Integral with two different answers using real and complex analysis
Clash Royale CLAN TAG#URR8PPP
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The integral is$$int_0^2pifracmathrm dθ2-cosθ.$$Just to skip time, the answer of the indefinite integral is $dfrac2sqrt3tan^-1left(sqrt3tanleft(dfracθ2right)right)$.
Evaluating it from $0$ to $ 2 pi$ yields$$frac2sqrt3tan^-1(sqrt3 tanπ)-frac2sqrt3tan^-1(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfracmathrm dzz^2-4z+1=2iint_Cfracmathrm dz(z-2+sqrt3)(z-2-sqrt3),$$
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:
$$2iint_Cfracmathrm dz(z-2+sqrt3)(z-2-sqrt3)=2πifrac2i2-sqrt3-2-sqrt3=2πifrac2i-2sqrt3=frac2πsqrt3.$$
Using real analysis I get $0$, using complex analysis I get $dfrac2πsqrt3$. What is wrong?
complex-analysis definite-integrals cauchy-integral-formula
edited Jan 2 at 1:22
Saad
19.7k92352
asked Jan 2 at 0:17
khaled014zkhaled014z
1107
$endgroup$
add a comment |
$begingroup$
The integral is$$int_0^2pifracmathrm dθ2-cosθ.$$Just to skip time, the answer of the indefinite integral is $dfrac2sqrt3tan^-1left(sqrt3tanleft(dfracθ2right)right)$.
Evaluating it from $0$ to $ 2 pi$ yields$$frac2sqrt3tan^-1(sqrt3 tanπ)-frac2sqrt3tan^-1(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfracmathrm dzz^2-4z+1=2iint_Cfracmathrm dz(z-2+sqrt3)(z-2-sqrt3),$$
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:
$$2iint_Cfracmathrm dz(z-2+sqrt3)(z-2-sqrt3)=2πifrac2i2-sqrt3-2-sqrt3=2πifrac2i-2sqrt3=frac2πsqrt3.$$
Using real analysis I get $0$, using complex analysis I get $dfrac2πsqrt3$. What is wrong?
complex-analysis definite-integrals cauchy-integral-formula
edited Jan 2 at 1:22
Saad
19.7k92352
asked Jan 2 at 0:17
khaled014zkhaled014z
1107
$endgroup$
2
$begingroup$
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.
$endgroup$
– Ben Millwood
Jan 2 at 12:56
add a comment |
$begingroup$
The integral is$$int_0^2pifracmathrm dθ2-cosθ.$$Just to skip time, the answer of the indefinite integral is $dfrac2sqrt3tan^-1left(sqrt3tanleft(dfracθ2right)right)$.
Evaluating it from $0$ to $ 2 pi$ yields$$frac2sqrt3tan^-1(sqrt3 tanπ)-frac2sqrt3tan^-1(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfracmathrm dzz^2-4z+1=2iint_Cfracmathrm dz(z-2+sqrt3)(z-2-sqrt3),$$
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:
$$2iint_Cfracmathrm dz(z-2+sqrt3)(z-2-sqrt3)=2πifrac2i2-sqrt3-2-sqrt3=2πifrac2i-2sqrt3=frac2πsqrt3.$$
Using real analysis I get $0$, using complex analysis I get $dfrac2πsqrt3$. What is wrong?
complex-analysis definite-integrals cauchy-integral-formula
edited Jan 2 at 1:22
Saad
19.7k92352
asked Jan 2 at 0:17
khaled014zkhaled014z
1107
$endgroup$
The integral is$$int_0^2pifracmathrm dθ2-cosθ.$$Just to skip time, the answer of the indefinite integral is $dfrac2sqrt3tan^-1left(sqrt3tanleft(dfracθ2right)right)$.
Evaluating it from $0$ to $ 2 pi$ yields$$frac2sqrt3tan^-1(sqrt3 tanπ)-frac2sqrt3tan^-1(sqrt3 tan0)=0-0=0.$$But using complex analysis, the integral is transformed into$$2iint_Cfracmathrm dzz^2-4z+1=2iint_Cfracmathrm dz(z-2+sqrt3)(z-2-sqrt3),$$
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:
$$2iint_Cfracmathrm dz(z-2+sqrt3)(z-2-sqrt3)=2πifrac2i2-sqrt3-2-sqrt3=2πifrac2i-2sqrt3=frac2πsqrt3.$$
Using real analysis I get $0$, using complex analysis I get $dfrac2πsqrt3$. What is wrong?
complex-analysis definite-integrals cauchy-integral-formula
complex-analysis definite-integrals cauchy-integral-formula
edited Jan 2 at 1:22
Saad
19.7k92352
asked Jan 2 at 0:17
khaled014zkhaled014z
1107
edited Jan 2 at 1:22
Saad
19.7k92352
asked Jan 2 at 0:17
khaled014zkhaled014z
1107
edited Jan 2 at 1:22
Saad
19.7k92352
edited Jan 2 at 1:22
Saad
19.7k92352
edited Jan 2 at 1:22
Saad
19.7k92352
19.7k92352
asked Jan 2 at 0:17
khaled014zkhaled014z
1107
asked Jan 2 at 0:17
khaled014zkhaled014z
1107
asked Jan 2 at 0:17
khaled014zkhaled014z
1107
1107
2
$begingroup$
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.
$endgroup$
– Ben Millwood
Jan 2 at 12:56
add a comment |
2
$begingroup$
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.
$endgroup$
– Ben Millwood
Jan 2 at 12:56
2
2
$begingroup$
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.
$endgroup$
– Ben Millwood
Jan 2 at 12:56
$begingroup$
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.
$endgroup$
– Ben Millwood
Jan 2 at 12:56
add a comment |
2 Answers
2
active
oldest
votes
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The problem with the real approach is that you make the change of variable $t=tanleft(dfractheta2right)$ 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 $arctan(+infty)$:
$$int_0^2 pi fracmathrmdθ2-cos theta=int_0^pi fracmathrmdθ2-cos theta+int_pi ^2 pi fracmathrmdθ2-cos theta\
= lim_r to pi_- int_0^r fracmathrmdθ2-cos theta+ lim_w to pi_+ int_w^2 pi fracmathrmdθ2-cos theta\= lim_r to pi_- left(frac2tan^-1( sqrt3 tan( frac r2)) sqrt3-0right)+ lim_w to pi_+left(0- frac2tan^-1( sqrt3 tan( frac r2)) sqrt3right).$$
edited Jan 2 at 0:55
Saad
19.7k92352
answered Jan 2 at 0:25
N. S.N. S.
103k6111208
$endgroup$
$begingroup$
Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
$endgroup$
– khaled014z
Jan 2 at 0:26
1
$begingroup$
@khaled014z See the edit. Let me know if you want more details.
$endgroup$
– N. S.
Jan 2 at 0:27
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Brilliant, that was kind of a tricky substitution, thank you
$endgroup$
– khaled014z
Jan 2 at 0:31
$begingroup$
When this is next edited, you want tan^-1 twice in the last line.
$endgroup$
– Teepeemm
Jan 2 at 14:12
add a comment |
$begingroup$
Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2sqrt3 arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.
Instead, we have
$$int_0^2pifrac12-cos(theta),dtheta=2int_0^pifrac12-cos(theta),dtheta=frac4sqrt3left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac2pisqrt3$$
answered Jan 2 at 0:28
Mark ViolaMark Viola
131k1275171
$endgroup$
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Don't you mean $x = fracpi2 + npi$?
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– DavidG
Jan 2 at 3:18
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Hi Mark ! Happy New Year !
$endgroup$
– Claude Leibovici
Jan 2 at 3:33
$begingroup$
@DavidG Yes, of course. Thank you for the comment.
$endgroup$
– Mark Viola
Jan 2 at 14:36
$begingroup$
@ClaudeLeibovici Hi Claude! Happy New Year to you too!
$endgroup$
– Mark Viola
Jan 2 at 14:36
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem with the real approach is that you make the change of variable $t=tanleft(dfractheta2right)$ 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 $arctan(+infty)$:
$$int_0^2 pi fracmathrmdθ2-cos theta=int_0^pi fracmathrmdθ2-cos theta+int_pi ^2 pi fracmathrmdθ2-cos theta\
= lim_r to pi_- int_0^r fracmathrmdθ2-cos theta+ lim_w to pi_+ int_w^2 pi fracmathrmdθ2-cos theta\= lim_r to pi_- left(frac2tan^-1( sqrt3 tan( frac r2)) sqrt3-0right)+ lim_w to pi_+left(0- frac2tan^-1( sqrt3 tan( frac r2)) sqrt3right).$$
edited Jan 2 at 0:55
Saad
19.7k92352
answered Jan 2 at 0:25
N. S.N. S.
103k6111208
$endgroup$
$begingroup$
Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
$endgroup$
– khaled014z
Jan 2 at 0:26
1
$begingroup$
@khaled014z See the edit. Let me know if you want more details.
$endgroup$
– N. S.
Jan 2 at 0:27
$begingroup$
Brilliant, that was kind of a tricky substitution, thank you
$endgroup$
– khaled014z
Jan 2 at 0:31
$begingroup$
When this is next edited, you want tan^-1 twice in the last line.
$endgroup$
– Teepeemm
Jan 2 at 14:12
add a comment |
$begingroup$
The problem with the real approach is that you make the change of variable $t=tanleft(dfractheta2right)$ 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 $arctan(+infty)$:
$$int_0^2 pi fracmathrmdθ2-cos theta=int_0^pi fracmathrmdθ2-cos theta+int_pi ^2 pi fracmathrmdθ2-cos theta\
= lim_r to pi_- int_0^r fracmathrmdθ2-cos theta+ lim_w to pi_+ int_w^2 pi fracmathrmdθ2-cos theta\= lim_r to pi_- left(frac2tan^-1( sqrt3 tan( frac r2)) sqrt3-0right)+ lim_w to pi_+left(0- frac2tan^-1( sqrt3 tan( frac r2)) sqrt3right).$$
edited Jan 2 at 0:55
Saad
19.7k92352
answered Jan 2 at 0:25
N. S.N. S.
103k6111208
$endgroup$
$begingroup$
Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
$endgroup$
– khaled014z
Jan 2 at 0:26
1
$begingroup$
@khaled014z See the edit. Let me know if you want more details.
$endgroup$
– N. S.
Jan 2 at 0:27
$begingroup$
Brilliant, that was kind of a tricky substitution, thank you
$endgroup$
– khaled014z
Jan 2 at 0:31
$begingroup$
When this is next edited, you want tan^-1 twice in the last line.
$endgroup$
– Teepeemm
Jan 2 at 14:12
add a comment |
$begingroup$
The problem with the real approach is that you make the change of variable $t=tanleft(dfractheta2right)$ 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 $arctan(+infty)$:
$$int_0^2 pi fracmathrmdθ2-cos theta=int_0^pi fracmathrmdθ2-cos theta+int_pi ^2 pi fracmathrmdθ2-cos theta\
= lim_r to pi_- int_0^r fracmathrmdθ2-cos theta+ lim_w to pi_+ int_w^2 pi fracmathrmdθ2-cos theta\= lim_r to pi_- left(frac2tan^-1( sqrt3 tan( frac r2)) sqrt3-0right)+ lim_w to pi_+left(0- frac2tan^-1( sqrt3 tan( frac r2)) sqrt3right).$$
edited Jan 2 at 0:55
Saad
19.7k92352
answered Jan 2 at 0:25
N. S.N. S.
103k6111208
$endgroup$
The problem with the real approach is that you make the change of variable $t=tanleft(dfractheta2right)$ 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 $arctan(+infty)$:
$$int_0^2 pi fracmathrmdθ2-cos theta=int_0^pi fracmathrmdθ2-cos theta+int_pi ^2 pi fracmathrmdθ2-cos theta\
= lim_r to pi_- int_0^r fracmathrmdθ2-cos theta+ lim_w to pi_+ int_w^2 pi fracmathrmdθ2-cos theta\= lim_r to pi_- left(frac2tan^-1( sqrt3 tan( frac r2)) sqrt3-0right)+ lim_w to pi_+left(0- frac2tan^-1( sqrt3 tan( frac r2)) sqrt3right).$$
edited Jan 2 at 0:55
Saad
19.7k92352
answered Jan 2 at 0:25
N. S.N. S.
103k6111208
edited Jan 2 at 0:55
Saad
19.7k92352
edited Jan 2 at 0:55
Saad
19.7k92352
edited Jan 2 at 0:55
Saad
19.7k92352
19.7k92352
answered Jan 2 at 0:25
N. S.N. S.
103k6111208
answered Jan 2 at 0:25
N. S.N. S.
103k6111208
answered Jan 2 at 0:25
N. S.N. S.
103k6111208
103k6111208
$begingroup$
Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
$endgroup$
– khaled014z
Jan 2 at 0:26
1
$begingroup$
@khaled014z See the edit. Let me know if you want more details.
$endgroup$
– N. S.
Jan 2 at 0:27
$begingroup$
Brilliant, that was kind of a tricky substitution, thank you
$endgroup$
– khaled014z
Jan 2 at 0:31
$begingroup$
When this is next edited, you want tan^-1 twice in the last line.
$endgroup$
– Teepeemm
Jan 2 at 14:12
add a comment |
$begingroup$
Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
$endgroup$
– khaled014z
Jan 2 at 0:26
1
$begingroup$
@khaled014z See the edit. Let me know if you want more details.
$endgroup$
– N. S.
Jan 2 at 0:27
$begingroup$
Brilliant, that was kind of a tricky substitution, thank you
$endgroup$
– khaled014z
Jan 2 at 0:31
$begingroup$
When this is next edited, you want tan^-1 twice in the last line.
$endgroup$
– Teepeemm
Jan 2 at 14:12
$begingroup$
Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
$endgroup$
– khaled014z
Jan 2 at 0:26
$begingroup$
Oh I see, so I have to solve it without this substitution? Or could I keep this substitution and change something else?
$endgroup$
– khaled014z
Jan 2 at 0:26
1
1
$begingroup$
@khaled014z See the edit. Let me know if you want more details.
$endgroup$
– N. S.
Jan 2 at 0:27
$begingroup$
@khaled014z See the edit. Let me know if you want more details.
$endgroup$
– N. S.
Jan 2 at 0:27
$begingroup$
Brilliant, that was kind of a tricky substitution, thank you
$endgroup$
– khaled014z
Jan 2 at 0:31
$begingroup$
Brilliant, that was kind of a tricky substitution, thank you
$endgroup$
– khaled014z
Jan 2 at 0:31
$begingroup$
When this is next edited, you want tan^-1 twice in the last line.
$endgroup$
– Teepeemm
Jan 2 at 14:12
$begingroup$
When this is next edited, you want tan^-1 twice in the last line.
$endgroup$
– Teepeemm
Jan 2 at 14:12
add a comment |
$begingroup$
Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2sqrt3 arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.
Instead, we have
$$int_0^2pifrac12-cos(theta),dtheta=2int_0^pifrac12-cos(theta),dtheta=frac4sqrt3left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac2pisqrt3$$
answered Jan 2 at 0:28
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Don't you mean $x = fracpi2 + npi$?
$endgroup$
– DavidG
Jan 2 at 3:18
$begingroup$
Hi Mark ! Happy New Year !
$endgroup$
– Claude Leibovici
Jan 2 at 3:33
$begingroup$
@DavidG Yes, of course. Thank you for the comment.
$endgroup$
– Mark Viola
Jan 2 at 14:36
$begingroup$
@ClaudeLeibovici Hi Claude! Happy New Year to you too!
$endgroup$
– Mark Viola
Jan 2 at 14:36
add a comment |
$begingroup$
Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2sqrt3 arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.
Instead, we have
$$int_0^2pifrac12-cos(theta),dtheta=2int_0^pifrac12-cos(theta),dtheta=frac4sqrt3left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac2pisqrt3$$
answered Jan 2 at 0:28
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Don't you mean $x = fracpi2 + npi$?
$endgroup$
– DavidG
Jan 2 at 3:18
$begingroup$
Hi Mark ! Happy New Year !
$endgroup$
– Claude Leibovici
Jan 2 at 3:33
$begingroup$
@DavidG Yes, of course. Thank you for the comment.
$endgroup$
– Mark Viola
Jan 2 at 14:36
$begingroup$
@ClaudeLeibovici Hi Claude! Happy New Year to you too!
$endgroup$
– Mark Viola
Jan 2 at 14:36
add a comment |
$begingroup$
Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2sqrt3 arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.
Instead, we have
$$int_0^2pifrac12-cos(theta),dtheta=2int_0^pifrac12-cos(theta),dtheta=frac4sqrt3left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac2pisqrt3$$
answered Jan 2 at 0:28
Mark ViolaMark Viola
131k1275171
$endgroup$
Note that that tangent function, $tan(x)$, is discontinuous when $x=pi/2+npi$. So, the antiderivative $frac2sqrt3 arctanleft(sqrt 3 tan(theta/2)right)$ is not valid over the interval $[0,2pi]$.
Instead, we have
$$int_0^2pifrac12-cos(theta),dtheta=2int_0^pifrac12-cos(theta),dtheta=frac4sqrt3left.left(arctanleft(sqrt 3 tan(theta/2)right)right)right|_0^pi=frac2pisqrt3$$
answered Jan 2 at 0:28
Mark ViolaMark Viola
131k1275171
edited Jan 2 at 14:35
answered Jan 2 at 0:28
Mark ViolaMark Viola
131k1275171
answered Jan 2 at 0:28
Mark ViolaMark Viola
131k1275171
answered Jan 2 at 0:28
Mark ViolaMark Viola
131k1275171
131k1275171
$begingroup$
Don't you mean $x = fracpi2 + npi$?
$endgroup$
– DavidG
Jan 2 at 3:18
$begingroup$
Hi Mark ! Happy New Year !
$endgroup$
– Claude Leibovici
Jan 2 at 3:33
$begingroup$
@DavidG Yes, of course. Thank you for the comment.
$endgroup$
– Mark Viola
Jan 2 at 14:36
$begingroup$
@ClaudeLeibovici Hi Claude! Happy New Year to you too!
$endgroup$
– Mark Viola
Jan 2 at 14:36
add a comment |
$begingroup$
Don't you mean $x = fracpi2 + npi$?
$endgroup$
– DavidG
Jan 2 at 3:18
$begingroup$
Hi Mark ! Happy New Year !
$endgroup$
– Claude Leibovici
Jan 2 at 3:33
$begingroup$
@DavidG Yes, of course. Thank you for the comment.
$endgroup$
– Mark Viola
Jan 2 at 14:36
$begingroup$
@ClaudeLeibovici Hi Claude! Happy New Year to you too!
$endgroup$
– Mark Viola
Jan 2 at 14:36
$begingroup$
Don't you mean $x = fracpi2 + npi$?
$endgroup$
– DavidG
Jan 2 at 3:18
$begingroup$
Don't you mean $x = fracpi2 + npi$?
$endgroup$
– DavidG
Jan 2 at 3:18
$begingroup$
Hi Mark ! Happy New Year !
$endgroup$
– Claude Leibovici
Jan 2 at 3:33
$begingroup$
Hi Mark ! Happy New Year !
$endgroup$
– Claude Leibovici
Jan 2 at 3:33
$begingroup$
@DavidG Yes, of course. Thank you for the comment.
$endgroup$
– Mark Viola
Jan 2 at 14:36
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@DavidG Yes, of course. Thank you for the comment.
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– Mark Viola
Jan 2 at 14:36
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@ClaudeLeibovici Hi Claude! Happy New Year to you too!
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– Mark Viola
Jan 2 at 14:36
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@ClaudeLeibovici Hi Claude! Happy New Year to you too!
$endgroup$
– Mark Viola
Jan 2 at 14:36
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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.
$endgroup$
– Ben Millwood
Jan 2 at 12:56