Does this integral diverge? $int_{-pi/2}^{pi/2}csc{x}dx$
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
150k22120221
asked Nov 27 at 11:53
GambitSquared
1,1581137
add a comment |
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
150k22120221
asked Nov 27 at 11:53
GambitSquared
1,1581137
add a comment |
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
150k22120221
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
150k22120221
asked Nov 27 at 11:53
GambitSquared
1,1581137
edited Nov 27 at 11:59
José Carlos Santos
150k22120221
asked Nov 27 at 11:53
GambitSquared
1,1581137
edited Nov 27 at 11:59
José Carlos Santos
150k22120221
edited Nov 27 at 11:59
José Carlos Santos
150k22120221
edited Nov 27 at 11:59
José Carlos Santos
150k22120221
150k22120221
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
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oldest
votes
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
50k31854
add a comment |
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
150k22120221
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
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
50k31854
add a comment |
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
50k31854
add a comment |
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
50k31854
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
50k31854
answered Nov 27 at 11:59
Kavi Rama Murthy
50k31854
answered Nov 27 at 11:59
Kavi Rama Murthy
50k31854
answered Nov 27 at 11:59
Kavi Rama Murthy
50k31854
50k31854
add a comment |
add a comment |
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
150k22120221
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
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
150k22120221
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
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
150k22120221
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
150k22120221
answered Nov 27 at 11:58
José Carlos Santos
150k22120221
answered Nov 27 at 11:58
José Carlos Santos
150k22120221
answered Nov 27 at 11:58
José Carlos Santos
150k22120221
150k22120221
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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