 at 11:00


















  • Just wondering: Why do you want to calculate this (by hand)
    – klirk
    Nov 20 at 20:38










  • @klirk Just an interest.
    – Kemono Chen
    Nov 21 at 0:18










  • $$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
    – user178256
    Nov 21 at 11:00
















Just wondering: Why do you want to calculate this (by hand)
– klirk
Nov 20 at 20:38




Just wondering: Why do you want to calculate this (by hand)
– klirk
Nov 20 at 20:38












@klirk Just an interest.
– Kemono Chen
Nov 21 at 0:18




@klirk Just an interest.
– Kemono Chen
Nov 21 at 0:18












$$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
– user178256
Nov 21 at 11:00




$$int_0^1frac{ln frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt=frac{pi^2}{48}$$
– user178256
Nov 21 at 11:00










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.






share|cite|improve this answer



















  • 1




    (+1) Very neat.
    – nospoon
    Nov 21 at 12:24


















up vote
0
down vote













Through the dilogarithm/trilogarithm machinery it can be shown that



$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.






share|cite|improve this answer





















  • Thank you for the great answer. :) But I prefer a solution without brute force.
    – Kemono Chen
    Nov 21 at 3:02











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',
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%2f3006106%2fan-amm-like-integral-int-01-frac-arctan-xx-ln-frac1x231x2dx%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
3
down vote



accepted










$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.






share|cite|improve this answer



















  • 1




    (+1) Very neat.
    – nospoon
    Nov 21 at 12:24















up vote
3
down vote



accepted










$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.






share|cite|improve this answer



















  • 1




    (+1) Very neat.
    – nospoon
    Nov 21 at 12:24













up vote
3
down vote



accepted







up vote
3
down vote



accepted






$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.






share|cite|improve this answer














$text{A solution by Cornel Ioan Valean.}$ The problem is similar to the problem $textbf{AMM 12054}.$ Using the well-known result in $textbf{4.535.1}$ from $text{Table of Integrals, Series and Products}$ by I.S. Gradshteyn and I.M. Ryzhik:
begin{equation*}
int_0^1 frac{arctan(y x)}{1+y^2x}textrm{d}x=frac{1}{2y^2}arctan(y)log(1+y^2),
end{equation*}
We have:
begin{equation*}
frac{1}{2}int_0^1frac{arctan(y)log(1+y^2)}{y}=int_0^1left(int_0^1 frac{yarctan(y x)}{1+y^2x}textrm{d}xright)textrm{d}yoverset{yx=t}{=}int_0^1left(int_0^y frac{arctan(t)}{1+y t}textrm{d}tright)textrm{d}y\
end{equation*}
begin{equation*}
=int_0^1left(int_t^1 frac{arctan(t)}{1+y t}textrm{d}yright)textrm{d}t=int_0^1frac{displaystyle arctan(y)logleft(frac{1+y}{1+y^2}right)}{y} textrm{d}y,
end{equation*}
And the result is proved.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 21 at 12:19

























answered Nov 21 at 11:52









Zacky

3,2981337




3,2981337








  • 1




    (+1) Very neat.
    – nospoon
    Nov 21 at 12:24














  • 1




    (+1) Very neat.
    – nospoon
    Nov 21 at 12:24








1




1




(+1) Very neat.
– nospoon
Nov 21 at 12:24




(+1) Very neat.
– nospoon
Nov 21 at 12:24










up vote
0
down vote













Through the dilogarithm/trilogarithm machinery it can be shown that



$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.






share|cite|improve this answer





















  • Thank you for the great answer. :) But I prefer a solution without brute force.
    – Kemono Chen
    Nov 21 at 3:02















up vote
0
down vote













Through the dilogarithm/trilogarithm machinery it can be shown that



$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.






share|cite|improve this answer





















  • Thank you for the great answer. :) But I prefer a solution without brute force.
    – Kemono Chen
    Nov 21 at 3:02













up vote
0
down vote










up vote
0
down vote









Through the dilogarithm/trilogarithm machinery it can be shown that



$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.






share|cite|improve this answer












Through the dilogarithm/trilogarithm machinery it can be shown that



$$ int_{0}^{1}frac{log(1+i x)log(1+x)}{x},dx=\frac{pi K}{2}-frac{9ipi^3}{64}+3iKlog(2)-frac{3pi i}{16}log^2(2)+frac{5pi^2}{32}log(2)-frac{log^3(2)}{8}-frac{69}{16}zeta(3)+6,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log^2(1+i x)}{x},dx=\
-frac{pi K}{2}-frac{3ipi^3}{64}+iKlog(2)-frac{pi i}{16}log^2(2)+frac{5pi^2}{96}log(2)-frac{log^3(2)}{24}-frac{3}{16}zeta(3)+2,text{Li}_3left(tfrac{1+i}{2}right) $$



$$ int_{0}^{1}frac{log(1+ix)log(1-ix)}{x},dx= frac{pi K}{2}-frac{27}{32}zeta(3)$$
hence the claim follows by $arctan x=text{Im},log(1+ix)$ and $log(1+x^2)=log(1+ix)+log(1-ix)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 at 20:20









Jack D'Aurizio

285k33275654




285k33275654












  • Thank you for the great answer. :) But I prefer a solution without brute force.
    – Kemono Chen
    Nov 21 at 3:02


















  • Thank you for the great answer. :) But I prefer a solution without brute force.
    – Kemono Chen
    Nov 21 at 3:02
















Thank you for the great answer. :) But I prefer a solution without brute force.
– Kemono Chen
Nov 21 at 3:02




Thank you for the great answer. :) But I prefer a solution without brute force.
– Kemono Chen
Nov 21 at 3:02


















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3006106%2fan-amm-like-integral-int-01-frac-arctan-xx-ln-frac1x231x2dx%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