Does this integral diverge? $int_{-pi/2}^{pi/2}csc{x}dx$
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According to my calculusbook the following integral diverges:
$$int_{-pi/2}^{pi/2}csc{x}dx$$
This is the case since $csc{x}gefrac{1}{x}$ on $(0,pi/2]$
My question is: I would have guessed the integral would be zero, since $csc{x}$ is an odd function. Why is this not the case?
On $(0,pi/2]$ the integral goes to $infty$ and on $[-pi/2,0)$ the integral goes to $-infty$ so we have $infty-infty$, which is usually undefined, but doesn't the fact that $csc{x}$ is odd imply that $infty-infty=0$ in this case?
calculus integration limits improper-integrals
edited Nov 27 at 11:59
José Carlos Santos
146k22116215
asked Nov 27 at 11:53
GambitSquared
1,1581137
add a comment |
up vote
-1
down vote
favorite
According to my calculusbook the following integral diverges:
$$int_{-pi/2}^{pi/2}csc{x}dx$$
This is the case since $csc{x}gefrac{1}{x}$ on $(0,pi/2]$
My question is: I would have guessed the integral would be zero, since $csc{x}$ is an odd function. Why is this not the case?
On $(0,pi/2]$ the integral goes to $infty$ and on $[-pi/2,0)$ the integral goes to $-infty$ so we have $infty-infty$, which is usually undefined, but doesn't the fact that $csc{x}$ is odd imply that $infty-infty=0$ in this case?
calculus integration limits improper-integrals
edited Nov 27 at 11:59
José Carlos Santos
146k22116215
asked Nov 27 at 11:53
GambitSquared
1,1581137
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
According to my calculusbook the following integral diverges:
$$int_{-pi/2}^{pi/2}csc{x}dx$$
This is the case since $csc{x}gefrac{1}{x}$ on $(0,pi/2]$
My question is: I would have guessed the integral would be zero, since $csc{x}$ is an odd function. Why is this not the case?
On $(0,pi/2]$ the integral goes to $infty$ and on $[-pi/2,0)$ the integral goes to $-infty$ so we have $infty-infty$, which is usually undefined, but doesn't the fact that $csc{x}$ is odd imply that $infty-infty=0$ in this case?
calculus integration limits improper-integrals
edited Nov 27 at 11:59
José Carlos Santos
146k22116215
asked Nov 27 at 11:53
GambitSquared
1,1581137
According to my calculusbook the following integral diverges:
$$int_{-pi/2}^{pi/2}csc{x}dx$$
This is the case since $csc{x}gefrac{1}{x}$ on $(0,pi/2]$
My question is: I would have guessed the integral would be zero, since $csc{x}$ is an odd function. Why is this not the case?
On $(0,pi/2]$ the integral goes to $infty$ and on $[-pi/2,0)$ the integral goes to $-infty$ so we have $infty-infty$, which is usually undefined, but doesn't the fact that $csc{x}$ is odd imply that $infty-infty=0$ in this case?
calculus integration limits improper-integrals
calculus integration limits improper-integrals
edited Nov 27 at 11:59
José Carlos Santos
146k22116215
asked Nov 27 at 11:53
GambitSquared
1,1581137
edited Nov 27 at 11:59
José Carlos Santos
146k22116215
asked Nov 27 at 11:53
GambitSquared
1,1581137
edited Nov 27 at 11:59
José Carlos Santos
146k22116215
edited Nov 27 at 11:59
José Carlos Santos
146k22116215
edited Nov 27 at 11:59
José Carlos Santos
146k22116215
146k22116215
asked Nov 27 at 11:53
GambitSquared
1,1581137
asked Nov 27 at 11:53
GambitSquared
1,1581137
asked Nov 27 at 11:53
GambitSquared
1,1581137
1,1581137
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.
answered Nov 27 at 11:59
Kavi Rama Murthy
46.6k31854
add a comment |
up vote
3
down vote
No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).
In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.
answered Nov 27 at 11:58
José Carlos Santos
146k22116215
But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
– GambitSquared
Nov 27 at 12:03
By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
– José Carlos Santos
Nov 27 at 12:04
@GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
– SinTan1729
Nov 27 at 12:04
@JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
– GambitSquared
Nov 27 at 12:22
@SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
– GambitSquared
Nov 27 at 12:23
|
show 3 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.
answered Nov 27 at 11:59
Kavi Rama Murthy
46.6k31854
add a comment |
up vote
2
down vote
accepted
The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.
answered Nov 27 at 11:59
Kavi Rama Murthy
46.6k31854
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.
answered Nov 27 at 11:59
Kavi Rama Murthy
46.6k31854
The integral does not exist as proper Riemann integral because the function is not bounded. If you interpret the integral as limit of $int_{epsilon <|x|<frac {pi} 2} csc (x) dx$ as $epsilon to 0$ then the integral exists and the value is $0$. As a Lebesgue integral it does not exist because the integrand behaves like $frac 1 x$ near $0$. So the answer depends very much what type of integral you want to consider.
answered Nov 27 at 11:59
Kavi Rama Murthy
46.6k31854
answered Nov 27 at 11:59
Kavi Rama Murthy
46.6k31854
answered Nov 27 at 11:59
Kavi Rama Murthy
46.6k31854
answered Nov 27 at 11:59
Kavi Rama Murthy
46.6k31854
46.6k31854
add a comment |
add a comment |
up vote
3
down vote
No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).
In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.
answered Nov 27 at 11:58
José Carlos Santos
146k22116215
But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
– GambitSquared
Nov 27 at 12:03
By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
– José Carlos Santos
Nov 27 at 12:04
@GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
– SinTan1729
Nov 27 at 12:04
@JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
– GambitSquared
Nov 27 at 12:22
@SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
– GambitSquared
Nov 27 at 12:23
|
show 3 more comments
up vote
3
down vote
No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).
In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.
answered Nov 27 at 11:58
José Carlos Santos
146k22116215
But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
– GambitSquared
Nov 27 at 12:03
By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
– José Carlos Santos
Nov 27 at 12:04
@GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
– SinTan1729
Nov 27 at 12:04
@JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
– GambitSquared
Nov 27 at 12:22
@SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
– GambitSquared
Nov 27 at 12:23
|
show 3 more comments
up vote
3
down vote
up vote
3
down vote
No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).
In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.
answered Nov 27 at 11:58
José Carlos Santos
146k22116215
No, it doesn't imply that. By definition, an improper integral $int_a^bf(x),mathrm dx$, where $f$ is a map from $[a,b]setminus{c}$ into $mathbb R$, converges if both integrals $int_c^bf(x),mathrm dx$ and $int_a^cf(x),mathrm dx$ converge (and, if they do, then $int_a^b f(x),mathrm dx$ is the sum of those two integrals).
In your case, none of the integrals converges. The fact that $csc$ is odd is irrelevant.
answered Nov 27 at 11:58
José Carlos Santos
146k22116215
answered Nov 27 at 11:58
José Carlos Santos
146k22116215
answered Nov 27 at 11:58
José Carlos Santos
146k22116215
answered Nov 27 at 11:58
José Carlos Santos
146k22116215
146k22116215
But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
– GambitSquared
Nov 27 at 12:03
By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
– José Carlos Santos
Nov 27 at 12:04
@GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
– SinTan1729
Nov 27 at 12:04
@JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
– GambitSquared
Nov 27 at 12:22
@SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
– GambitSquared
Nov 27 at 12:23
|
show 3 more comments
But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
– GambitSquared
Nov 27 at 12:03
By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
– José Carlos Santos
Nov 27 at 12:04
@GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
– SinTan1729
Nov 27 at 12:04
@JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
– GambitSquared
Nov 27 at 12:22
@SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
– GambitSquared
Nov 27 at 12:23
But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
– GambitSquared
Nov 27 at 12:03
But isn't it "clear", also from the graph that the two surfaces on both sides of $0$ must be equal?
– GambitSquared
Nov 27 at 12:03
By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
– José Carlos Santos
Nov 27 at 12:04
By the same approach $int_{-infty}^{+infty}x,mathrm dx=0$.
– José Carlos Santos
Nov 27 at 12:04
@GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
– SinTan1729
Nov 27 at 12:04
@GambitSquared You cannot say that. Both surfaces have infinite area, it's not necessary that they cancel out.
– SinTan1729
Nov 27 at 12:04
@JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
– GambitSquared
Nov 27 at 12:22
@JoséCarlosSantos Hmmm, yes, would that be an unreasonable claim though?
– GambitSquared
Nov 27 at 12:22
@SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
– GambitSquared
Nov 27 at 12:23
@SinTan1729 Why don't they cancel out if the function is odd? What would be an argument to assume they do not cancel out?
– GambitSquared
Nov 27 at 12:23
|
show 3 more comments
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