Integral involving hypergeometric function












2












$begingroup$


I've worked out the projection of a spherically symmetric power law volume density profile $rho(r)=br^a$, i.e. its surface density $sigma(R)$, and am now trying to integrate this in a series of annular bins to get the total mass in each, i.e. $int_A^Bsigma(R)R,{rm d}R$. (Actually the profile is a continuous series of power laws, but that's more of a footnote.)



I've worked just about everything out, but am stuck on an integral of the form:



$$I(r,R,a)=intleft(left(frac{r}{R}right)^2-1right)^{frac{1}{2}}R^{a+2} {}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};-left(left(frac{r}{R}right)^2-1right)right){rm d}R$$



${}_2{rm F}_1$ is the hypergeometric function, as defined here. $a$ and $r$ are constants, in this context.



Given that the solution of this integral will be used to calculate the elements of a big-ish matrix and I have some iteration in mind, integrating numerically is... inconvenient, but I'm stumped on finding an analytic solution. Is there any hope here, or should I start thinking about efficient numerical solutions?






calculus integration indefinite-integrals hypergeometric-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try expanding the Hypergeometric then integrating term by term
    $endgroup$
    – clathratus
    Jan 10 at 19:34










  • $begingroup$
    @Kyle thank you very much for your generous bounty!
    $endgroup$
    – Paul Enta
    Jan 22 at 9:07










  • $begingroup$
    @PaulEnta thanks for the answer! The solution of the integral goes into a computation for a measurement in astrophysics/cosmology I'm collaborating on at the moment, which we should start writing up for a journal soon. If you would like, I'd be happy to add an acknowledgement of your contribution along the lines of "The authors thank the math.stackexchange.com user Paul Enta for providing an analytic expression for eq X", or similar. There is usually a short section with such notes anyway. You may reach me by email at either of the addresses listed here.
    $endgroup$
    – Kyle
    Jan 22 at 9:20










  • $begingroup$
    I am happy to see that this answer can be useful to you. Thank you for proposing to acknowledge this contribution in your future paper. I think that it can be a good advertisment for MSE in your community. I wish you the best for this article and for the next ones!
    $endgroup$
    – Paul Enta
    Jan 22 at 9:59
















2












$begingroup$


I've worked out the projection of a spherically symmetric power law volume density profile $rho(r)=br^a$, i.e. its surface density $sigma(R)$, and am now trying to integrate this in a series of annular bins to get the total mass in each, i.e. $int_A^Bsigma(R)R,{rm d}R$. (Actually the profile is a continuous series of power laws, but that's more of a footnote.)



I've worked just about everything out, but am stuck on an integral of the form:



$$I(r,R,a)=intleft(left(frac{r}{R}right)^2-1right)^{frac{1}{2}}R^{a+2} {}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};-left(left(frac{r}{R}right)^2-1right)right){rm d}R$$



${}_2{rm F}_1$ is the hypergeometric function, as defined here. $a$ and $r$ are constants, in this context.



Given that the solution of this integral will be used to calculate the elements of a big-ish matrix and I have some iteration in mind, integrating numerically is... inconvenient, but I'm stumped on finding an analytic solution. Is there any hope here, or should I start thinking about efficient numerical solutions?






calculus integration indefinite-integrals hypergeometric-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try expanding the Hypergeometric then integrating term by term
    $endgroup$
    – clathratus
    Jan 10 at 19:34










  • $begingroup$
    @Kyle thank you very much for your generous bounty!
    $endgroup$
    – Paul Enta
    Jan 22 at 9:07










  • $begingroup$
    @PaulEnta thanks for the answer! The solution of the integral goes into a computation for a measurement in astrophysics/cosmology I'm collaborating on at the moment, which we should start writing up for a journal soon. If you would like, I'd be happy to add an acknowledgement of your contribution along the lines of "The authors thank the math.stackexchange.com user Paul Enta for providing an analytic expression for eq X", or similar. There is usually a short section with such notes anyway. You may reach me by email at either of the addresses listed here.
    $endgroup$
    – Kyle
    Jan 22 at 9:20










  • $begingroup$
    I am happy to see that this answer can be useful to you. Thank you for proposing to acknowledge this contribution in your future paper. I think that it can be a good advertisment for MSE in your community. I wish you the best for this article and for the next ones!
    $endgroup$
    – Paul Enta
    Jan 22 at 9:59














2












2








2





$begingroup$


I've worked out the projection of a spherically symmetric power law volume density profile $rho(r)=br^a$, i.e. its surface density $sigma(R)$, and am now trying to integrate this in a series of annular bins to get the total mass in each, i.e. $int_A^Bsigma(R)R,{rm d}R$. (Actually the profile is a continuous series of power laws, but that's more of a footnote.)



I've worked just about everything out, but am stuck on an integral of the form:



$$I(r,R,a)=intleft(left(frac{r}{R}right)^2-1right)^{frac{1}{2}}R^{a+2} {}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};-left(left(frac{r}{R}right)^2-1right)right){rm d}R$$



${}_2{rm F}_1$ is the hypergeometric function, as defined here. $a$ and $r$ are constants, in this context.



Given that the solution of this integral will be used to calculate the elements of a big-ish matrix and I have some iteration in mind, integrating numerically is... inconvenient, but I'm stumped on finding an analytic solution. Is there any hope here, or should I start thinking about efficient numerical solutions?






calculus integration indefinite-integrals hypergeometric-function







share|cite|improve this question









$endgroup$




I've worked out the projection of a spherically symmetric power law volume density profile $rho(r)=br^a$, i.e. its surface density $sigma(R)$, and am now trying to integrate this in a series of annular bins to get the total mass in each, i.e. $int_A^Bsigma(R)R,{rm d}R$. (Actually the profile is a continuous series of power laws, but that's more of a footnote.)



I've worked just about everything out, but am stuck on an integral of the form:



$$I(r,R,a)=intleft(left(frac{r}{R}right)^2-1right)^{frac{1}{2}}R^{a+2} {}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};-left(left(frac{r}{R}right)^2-1right)right){rm d}R$$



${}_2{rm F}_1$ is the hypergeometric function, as defined here. $a$ and $r$ are constants, in this context.



Given that the solution of this integral will be used to calculate the elements of a big-ish matrix and I have some iteration in mind, integrating numerically is... inconvenient, but I'm stumped on finding an analytic solution. Is there any hope here, or should I start thinking about efficient numerical solutions?






calculus integration indefinite-integrals hypergeometric-function




calculus integration indefinite-integrals hypergeometric-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 15:43









KyleKyle

18328




18328












  • $begingroup$
    Try expanding the Hypergeometric then integrating term by term
    $endgroup$
    – clathratus
    Jan 10 at 19:34










  • $begingroup$
    @Kyle thank you very much for your generous bounty!
    $endgroup$
    – Paul Enta
    Jan 22 at 9:07










  • $begingroup$
    @PaulEnta thanks for the answer! The solution of the integral goes into a computation for a measurement in astrophysics/cosmology I'm collaborating on at the moment, which we should start writing up for a journal soon. If you would like, I'd be happy to add an acknowledgement of your contribution along the lines of "The authors thank the math.stackexchange.com user Paul Enta for providing an analytic expression for eq X", or similar. There is usually a short section with such notes anyway. You may reach me by email at either of the addresses listed here.
    $endgroup$
    – Kyle
    Jan 22 at 9:20










  • $begingroup$
    I am happy to see that this answer can be useful to you. Thank you for proposing to acknowledge this contribution in your future paper. I think that it can be a good advertisment for MSE in your community. I wish you the best for this article and for the next ones!
    $endgroup$
    – Paul Enta
    Jan 22 at 9:59


















  • $begingroup$
    Try expanding the Hypergeometric then integrating term by term
    $endgroup$
    – clathratus
    Jan 10 at 19:34










  • $begingroup$
    @Kyle thank you very much for your generous bounty!
    $endgroup$
    – Paul Enta
    Jan 22 at 9:07










  • $begingroup$
    @PaulEnta thanks for the answer! The solution of the integral goes into a computation for a measurement in astrophysics/cosmology I'm collaborating on at the moment, which we should start writing up for a journal soon. If you would like, I'd be happy to add an acknowledgement of your contribution along the lines of "The authors thank the math.stackexchange.com user Paul Enta for providing an analytic expression for eq X", or similar. There is usually a short section with such notes anyway. You may reach me by email at either of the addresses listed here.
    $endgroup$
    – Kyle
    Jan 22 at 9:20










  • $begingroup$
    I am happy to see that this answer can be useful to you. Thank you for proposing to acknowledge this contribution in your future paper. I think that it can be a good advertisment for MSE in your community. I wish you the best for this article and for the next ones!
    $endgroup$
    – Paul Enta
    Jan 22 at 9:59
















$begingroup$
Try expanding the Hypergeometric then integrating term by term
$endgroup$
– clathratus
Jan 10 at 19:34




$begingroup$
Try expanding the Hypergeometric then integrating term by term
$endgroup$
– clathratus
Jan 10 at 19:34












$begingroup$
@Kyle thank you very much for your generous bounty!
$endgroup$
– Paul Enta
Jan 22 at 9:07




$begingroup$
@Kyle thank you very much for your generous bounty!
$endgroup$
– Paul Enta
Jan 22 at 9:07












$begingroup$
@PaulEnta thanks for the answer! The solution of the integral goes into a computation for a measurement in astrophysics/cosmology I'm collaborating on at the moment, which we should start writing up for a journal soon. If you would like, I'd be happy to add an acknowledgement of your contribution along the lines of "The authors thank the math.stackexchange.com user Paul Enta for providing an analytic expression for eq X", or similar. There is usually a short section with such notes anyway. You may reach me by email at either of the addresses listed here.
$endgroup$
– Kyle
Jan 22 at 9:20




$begingroup$
@PaulEnta thanks for the answer! The solution of the integral goes into a computation for a measurement in astrophysics/cosmology I'm collaborating on at the moment, which we should start writing up for a journal soon. If you would like, I'd be happy to add an acknowledgement of your contribution along the lines of "The authors thank the math.stackexchange.com user Paul Enta for providing an analytic expression for eq X", or similar. There is usually a short section with such notes anyway. You may reach me by email at either of the addresses listed here.
$endgroup$
– Kyle
Jan 22 at 9:20












$begingroup$
I am happy to see that this answer can be useful to you. Thank you for proposing to acknowledge this contribution in your future paper. I think that it can be a good advertisment for MSE in your community. I wish you the best for this article and for the next ones!
$endgroup$
– Paul Enta
Jan 22 at 9:59




$begingroup$
I am happy to see that this answer can be useful to you. Thank you for proposing to acknowledge this contribution in your future paper. I think that it can be a good advertisment for MSE in your community. I wish you the best for this article and for the next ones!
$endgroup$
– Paul Enta
Jan 22 at 9:59










1 Answer
1






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oldest

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5





+50







$begingroup$

In the following, we assume $R>r$.
begin{equation}
I(r,R,a)=intleft(frac{r^2}{R^2}-1right)^{frac{1}{2}}R^{a+2} {}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};1-frac{r^2}{R^2}right){rm d}R
end{equation}
changing $u=1-r^2/R^2$, we obtain
begin{equation}
I(r,R,a)=ifrac{r^{a+3}}{2}int u^{frac12}left( 1-u right)^{-frac{a}{2}-frac{5}{2}}{}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};uright){rm d}u
end{equation}
where the principal determination for the square root was taken. Using the indefinite integration formula:
begin{equation}
int z^{c-1 }(1 - z)^{b - c-1}{}_2{rm F}_1left( a,b;c;z right)= z^c(1 - z)^{b - c} frac{Gamma(c)}{Gamma(c+1)} {}_2{rm F}_1left( 1+a,b;1+c;z right)
end{equation}
with $a=1/2,b=-a/2,c=3/2$, we find
begin{equation}
I(r,R,a)=-frac{1}{3}R^{a+3}left( frac{r^2}{R^2}-1 right)^{3/2} {}_2{rm F}_1left(frac{3}{2},-frac{a}{2};frac{5}{2};1-frac{r^2}{R^2}right)
end{equation}
which seems to be numerically correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Checks out, many thanks! Out of curiosity, and for future use, what led you to that integral identity for the hypergeometric function?
    $endgroup$
    – Kyle
    Jan 11 at 9:29






  • 1




    $begingroup$
    There are not so many indefinite integrals for the hypergeometric functions and yours seemed to be "well conditioned".
    $endgroup$
    – Paul Enta
    Jan 11 at 9:33











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5





+50







$begingroup$

In the following, we assume $R>r$.
begin{equation}
I(r,R,a)=intleft(frac{r^2}{R^2}-1right)^{frac{1}{2}}R^{a+2} {}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};1-frac{r^2}{R^2}right){rm d}R
end{equation}
changing $u=1-r^2/R^2$, we obtain
begin{equation}
I(r,R,a)=ifrac{r^{a+3}}{2}int u^{frac12}left( 1-u right)^{-frac{a}{2}-frac{5}{2}}{}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};uright){rm d}u
end{equation}
where the principal determination for the square root was taken. Using the indefinite integration formula:
begin{equation}
int z^{c-1 }(1 - z)^{b - c-1}{}_2{rm F}_1left( a,b;c;z right)= z^c(1 - z)^{b - c} frac{Gamma(c)}{Gamma(c+1)} {}_2{rm F}_1left( 1+a,b;1+c;z right)
end{equation}
with $a=1/2,b=-a/2,c=3/2$, we find
begin{equation}
I(r,R,a)=-frac{1}{3}R^{a+3}left( frac{r^2}{R^2}-1 right)^{3/2} {}_2{rm F}_1left(frac{3}{2},-frac{a}{2};frac{5}{2};1-frac{r^2}{R^2}right)
end{equation}
which seems to be numerically correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Checks out, many thanks! Out of curiosity, and for future use, what led you to that integral identity for the hypergeometric function?
    $endgroup$
    – Kyle
    Jan 11 at 9:29






  • 1




    $begingroup$
    There are not so many indefinite integrals for the hypergeometric functions and yours seemed to be "well conditioned".
    $endgroup$
    – Paul Enta
    Jan 11 at 9:33
















5





+50







$begingroup$

In the following, we assume $R>r$.
begin{equation}
I(r,R,a)=intleft(frac{r^2}{R^2}-1right)^{frac{1}{2}}R^{a+2} {}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};1-frac{r^2}{R^2}right){rm d}R
end{equation}
changing $u=1-r^2/R^2$, we obtain
begin{equation}
I(r,R,a)=ifrac{r^{a+3}}{2}int u^{frac12}left( 1-u right)^{-frac{a}{2}-frac{5}{2}}{}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};uright){rm d}u
end{equation}
where the principal determination for the square root was taken. Using the indefinite integration formula:
begin{equation}
int z^{c-1 }(1 - z)^{b - c-1}{}_2{rm F}_1left( a,b;c;z right)= z^c(1 - z)^{b - c} frac{Gamma(c)}{Gamma(c+1)} {}_2{rm F}_1left( 1+a,b;1+c;z right)
end{equation}
with $a=1/2,b=-a/2,c=3/2$, we find
begin{equation}
I(r,R,a)=-frac{1}{3}R^{a+3}left( frac{r^2}{R^2}-1 right)^{3/2} {}_2{rm F}_1left(frac{3}{2},-frac{a}{2};frac{5}{2};1-frac{r^2}{R^2}right)
end{equation}
which seems to be numerically correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Checks out, many thanks! Out of curiosity, and for future use, what led you to that integral identity for the hypergeometric function?
    $endgroup$
    – Kyle
    Jan 11 at 9:29






  • 1




    $begingroup$
    There are not so many indefinite integrals for the hypergeometric functions and yours seemed to be "well conditioned".
    $endgroup$
    – Paul Enta
    Jan 11 at 9:33














5





+50







5





+50



5




+50



$begingroup$

In the following, we assume $R>r$.
begin{equation}
I(r,R,a)=intleft(frac{r^2}{R^2}-1right)^{frac{1}{2}}R^{a+2} {}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};1-frac{r^2}{R^2}right){rm d}R
end{equation}
changing $u=1-r^2/R^2$, we obtain
begin{equation}
I(r,R,a)=ifrac{r^{a+3}}{2}int u^{frac12}left( 1-u right)^{-frac{a}{2}-frac{5}{2}}{}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};uright){rm d}u
end{equation}
where the principal determination for the square root was taken. Using the indefinite integration formula:
begin{equation}
int z^{c-1 }(1 - z)^{b - c-1}{}_2{rm F}_1left( a,b;c;z right)= z^c(1 - z)^{b - c} frac{Gamma(c)}{Gamma(c+1)} {}_2{rm F}_1left( 1+a,b;1+c;z right)
end{equation}
with $a=1/2,b=-a/2,c=3/2$, we find
begin{equation}
I(r,R,a)=-frac{1}{3}R^{a+3}left( frac{r^2}{R^2}-1 right)^{3/2} {}_2{rm F}_1left(frac{3}{2},-frac{a}{2};frac{5}{2};1-frac{r^2}{R^2}right)
end{equation}
which seems to be numerically correct.






share|cite|improve this answer









$endgroup$



In the following, we assume $R>r$.
begin{equation}
I(r,R,a)=intleft(frac{r^2}{R^2}-1right)^{frac{1}{2}}R^{a+2} {}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};1-frac{r^2}{R^2}right){rm d}R
end{equation}
changing $u=1-r^2/R^2$, we obtain
begin{equation}
I(r,R,a)=ifrac{r^{a+3}}{2}int u^{frac12}left( 1-u right)^{-frac{a}{2}-frac{5}{2}}{}_2{rm F}_1left(frac{1}{2},-frac{a}{2};frac{3}{2};uright){rm d}u
end{equation}
where the principal determination for the square root was taken. Using the indefinite integration formula:
begin{equation}
int z^{c-1 }(1 - z)^{b - c-1}{}_2{rm F}_1left( a,b;c;z right)= z^c(1 - z)^{b - c} frac{Gamma(c)}{Gamma(c+1)} {}_2{rm F}_1left( 1+a,b;1+c;z right)
end{equation}
with $a=1/2,b=-a/2,c=3/2$, we find
begin{equation}
I(r,R,a)=-frac{1}{3}R^{a+3}left( frac{r^2}{R^2}-1 right)^{3/2} {}_2{rm F}_1left(frac{3}{2},-frac{a}{2};frac{5}{2};1-frac{r^2}{R^2}right)
end{equation}
which seems to be numerically correct.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 21:25









Paul EntaPaul Enta

5,18111334




5,18111334












  • $begingroup$
    Checks out, many thanks! Out of curiosity, and for future use, what led you to that integral identity for the hypergeometric function?
    $endgroup$
    – Kyle
    Jan 11 at 9:29






  • 1




    $begingroup$
    There are not so many indefinite integrals for the hypergeometric functions and yours seemed to be "well conditioned".
    $endgroup$
    – Paul Enta
    Jan 11 at 9:33


















  • $begingroup$
    Checks out, many thanks! Out of curiosity, and for future use, what led you to that integral identity for the hypergeometric function?
    $endgroup$
    – Kyle
    Jan 11 at 9:29






  • 1




    $begingroup$
    There are not so many indefinite integrals for the hypergeometric functions and yours seemed to be "well conditioned".
    $endgroup$
    – Paul Enta
    Jan 11 at 9:33
















$begingroup$
Checks out, many thanks! Out of curiosity, and for future use, what led you to that integral identity for the hypergeometric function?
$endgroup$
– Kyle
Jan 11 at 9:29




$begingroup$
Checks out, many thanks! Out of curiosity, and for future use, what led you to that integral identity for the hypergeometric function?
$endgroup$
– Kyle
Jan 11 at 9:29




1




1




$begingroup$
There are not so many indefinite integrals for the hypergeometric functions and yours seemed to be "well conditioned".
$endgroup$
– Paul Enta
Jan 11 at 9:33




$begingroup$
There are not so many indefinite integrals for the hypergeometric functions and yours seemed to be "well conditioned".
$endgroup$
– Paul Enta
Jan 11 at 9:33


















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Human spaceflight

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Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
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Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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