Integral with two different answers using real and complex analysis












9














The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
$$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






complex-analysis definite-integrals cauchy-integral-formula







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  • 2




    OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
    – Ben Millwood
    Jan 2 at 12:56
















9














The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
$$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






complex-analysis definite-integrals cauchy-integral-formula







share|cite|improve this question




















  • 2




    OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
    – Ben Millwood
    Jan 2 at 12:56














9












9








9


2





The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
$$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






complex-analysis definite-integrals cauchy-integral-formula







share|cite|improve this question















The integral is$$int_0^{2pi}frac{mathrm dθ}{2-cosθ}.$$Just to skip time, the answer of the indefinite integral is $dfrac2{sqrt{3}}tan^{-1}left(sqrt3tanleft(dfracθ2right)right)$.



Evaluating it from $0$ to $ 2 pi$ yields$$frac2{sqrt3}tan^{-1}(sqrt3 tanπ)-frac2{sqrt3}tan^{-1}(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfrac{mathrm dz}{z^2-4z+1}=2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)},$$
where $C$ is the boundary of the circle $|z|=1$. Then by Cauchy's integral formula, since $z=2-sqrt3$ is inside the domain of the region bounded by $C$, then:
$$2iint_Cfrac{mathrm dz}{(z-2+sqrt3)(z-2-sqrt3)}=2πifrac{2i}{2-sqrt3-2-sqrt3}=2πifrac{2i}{-2sqrt3}=frac{2π}{sqrt3}.$$



Using real analysis I get $0$, using complex analysis I get $dfrac{2π}{sqrt3}$. What is wrong?






complex-analysis definite-integrals cauchy-integral-formula




complex-analysis definite-integrals cauchy-integral-formula






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 1:22









Saad

19.7k92352




19.7k92352










asked Jan 2 at 0:17









khaled014zkhaled014z

1107




1107








  • 2




    OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
    – Ben Millwood
    Jan 2 at 12:56














  • 2




    OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
    – Ben Millwood
    Jan 2 at 12:56








2




2




OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
– Ben Millwood
Jan 2 at 12:56




OK, so I only spotted this after I already knew what the answer was, but nonetheless: notice that the integrand is always strictly positive, so the integral can't possibly be 0.
– Ben Millwood
Jan 2 at 12:56










2 Answers
2






active

oldest

votes


















13














The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






share|cite|improve this answer























  • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    – khaled014z
    Jan 2 at 0:26






  • 1




    @khaled014z See the edit. Let me know if you want more details.
    – N. S.
    Jan 2 at 0:27












  • Brilliant, that was kind of a tricky substitution, thank you
    – khaled014z
    Jan 2 at 0:31










  • When this is next edited, you want tan^{-1} twice in the last line.
    – Teepeemm
    Jan 2 at 14:12



















3














Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






share|cite|improve this answer























  • Don't you mean $x = frac{pi}{2} + npi$?
    – DavidG
    Jan 2 at 3:18










  • Hi Mark ! Happy New Year !
    – Claude Leibovici
    Jan 2 at 3:33










  • @DavidG Yes, of course. Thank you for the comment.
    – Mark Viola
    Jan 2 at 14:36










  • @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    – Mark Viola
    Jan 2 at 14:36











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2 Answers
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2 Answers
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13














The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






share|cite|improve this answer























  • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    – khaled014z
    Jan 2 at 0:26






  • 1




    @khaled014z See the edit. Let me know if you want more details.
    – N. S.
    Jan 2 at 0:27












  • Brilliant, that was kind of a tricky substitution, thank you
    – khaled014z
    Jan 2 at 0:31










  • When this is next edited, you want tan^{-1} twice in the last line.
    – Teepeemm
    Jan 2 at 14:12
















13














The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






share|cite|improve this answer























  • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    – khaled014z
    Jan 2 at 0:26






  • 1




    @khaled014z See the edit. Let me know if you want more details.
    – N. S.
    Jan 2 at 0:27












  • Brilliant, that was kind of a tricky substitution, thank you
    – khaled014z
    Jan 2 at 0:31










  • When this is next edited, you want tan^{-1} twice in the last line.
    – Teepeemm
    Jan 2 at 14:12














13












13








13






The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$






share|cite|improve this answer














The problem with the real approach is that you make the change of variable $t=tanleft(dfrac{theta}{2}right)$ for $0 < theta < 2 pi$.



This is problematic since your substitution need to be defined and continuous for all $theta$, but you have a problem when $theta=pi$.



Edit: Note that if you split the integral into $int_0^pi+int_pi^{2 pi}$, you are going to get the right answer, as for one integral you are going to get $arctan(- infty)$ and for the other $arctan(+infty)$:



$$int_0^{2 pi} frac{mathrm{d}θ}{2-cos theta}=int_0^pi frac{mathrm{d}θ}{2-cos theta}+int_pi ^{2 pi} frac{mathrm{d}θ}{2-cos theta}\
= lim_{r to pi_-} int_0^r frac{mathrm{d}θ}{2-cos theta}+ lim_{w to pi_+} int_w^{2 pi} frac{mathrm{d}θ}{2-cos theta}\= lim_{r to pi_-} left(frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}-0right)+ lim_{w to pi_+}left(0- frac{2tan^-1( sqrt{3} tan( frac{ r}{2}))}{ sqrt{3}}right).$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 0:55









Saad

19.7k92352




19.7k92352










answered Jan 2 at 0:25









N. S.N. S.

102k6111207




102k6111207












  • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    – khaled014z
    Jan 2 at 0:26






  • 1




    @khaled014z See the edit. Let me know if you want more details.
    – N. S.
    Jan 2 at 0:27












  • Brilliant, that was kind of a tricky substitution, thank you
    – khaled014z
    Jan 2 at 0:31










  • When this is next edited, you want tan^{-1} twice in the last line.
    – Teepeemm
    Jan 2 at 14:12


















  • Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
    – khaled014z
    Jan 2 at 0:26






  • 1




    @khaled014z See the edit. Let me know if you want more details.
    – N. S.
    Jan 2 at 0:27












  • Brilliant, that was kind of a tricky substitution, thank you
    – khaled014z
    Jan 2 at 0:31










  • When this is next edited, you want tan^{-1} twice in the last line.
    – Teepeemm
    Jan 2 at 14:12
















Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
– khaled014z
Jan 2 at 0:26




Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
– khaled014z
Jan 2 at 0:26




1




1




@khaled014z See the edit. Let me know if you want more details.
– N. S.
Jan 2 at 0:27






@khaled014z See the edit. Let me know if you want more details.
– N. S.
Jan 2 at 0:27














Brilliant, that was kind of a tricky substitution, thank you
– khaled014z
Jan 2 at 0:31




Brilliant, that was kind of a tricky substitution, thank you
– khaled014z
Jan 2 at 0:31












When this is next edited, you want tan^{-1} twice in the last line.
– Teepeemm
Jan 2 at 14:12




When this is next edited, you want tan^{-1} twice in the last line.
– Teepeemm
Jan 2 at 14:12











3














Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






share|cite|improve this answer























  • Don't you mean $x = frac{pi}{2} + npi$?
    – DavidG
    Jan 2 at 3:18










  • Hi Mark ! Happy New Year !
    – Claude Leibovici
    Jan 2 at 3:33










  • @DavidG Yes, of course. Thank you for the comment.
    – Mark Viola
    Jan 2 at 14:36










  • @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    – Mark Viola
    Jan 2 at 14:36
















3














Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






share|cite|improve this answer























  • Don't you mean $x = frac{pi}{2} + npi$?
    – DavidG
    Jan 2 at 3:18










  • Hi Mark ! Happy New Year !
    – Claude Leibovici
    Jan 2 at 3:33










  • @DavidG Yes, of course. Thank you for the comment.
    – Mark Viola
    Jan 2 at 14:36










  • @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    – Mark Viola
    Jan 2 at 14:36














3












3








3






Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$






share|cite|improve this answer














Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2{sqrt{3}} arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.



Instead, we have



$$int_0^{2pi}frac{1}{2-cos(theta)},dtheta=2int_0^pifrac{1}{2-cos(theta)},dtheta=frac{4}{sqrt3}left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac{2pi}{sqrt3}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 2 at 14:35

























answered Jan 2 at 0:28









Mark ViolaMark Viola

130k1274170




130k1274170












  • Don't you mean $x = frac{pi}{2} + npi$?
    – DavidG
    Jan 2 at 3:18










  • Hi Mark ! Happy New Year !
    – Claude Leibovici
    Jan 2 at 3:33










  • @DavidG Yes, of course. Thank you for the comment.
    – Mark Viola
    Jan 2 at 14:36










  • @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    – Mark Viola
    Jan 2 at 14:36


















  • Don't you mean $x = frac{pi}{2} + npi$?
    – DavidG
    Jan 2 at 3:18










  • Hi Mark ! Happy New Year !
    – Claude Leibovici
    Jan 2 at 3:33










  • @DavidG Yes, of course. Thank you for the comment.
    – Mark Viola
    Jan 2 at 14:36










  • @ClaudeLeibovici Hi Claude! Happy New Year to you too!
    – Mark Viola
    Jan 2 at 14:36
















Don't you mean $x = frac{pi}{2} + npi$?
– DavidG
Jan 2 at 3:18




Don't you mean $x = frac{pi}{2} + npi$?
– DavidG
Jan 2 at 3:18












Hi Mark ! Happy New Year !
– Claude Leibovici
Jan 2 at 3:33




Hi Mark ! Happy New Year !
– Claude Leibovici
Jan 2 at 3:33












@DavidG Yes, of course. Thank you for the comment.
– Mark Viola
Jan 2 at 14:36




@DavidG Yes, of course. Thank you for the comment.
– Mark Viola
Jan 2 at 14:36












@ClaudeLeibovici Hi Claude! Happy New Year to you too!
– Mark Viola
Jan 2 at 14:36




@ClaudeLeibovici Hi Claude! Happy New Year to you too!
– Mark Viola
Jan 2 at 14:36


















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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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