Find using Residue Theorem the following integral












0












$begingroup$


Find using Residue Theorem



$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



My try:



I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.



and $C_2$ is the line joining $-1$ to $1$.



I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.



Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Any help from someone here?



Thanks for reading my post






complex-analysis residue-calculus







share|cite|improve this question











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  • $begingroup$
    Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 17:02












  • $begingroup$
    @DonAntonio;edited my question,can u have a look now
    $endgroup$
    – Join_PhD
    Dec 4 '18 at 17:07










  • $begingroup$
    Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 18:14


















0












$begingroup$


Find using Residue Theorem



$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



My try:



I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.



and $C_2$ is the line joining $-1$ to $1$.



I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.



Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Any help from someone here?



Thanks for reading my post






complex-analysis residue-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 17:02












  • $begingroup$
    @DonAntonio;edited my question,can u have a look now
    $endgroup$
    – Join_PhD
    Dec 4 '18 at 17:07










  • $begingroup$
    Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 18:14
















0












0








0


0



$begingroup$


Find using Residue Theorem



$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



My try:



I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.



and $C_2$ is the line joining $-1$ to $1$.



I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.



Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Any help from someone here?



Thanks for reading my post






complex-analysis residue-calculus







share|cite|improve this question











$endgroup$




Find using Residue Theorem



$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



My try:



I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.



and $C_2$ is the line joining $-1$ to $1$.



I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$



But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.



Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$



Any help from someone here?



Thanks for reading my post






complex-analysis residue-calculus




complex-analysis residue-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 17:06







Join_PhD

















asked Dec 4 '18 at 16:57









Join_PhDJoin_PhD

3898




3898












  • $begingroup$
    Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 17:02












  • $begingroup$
    @DonAntonio;edited my question,can u have a look now
    $endgroup$
    – Join_PhD
    Dec 4 '18 at 17:07










  • $begingroup$
    Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 18:14




















  • $begingroup$
    Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 17:02












  • $begingroup$
    @DonAntonio;edited my question,can u have a look now
    $endgroup$
    – Join_PhD
    Dec 4 '18 at 17:07










  • $begingroup$
    Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
    $endgroup$
    – DonAntonio
    Dec 4 '18 at 18:14


















$begingroup$
Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
$endgroup$
– DonAntonio
Dec 4 '18 at 17:02






$begingroup$
Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
$endgroup$
– DonAntonio
Dec 4 '18 at 17:02














$begingroup$
@DonAntonio;edited my question,can u have a look now
$endgroup$
– Join_PhD
Dec 4 '18 at 17:07




$begingroup$
@DonAntonio;edited my question,can u have a look now
$endgroup$
– Join_PhD
Dec 4 '18 at 17:07












$begingroup$
Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
$endgroup$
– DonAntonio
Dec 4 '18 at 18:14






$begingroup$
Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
$endgroup$
– DonAntonio
Dec 4 '18 at 18:14












1 Answer
1






active

oldest

votes


















0












$begingroup$

The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.






share|cite|improve this answer









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  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09
















0












$begingroup$

The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09














0












0








0





$begingroup$

The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.






share|cite|improve this answer









$endgroup$



The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.



Let's pick a contour similar to the one seen in this example.



We pick the branch cut on $[-1,1]$ on the real line such that



begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}



The contour consists of:





  • $C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,


  • $C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$


  • $C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$


  • $C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$



In the limit of $epsilon to 0$, we have



$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$



$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$




Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality



$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1-z| ge 2$



$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$



Since $|z| ge 1$ and $|1+z| ge 2$




Finally, use residues to finish the rest. You may also need to find the residue at infinity.







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share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 10:07









DylanDylan

12.7k31026




12.7k31026












  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09


















  • $begingroup$
    Thanks for an answer,I will need some time and patience to understand it
    $endgroup$
    – Join_PhD
    Dec 5 '18 at 10:09
















$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09




$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09


















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Plaza Victoria

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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...
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Puebla de Zaragoza

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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...
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Musa

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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...
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