Interesting Integral $int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx$












21












$begingroup$


I am asking this question out of curiosity.



$$int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx = frac{ left(2cos frac{n}{2} right)^{alpha +beta-2}}{Gamma(alpha+beta-1)}e^{frac{in}{2}(beta - alpha)} quad |n|<pi quad text{and} quad Re(alpha+beta)>1$$




  • How did Ramanujan derive this formula?

  • I have noticed that Ramanujan has discovered many integrals involving gamma function. Is there a general method to deal with such integrals?






integration special-functions improper-integrals gamma-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
    $endgroup$
    – daniel
    Mar 15 '13 at 13:44












  • $begingroup$
    @daniel: I will try to find the book you suggested.
    $endgroup$
    – Shobhit Bhatnagar
    Mar 15 '13 at 15:12
















21












$begingroup$


I am asking this question out of curiosity.



$$int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx = frac{ left(2cos frac{n}{2} right)^{alpha +beta-2}}{Gamma(alpha+beta-1)}e^{frac{in}{2}(beta - alpha)} quad |n|<pi quad text{and} quad Re(alpha+beta)>1$$




  • How did Ramanujan derive this formula?

  • I have noticed that Ramanujan has discovered many integrals involving gamma function. Is there a general method to deal with such integrals?






integration special-functions improper-integrals gamma-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
    $endgroup$
    – daniel
    Mar 15 '13 at 13:44












  • $begingroup$
    @daniel: I will try to find the book you suggested.
    $endgroup$
    – Shobhit Bhatnagar
    Mar 15 '13 at 15:12














21












21








21


14



$begingroup$


I am asking this question out of curiosity.



$$int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx = frac{ left(2cos frac{n}{2} right)^{alpha +beta-2}}{Gamma(alpha+beta-1)}e^{frac{in}{2}(beta - alpha)} quad |n|<pi quad text{and} quad Re(alpha+beta)>1$$




  • How did Ramanujan derive this formula?

  • I have noticed that Ramanujan has discovered many integrals involving gamma function. Is there a general method to deal with such integrals?






integration special-functions improper-integrals gamma-function







share|cite|improve this question











$endgroup$




I am asking this question out of curiosity.



$$int_{-infty}^{infty}frac{e^{i nx}}{Gamma(alpha+x) Gamma(beta -x)}dx = frac{ left(2cos frac{n}{2} right)^{alpha +beta-2}}{Gamma(alpha+beta-1)}e^{frac{in}{2}(beta - alpha)} quad |n|<pi quad text{and} quad Re(alpha+beta)>1$$




  • How did Ramanujan derive this formula?

  • I have noticed that Ramanujan has discovered many integrals involving gamma function. Is there a general method to deal with such integrals?






integration special-functions improper-integrals gamma-function




integration special-functions improper-integrals gamma-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 15 '13 at 14:46







Shobhit Bhatnagar

















asked Mar 15 '13 at 11:28









Shobhit BhatnagarShobhit Bhatnagar

6,32053274




6,32053274








  • 1




    $begingroup$
    Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
    $endgroup$
    – daniel
    Mar 15 '13 at 13:44












  • $begingroup$
    @daniel: I will try to find the book you suggested.
    $endgroup$
    – Shobhit Bhatnagar
    Mar 15 '13 at 15:12














  • 1




    $begingroup$
    Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
    $endgroup$
    – daniel
    Mar 15 '13 at 13:44












  • $begingroup$
    @daniel: I will try to find the book you suggested.
    $endgroup$
    – Shobhit Bhatnagar
    Mar 15 '13 at 15:12








1




1




$begingroup$
Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
$endgroup$
– daniel
Mar 15 '13 at 13:44






$begingroup$
Just a comment really. Ramanujan derives several very similar equations at ch. 27 of Collected works. I think you can derive your integral from eq. (7.11). If I find time I will try to work it out and post as an answer.
$endgroup$
– daniel
Mar 15 '13 at 13:44














$begingroup$
@daniel: I will try to find the book you suggested.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:12




$begingroup$
@daniel: I will try to find the book you suggested.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:12










2 Answers
2






active

oldest

votes


















10












$begingroup$

This is not a complete answer but maybe it will save a trip to the library.



In Ramanujan's Collected Works at ch. 27 he says it is "well known" that



$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$



He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.



Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...



Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
    $endgroup$
    – Sangchul Lee
    Mar 15 '13 at 15:42










  • $begingroup$
    @daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
    $endgroup$
    – Shobhit Bhatnagar
    Mar 15 '13 at 15:53










  • $begingroup$
    I think I understood it!
    $endgroup$
    – Shobhit Bhatnagar
    Mar 17 '13 at 3:20






  • 2




    $begingroup$
    Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
    $endgroup$
    – J. M. is not a mathematician
    Apr 9 '13 at 1:41










  • $begingroup$
    Using (1.1) implies $alpha=beta$
    $endgroup$
    – Anastasiya-Romanova 秀
    Sep 26 '14 at 9:58



















13





+100







$begingroup$

$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$


$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.




Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}




begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}




begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}




$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$




Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}







share|cite|improve this answer











$endgroup$













  • $begingroup$
    The manipulations with the delta function were really helpful to me.
    $endgroup$
    – daniel
    Sep 25 '14 at 7:54










  • $begingroup$
    @daniel It's nice it was helpful for you. Thanks.
    $endgroup$
    – Felix Marin
    Sep 25 '14 at 21:25










  • $begingroup$
    @daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
    $endgroup$
    – Did
    Dec 17 '18 at 0:22










  • $begingroup$
    @Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
    $endgroup$
    – daniel
    Dec 17 '18 at 7:06













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









10












$begingroup$

This is not a complete answer but maybe it will save a trip to the library.



In Ramanujan's Collected Works at ch. 27 he says it is "well known" that



$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$



He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.



Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...



Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
    $endgroup$
    – Sangchul Lee
    Mar 15 '13 at 15:42










  • $begingroup$
    @daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
    $endgroup$
    – Shobhit Bhatnagar
    Mar 15 '13 at 15:53










  • $begingroup$
    I think I understood it!
    $endgroup$
    – Shobhit Bhatnagar
    Mar 17 '13 at 3:20






  • 2




    $begingroup$
    Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
    $endgroup$
    – J. M. is not a mathematician
    Apr 9 '13 at 1:41










  • $begingroup$
    Using (1.1) implies $alpha=beta$
    $endgroup$
    – Anastasiya-Romanova 秀
    Sep 26 '14 at 9:58
















10












$begingroup$

This is not a complete answer but maybe it will save a trip to the library.



In Ramanujan's Collected Works at ch. 27 he says it is "well known" that



$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$



He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.



Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...



Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
    $endgroup$
    – Sangchul Lee
    Mar 15 '13 at 15:42










  • $begingroup$
    @daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
    $endgroup$
    – Shobhit Bhatnagar
    Mar 15 '13 at 15:53










  • $begingroup$
    I think I understood it!
    $endgroup$
    – Shobhit Bhatnagar
    Mar 17 '13 at 3:20






  • 2




    $begingroup$
    Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
    $endgroup$
    – J. M. is not a mathematician
    Apr 9 '13 at 1:41










  • $begingroup$
    Using (1.1) implies $alpha=beta$
    $endgroup$
    – Anastasiya-Romanova 秀
    Sep 26 '14 at 9:58














10












10








10





$begingroup$

This is not a complete answer but maybe it will save a trip to the library.



In Ramanujan's Collected Works at ch. 27 he says it is "well known" that



$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$



He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.



Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...



Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...






share|cite|improve this answer











$endgroup$



This is not a complete answer but maybe it will save a trip to the library.



In Ramanujan's Collected Works at ch. 27 he says it is "well known" that



$$ intlimits_{-pi/2}^{pi/2}(cos x)^m e^{inx}dx = frac{pi}{2^m}frac{Gamma(1+m)}{Gammaleft(1+ frac{1}{2}(m+n)right)Gammaleft(1+frac{1}{2}(m-n)right)}. tag{1.1} $$



He says that from this and Fourier's Theorem we can derive the relation in your question. He may work out some details in the chapter (I haven't looked). If you accept or can prove (1.1) this might be enough to solve the problem.



Edit in response to comment: All he is saying is that a function can sometimes be represented as a Fourier series. I suspect the proof is in an earlier paper and ch. 27 is building on your relation. But he clearly says that "it follows" from (1.1) so it's a start. There are 4 papers cited in the notes...



Edit: Here is one way to do this proof. I thought we could take advantage of the (visually obvious) F-transform aspect of the relation, but this proof is very direct and maybe there is no simpler way...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 13 '17 at 12:21









Community♦

1




1










answered Mar 15 '13 at 13:25









danieldaniel

6,29222259




6,29222259








  • 2




    $begingroup$
    You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
    $endgroup$
    – Sangchul Lee
    Mar 15 '13 at 15:42










  • $begingroup$
    @daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
    $endgroup$
    – Shobhit Bhatnagar
    Mar 15 '13 at 15:53










  • $begingroup$
    I think I understood it!
    $endgroup$
    – Shobhit Bhatnagar
    Mar 17 '13 at 3:20






  • 2




    $begingroup$
    Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
    $endgroup$
    – J. M. is not a mathematician
    Apr 9 '13 at 1:41










  • $begingroup$
    Using (1.1) implies $alpha=beta$
    $endgroup$
    – Anastasiya-Romanova 秀
    Sep 26 '14 at 9:58














  • 2




    $begingroup$
    You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
    $endgroup$
    – Sangchul Lee
    Mar 15 '13 at 15:42










  • $begingroup$
    @daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
    $endgroup$
    – Shobhit Bhatnagar
    Mar 15 '13 at 15:53










  • $begingroup$
    I think I understood it!
    $endgroup$
    – Shobhit Bhatnagar
    Mar 17 '13 at 3:20






  • 2




    $begingroup$
    Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
    $endgroup$
    – J. M. is not a mathematician
    Apr 9 '13 at 1:41










  • $begingroup$
    Using (1.1) implies $alpha=beta$
    $endgroup$
    – Anastasiya-Romanova 秀
    Sep 26 '14 at 9:58








2




2




$begingroup$
You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
$endgroup$
– Sangchul Lee
Mar 15 '13 at 15:42




$begingroup$
You may check here for the proof of $(1.1)$, though I don't know if it follows the same way Ramanujan reached his conclusion.
$endgroup$
– Sangchul Lee
Mar 15 '13 at 15:42












$begingroup$
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:53




$begingroup$
@daniel: Thanks! But, I don't understand how fourier's theorem can be used to produce the result.
$endgroup$
– Shobhit Bhatnagar
Mar 15 '13 at 15:53












$begingroup$
I think I understood it!
$endgroup$
– Shobhit Bhatnagar
Mar 17 '13 at 3:20




$begingroup$
I think I understood it!
$endgroup$
– Shobhit Bhatnagar
Mar 17 '13 at 3:20




2




2




$begingroup$
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
$endgroup$
– J. M. is not a mathematician
Apr 9 '13 at 1:41




$begingroup$
Your $(1.1)$ can of course be shown (tediously, tho) using the Wallis formulae...
$endgroup$
– J. M. is not a mathematician
Apr 9 '13 at 1:41












$begingroup$
Using (1.1) implies $alpha=beta$
$endgroup$
– Anastasiya-Romanova 秀
Sep 26 '14 at 9:58




$begingroup$
Using (1.1) implies $alpha=beta$
$endgroup$
– Anastasiya-Romanova 秀
Sep 26 '14 at 9:58











13





+100







$begingroup$

$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$


$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.




Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}




begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}




begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}




$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$




Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}







share|cite|improve this answer











$endgroup$













  • $begingroup$
    The manipulations with the delta function were really helpful to me.
    $endgroup$
    – daniel
    Sep 25 '14 at 7:54










  • $begingroup$
    @daniel It's nice it was helpful for you. Thanks.
    $endgroup$
    – Felix Marin
    Sep 25 '14 at 21:25










  • $begingroup$
    @daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
    $endgroup$
    – Did
    Dec 17 '18 at 0:22










  • $begingroup$
    @Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
    $endgroup$
    – daniel
    Dec 17 '18 at 7:06


















13





+100







$begingroup$

$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$


$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.




Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}




begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}




begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}




$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$




Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}







share|cite|improve this answer











$endgroup$













  • $begingroup$
    The manipulations with the delta function were really helpful to me.
    $endgroup$
    – daniel
    Sep 25 '14 at 7:54










  • $begingroup$
    @daniel It's nice it was helpful for you. Thanks.
    $endgroup$
    – Felix Marin
    Sep 25 '14 at 21:25










  • $begingroup$
    @daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
    $endgroup$
    – Did
    Dec 17 '18 at 0:22










  • $begingroup$
    @Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
    $endgroup$
    – daniel
    Dec 17 '18 at 7:06
















13





+100







13





+100



13




+100



$begingroup$

$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$


$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.




Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}




begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}




begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}




$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$




Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}







share|cite|improve this answer











$endgroup$



$newcommand{+}{^{dagger}}
newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{down}{downarrow}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{isdiv}{,left.rightvert,}
newcommand{ket}[1]{leftvert #1rightrangle}
newcommand{ol}[1]{overline{#1}}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{ul}[1]{underline{#1}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}
newcommand{wt}[1]{widetilde{#1}}$
$ds{int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x
={bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}}
,expo{ic npars{beta - alpha}/2}}$


$ds{verts{n} < pi,, Repars{alpha + beta} > 1}$.




Note that
begin{align}
&{1 over Gammapars{alpha + x}Gammapars{beta - x}}
={1 over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm]&={1 over Gammapars{alpha + beta - 1}},
{pars{alpha + beta - 2}! over pars{alpha + x - 1}!pars{beta - x - 1}!}
\[3mm] & ={1 over Gammapars{alpha + beta - 1}},
{alpha + beta - 2 choose alpha + x - 1}
end{align}




begin{align}&color{#c00000}{%
int_{-infty}^{infty}%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = & {1 over Gammapars{alpha + beta - 1}}
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}tag{1}
end{align}




begin{align}&color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=int_{-infty}^{infty}bracks{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha + x}},{dd z over 2piic}}expo{ic n x},dd x
\[3mm]&=-icoint_{verts{z} = 1}
{pars{1 + z}^{alpha + beta - 2} over z^{alpha}}braces{%
int_{-infty}^{infty}expo{icbracks{n - {rm Arg}pars{z}}x}
,{dd x over 2pi}},dd z
\[3mm]&=-ic overbrace{oint_{verts{z} = 1}{pars{1 + z}^{alpha + beta - 2} over z^{alpha}},deltapars{n - {rm Arg}pars{z}},dd z}
^{ds{mbox{Set} z equiv expo{ic theta},,quadverts{theta} < pi}}
\[3mm]&=-icint_{-pi}^{pi}
{pars{1 + expo{ictheta}}^{alpha + beta - 2} over
expo{icalphatheta}},deltapars{n - theta},expo{ictheta}ic,ddtheta
=pars{1 + expo{ic n}}^{alpha + beta - 2}expo{icpars{1 - alpha}n}
\[3mm]&=expo{icpars{alpha + beta - 2}n/2}
pars{expo{-ic n/2} + expo{ic n/2}}^{alpha + beta - 2}
expo{icpars{1 - alpha}n}
\[3mm] & =
bracks{2cospars{{n over 2}}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
end{align}




$$
color{#00f}{int_{-infty}^{infty}
{alpha + beta - 2 choose alpha + x - 1}expo{ic nx},dd x}
=bracks{2cospars{n over 2}}^{alpha + beta - 2}
expo{icpars{beta - alpha}n/2}
$$




Replace this result in $pars{1}$:
begin{align}
&color{#66f}{large%
int_{-infty}^{infty}!!%
{expo{ic nx} over Gammapars{alpha + x}Gammapars{beta - x}},dd x}
\[5mm] = &
{bracks{2cospars{n/2}}^{alpha + beta - 2}over
Gammapars{alpha + beta - 1}},expo{ic npars{beta - alpha}/2}
end{align}








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 0:16

























answered Jul 12 '14 at 7:50









Felix MarinFelix Marin

68.6k7109145




68.6k7109145












  • $begingroup$
    The manipulations with the delta function were really helpful to me.
    $endgroup$
    – daniel
    Sep 25 '14 at 7:54










  • $begingroup$
    @daniel It's nice it was helpful for you. Thanks.
    $endgroup$
    – Felix Marin
    Sep 25 '14 at 21:25










  • $begingroup$
    @daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
    $endgroup$
    – Did
    Dec 17 '18 at 0:22










  • $begingroup$
    @Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
    $endgroup$
    – daniel
    Dec 17 '18 at 7:06




















  • $begingroup$
    The manipulations with the delta function were really helpful to me.
    $endgroup$
    – daniel
    Sep 25 '14 at 7:54










  • $begingroup$
    @daniel It's nice it was helpful for you. Thanks.
    $endgroup$
    – Felix Marin
    Sep 25 '14 at 21:25










  • $begingroup$
    @daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
    $endgroup$
    – Did
    Dec 17 '18 at 0:22










  • $begingroup$
    @Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
    $endgroup$
    – daniel
    Dec 17 '18 at 7:06


















$begingroup$
The manipulations with the delta function were really helpful to me.
$endgroup$
– daniel
Sep 25 '14 at 7:54




$begingroup$
The manipulations with the delta function were really helpful to me.
$endgroup$
– daniel
Sep 25 '14 at 7:54












$begingroup$
@daniel It's nice it was helpful for you. Thanks.
$endgroup$
– Felix Marin
Sep 25 '14 at 21:25




$begingroup$
@daniel It's nice it was helpful for you. Thanks.
$endgroup$
– Felix Marin
Sep 25 '14 at 21:25












$begingroup$
@daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
$endgroup$
– Did
Dec 17 '18 at 0:22




$begingroup$
@daniel Be aware then, that these steps are usually considered to be lacking mathematical rigor. Depending on the context where you would want to rely on such tools, this could suffice to disqualify any method based on them.
$endgroup$
– Did
Dec 17 '18 at 0:22












$begingroup$
@Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
$endgroup$
– daniel
Dec 17 '18 at 7:06






$begingroup$
@Did: Thank you Did! I will think about this and try to see the weak links. I was not aware. I know in engineering the delta function is abused quite a lot.
$endgroup$
– daniel
Dec 17 '18 at 7:06




















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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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