$int_{0}^{2pi}frac{dtheta}{a+bsintheta}$ where $a,b>0$ [duplicate]
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This question already has an answer here:
Evaluating $int_0^{2pi} frac {cos(theta)}{5-3cos(theta)}$ using residue theorem
1 answer
$int_{0}^{2pi}dfrac{dtheta}{a+bsintheta}$ where $a,b>0$
Supposedly a very simple residue theorem problem but I'm stuck with the pole(s).
Let $z=e^{itheta}$ and $dtheta=frac{dz}{iz}$. Then,
$$begin{align*}
int_{mathbb{D}}frac{dz}{left[a+bleft(frac{z-z^{-1}}{2i}right)right]iz}&=int_{mathbb{D}}frac{2idz}{(2ai+bz-bz^{-1})iz}\
&=int_{mathbb{D}}frac{2dz}{bz^2+2aiz-b}
end{align*}$$
By quadratic formula, we can deduce that we have pole at $$z=frac{-aipmsqrt{b^2-a^2}}{b}$$
But I'm having trouble figuring out which, of the two is in fact in the unit circle. All we have is $a>0$ and $b>0$ and I feel like it could go so many different ways depending on the actual values that they take. Can anyone help me to proceed?
complex-analysis residue-calculus
edited Dec 28 '18 at 16:50
Martin Sleziak
44.9k10122277
asked Dec 23 '18 at 17:35
Ya GYa G
536211
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marked as duplicate by Jack D'Aurizio complex-analysis
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Dec 29 '18 at 21:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Evaluating $int_0^{2pi} frac {cos(theta)}{5-3cos(theta)}$ using residue theorem
1 answer
$int_{0}^{2pi}dfrac{dtheta}{a+bsintheta}$ where $a,b>0$
Supposedly a very simple residue theorem problem but I'm stuck with the pole(s).
Let $z=e^{itheta}$ and $dtheta=frac{dz}{iz}$. Then,
$$begin{align*}
int_{mathbb{D}}frac{dz}{left[a+bleft(frac{z-z^{-1}}{2i}right)right]iz}&=int_{mathbb{D}}frac{2idz}{(2ai+bz-bz^{-1})iz}\
&=int_{mathbb{D}}frac{2dz}{bz^2+2aiz-b}
end{align*}$$
By quadratic formula, we can deduce that we have pole at $$z=frac{-aipmsqrt{b^2-a^2}}{b}$$
But I'm having trouble figuring out which, of the two is in fact in the unit circle. All we have is $a>0$ and $b>0$ and I feel like it could go so many different ways depending on the actual values that they take. Can anyone help me to proceed?
complex-analysis residue-calculus
edited Dec 28 '18 at 16:50
Martin Sleziak
44.9k10122277
asked Dec 23 '18 at 17:35
Ya GYa G
536211
$endgroup$
marked as duplicate by Jack D'Aurizio complex-analysis
Users with the complex-analysis badge can single-handedly close complex-analysis questions as duplicates and reopen them as needed.
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Dec 29 '18 at 21:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
@metamorphy right! That was my initial thought too, until I actually tried to get my hands dirty with it. May be I’m doing something wrong.
$endgroup$
– Ya G
Dec 23 '18 at 17:46
$begingroup$
(Oops.) For $aleqslant b$ the integral diverges. Otherwise you're doing right.
$endgroup$
– metamorphy
Dec 23 '18 at 17:50
$begingroup$
@metamorphy edited!
$endgroup$
– Ya G
Dec 23 '18 at 17:54
$begingroup$
Now it should be easy for you (the poles are purely imaginary, one inside, the other outside).
$endgroup$
– metamorphy
Dec 23 '18 at 18:00
$begingroup$
@metamorphy I feel like I’m still having trouble computing for different relations between $a$ and $b$ these for some reason.
$endgroup$
– Ya G
Dec 23 '18 at 18:02
add a comment |
$begingroup$
This question already has an answer here:
Evaluating $int_0^{2pi} frac {cos(theta)}{5-3cos(theta)}$ using residue theorem
1 answer
$int_{0}^{2pi}dfrac{dtheta}{a+bsintheta}$ where $a,b>0$
Supposedly a very simple residue theorem problem but I'm stuck with the pole(s).
Let $z=e^{itheta}$ and $dtheta=frac{dz}{iz}$. Then,
$$begin{align*}
int_{mathbb{D}}frac{dz}{left[a+bleft(frac{z-z^{-1}}{2i}right)right]iz}&=int_{mathbb{D}}frac{2idz}{(2ai+bz-bz^{-1})iz}\
&=int_{mathbb{D}}frac{2dz}{bz^2+2aiz-b}
end{align*}$$
By quadratic formula, we can deduce that we have pole at $$z=frac{-aipmsqrt{b^2-a^2}}{b}$$
But I'm having trouble figuring out which, of the two is in fact in the unit circle. All we have is $a>0$ and $b>0$ and I feel like it could go so many different ways depending on the actual values that they take. Can anyone help me to proceed?
complex-analysis residue-calculus
edited Dec 28 '18 at 16:50
Martin Sleziak
44.9k10122277
asked Dec 23 '18 at 17:35
Ya GYa G
536211
$endgroup$
This question already has an answer here:
Evaluating $int_0^{2pi} frac {cos(theta)}{5-3cos(theta)}$ using residue theorem
1 answer
$int_{0}^{2pi}dfrac{dtheta}{a+bsintheta}$ where $a,b>0$
Supposedly a very simple residue theorem problem but I'm stuck with the pole(s).
Let $z=e^{itheta}$ and $dtheta=frac{dz}{iz}$. Then,
$$begin{align*}
int_{mathbb{D}}frac{dz}{left[a+bleft(frac{z-z^{-1}}{2i}right)right]iz}&=int_{mathbb{D}}frac{2idz}{(2ai+bz-bz^{-1})iz}\
&=int_{mathbb{D}}frac{2dz}{bz^2+2aiz-b}
end{align*}$$
By quadratic formula, we can deduce that we have pole at $$z=frac{-aipmsqrt{b^2-a^2}}{b}$$
But I'm having trouble figuring out which, of the two is in fact in the unit circle. All we have is $a>0$ and $b>0$ and I feel like it could go so many different ways depending on the actual values that they take. Can anyone help me to proceed?
This question already has an answer here:
Evaluating $int_0^{2pi} frac {cos(theta)}{5-3cos(theta)}$ using residue theorem
1 answer
complex-analysis residue-calculus
complex-analysis residue-calculus
edited Dec 28 '18 at 16:50
Martin Sleziak
44.9k10122277
asked Dec 23 '18 at 17:35
Ya GYa G
536211
edited Dec 28 '18 at 16:50
Martin Sleziak
44.9k10122277
asked Dec 23 '18 at 17:35
Ya GYa G
536211
edited Dec 28 '18 at 16:50
Martin Sleziak
44.9k10122277
edited Dec 28 '18 at 16:50
Martin Sleziak
44.9k10122277
edited Dec 28 '18 at 16:50
Martin Sleziak
44.9k10122277
44.9k10122277
asked Dec 23 '18 at 17:35
Ya GYa G
536211
asked Dec 23 '18 at 17:35
Ya GYa G
536211
asked Dec 23 '18 at 17:35
Ya GYa G
536211
536211
marked as duplicate by Jack D'Aurizio complex-analysis
Users with the complex-analysis badge can single-handedly close complex-analysis questions as duplicates and reopen them as needed.
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Dec 29 '18 at 21:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jack D'Aurizio complex-analysis
Users with the complex-analysis badge can single-handedly close complex-analysis questions as duplicates and reopen them as needed.
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Dec 29 '18 at 21:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
@metamorphy right! That was my initial thought too, until I actually tried to get my hands dirty with it. May be I’m doing something wrong.
$endgroup$
– Ya G
Dec 23 '18 at 17:46
$begingroup$
(Oops.) For $aleqslant b$ the integral diverges. Otherwise you're doing right.
$endgroup$
– metamorphy
Dec 23 '18 at 17:50
$begingroup$
@metamorphy edited!
$endgroup$
– Ya G
Dec 23 '18 at 17:54
$begingroup$
Now it should be easy for you (the poles are purely imaginary, one inside, the other outside).
$endgroup$
– metamorphy
Dec 23 '18 at 18:00
$begingroup$
@metamorphy I feel like I’m still having trouble computing for different relations between $a$ and $b$ these for some reason.
$endgroup$
– Ya G
Dec 23 '18 at 18:02
add a comment |
$begingroup$
@metamorphy right! That was my initial thought too, until I actually tried to get my hands dirty with it. May be I’m doing something wrong.
$endgroup$
– Ya G
Dec 23 '18 at 17:46
$begingroup$
(Oops.) For $aleqslant b$ the integral diverges. Otherwise you're doing right.
$endgroup$
– metamorphy
Dec 23 '18 at 17:50
$begingroup$
@metamorphy edited!
$endgroup$
– Ya G
Dec 23 '18 at 17:54
$begingroup$
Now it should be easy for you (the poles are purely imaginary, one inside, the other outside).
$endgroup$
– metamorphy
Dec 23 '18 at 18:00
$begingroup$
@metamorphy I feel like I’m still having trouble computing for different relations between $a$ and $b$ these for some reason.
$endgroup$
– Ya G
Dec 23 '18 at 18:02
$begingroup$
@metamorphy right! That was my initial thought too, until I actually tried to get my hands dirty with it. May be I’m doing something wrong.
$endgroup$
– Ya G
Dec 23 '18 at 17:46
$begingroup$
@metamorphy right! That was my initial thought too, until I actually tried to get my hands dirty with it. May be I’m doing something wrong.
$endgroup$
– Ya G
Dec 23 '18 at 17:46
$begingroup$
(Oops.) For $aleqslant b$ the integral diverges. Otherwise you're doing right.
$endgroup$
– metamorphy
Dec 23 '18 at 17:50
$begingroup$
(Oops.) For $aleqslant b$ the integral diverges. Otherwise you're doing right.
$endgroup$
– metamorphy
Dec 23 '18 at 17:50
$begingroup$
@metamorphy edited!
$endgroup$
– Ya G
Dec 23 '18 at 17:54
$begingroup$
@metamorphy edited!
$endgroup$
– Ya G
Dec 23 '18 at 17:54
$begingroup$
Now it should be easy for you (the poles are purely imaginary, one inside, the other outside).
$endgroup$
– metamorphy
Dec 23 '18 at 18:00
$begingroup$
Now it should be easy for you (the poles are purely imaginary, one inside, the other outside).
$endgroup$
– metamorphy
Dec 23 '18 at 18:00
$begingroup$
@metamorphy I feel like I’m still having trouble computing for different relations between $a$ and $b$ these for some reason.
$endgroup$
– Ya G
Dec 23 '18 at 18:02
$begingroup$
@metamorphy I feel like I’m still having trouble computing for different relations between $a$ and $b$ these for some reason.
$endgroup$
– Ya G
Dec 23 '18 at 18:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $aleqslant b$ the integral clearly diverges.
Otherwise, as you've shown, the poles are
$$z=frac{-apmsqrt{a^2-b^2}}{b}i,$$
and the pole with "$+$" is inside the unit circle. It remains to compute the residue.
answered Dec 23 '18 at 18:05
metamorphymetamorphy
3,8421721
$endgroup$
$begingroup$
What does it mean by integral diverges? And how do we know that?
$endgroup$
– Ya G
Dec 23 '18 at 18:07
$begingroup$
It means "the integrand is not integrable" - because $a+bsintheta$ reaches $0$ in this case.
$endgroup$
– metamorphy
Dec 23 '18 at 18:21
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For $aleqslant b$ the integral clearly diverges.
Otherwise, as you've shown, the poles are
$$z=frac{-apmsqrt{a^2-b^2}}{b}i,$$
and the pole with "$+$" is inside the unit circle. It remains to compute the residue.
answered Dec 23 '18 at 18:05
metamorphymetamorphy
3,8421721
$endgroup$
$begingroup$
What does it mean by integral diverges? And how do we know that?
$endgroup$
– Ya G
Dec 23 '18 at 18:07
$begingroup$
It means "the integrand is not integrable" - because $a+bsintheta$ reaches $0$ in this case.
$endgroup$
– metamorphy
Dec 23 '18 at 18:21
add a comment |
$begingroup$
For $aleqslant b$ the integral clearly diverges.
Otherwise, as you've shown, the poles are
$$z=frac{-apmsqrt{a^2-b^2}}{b}i,$$
and the pole with "$+$" is inside the unit circle. It remains to compute the residue.
answered Dec 23 '18 at 18:05
metamorphymetamorphy
3,8421721
$endgroup$
$begingroup$
What does it mean by integral diverges? And how do we know that?
$endgroup$
– Ya G
Dec 23 '18 at 18:07
$begingroup$
It means "the integrand is not integrable" - because $a+bsintheta$ reaches $0$ in this case.
$endgroup$
– metamorphy
Dec 23 '18 at 18:21
add a comment |
$begingroup$
For $aleqslant b$ the integral clearly diverges.
Otherwise, as you've shown, the poles are
$$z=frac{-apmsqrt{a^2-b^2}}{b}i,$$
and the pole with "$+$" is inside the unit circle. It remains to compute the residue.
answered Dec 23 '18 at 18:05
metamorphymetamorphy
3,8421721
$endgroup$
For $aleqslant b$ the integral clearly diverges.
Otherwise, as you've shown, the poles are
$$z=frac{-apmsqrt{a^2-b^2}}{b}i,$$
and the pole with "$+$" is inside the unit circle. It remains to compute the residue.
answered Dec 23 '18 at 18:05
metamorphymetamorphy
3,8421721
answered Dec 23 '18 at 18:05
metamorphymetamorphy
3,8421721
answered Dec 23 '18 at 18:05
metamorphymetamorphy
3,8421721
answered Dec 23 '18 at 18:05
metamorphymetamorphy
3,8421721
3,8421721
$begingroup$
What does it mean by integral diverges? And how do we know that?
$endgroup$
– Ya G
Dec 23 '18 at 18:07
$begingroup$
It means "the integrand is not integrable" - because $a+bsintheta$ reaches $0$ in this case.
$endgroup$
– metamorphy
Dec 23 '18 at 18:21
add a comment |
$begingroup$
What does it mean by integral diverges? And how do we know that?
$endgroup$
– Ya G
Dec 23 '18 at 18:07
$begingroup$
It means "the integrand is not integrable" - because $a+bsintheta$ reaches $0$ in this case.
$endgroup$
– metamorphy
Dec 23 '18 at 18:21
$begingroup$
What does it mean by integral diverges? And how do we know that?
$endgroup$
– Ya G
Dec 23 '18 at 18:07
$begingroup$
What does it mean by integral diverges? And how do we know that?
$endgroup$
– Ya G
Dec 23 '18 at 18:07
$begingroup$
It means "the integrand is not integrable" - because $a+bsintheta$ reaches $0$ in this case.
$endgroup$
– metamorphy
Dec 23 '18 at 18:21
$begingroup$
It means "the integrand is not integrable" - because $a+bsintheta$ reaches $0$ in this case.
$endgroup$
– metamorphy
Dec 23 '18 at 18:21
add a comment |
$begingroup$
@metamorphy right! That was my initial thought too, until I actually tried to get my hands dirty with it. May be I’m doing something wrong.
$endgroup$
– Ya G
Dec 23 '18 at 17:46
$begingroup$
(Oops.) For $aleqslant b$ the integral diverges. Otherwise you're doing right.
$endgroup$
– metamorphy
Dec 23 '18 at 17:50
$begingroup$
@metamorphy edited!
$endgroup$
– Ya G
Dec 23 '18 at 17:54
$begingroup$
Now it should be easy for you (the poles are purely imaginary, one inside, the other outside).
$endgroup$
– metamorphy
Dec 23 '18 at 18:00
$begingroup$
@metamorphy I feel like I’m still having trouble computing for different relations between $a$ and $b$ these for some reason.
$endgroup$
– Ya G
Dec 23 '18 at 18:02