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Area under a curve of an odd function from negative infinity to positive infinity
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In integration, there is a property that says: If you're integrating from -a to a some odd function f(x), then the area under the curve between -a and a is zero.
I was listening to this in class , and then I thought about integrating some odd function, like x^3, from negative infinity to positive infinity.
But, if you integrate x^3 and then solve it from negative infinity to positive infinity, wouldn't you end up subtracting infinity from infinity, which is undefined?
Given this, which answer is the correct one: is the area 0 or is it undefined?
integration improper-integrals even-and-odd-functions
asked Aug 9 at 15:23
Skatinima
475
 |Â
show 1 more comment
up vote
5
down vote
favorite
In integration, there is a property that says: If you're integrating from -a to a some odd function f(x), then the area under the curve between -a and a is zero.
I was listening to this in class , and then I thought about integrating some odd function, like x^3, from negative infinity to positive infinity.
But, if you integrate x^3 and then solve it from negative infinity to positive infinity, wouldn't you end up subtracting infinity from infinity, which is undefined?
Given this, which answer is the correct one: is the area 0 or is it undefined?
integration improper-integrals even-and-odd-functions
asked Aug 9 at 15:23
Skatinima
475
2
It is only zero if the integral is defined. The function $x mapsto x^3$ is not integrable on the real line, so the integral is not defined. You can define an improper integral as the limit, but the integral per so is not defined.
– copper.hat
Aug 9 at 15:31
In general when you have improper integrals you consider the bounds as limits (ie $int_0^infty$ actually represents $lim_n to infty int_0^n$)(same principle if both bounds are infinite)
– aidangallagher4
Aug 9 at 15:32
@copper.hat Not sure what you are implying about $x^3$ not being integratable.
– Rushabh Mehta
Aug 9 at 15:32
@RushabhMehta: Integrable usually means that the integral is finite.
– Clayton
Aug 9 at 15:32
@aidangallagher4 that's correct, which is why what TheSimpliFire did is quite incorrect.
– Rushabh Mehta
Aug 9 at 15:32
 |Â
show 1 more comment
up vote
5
down vote
favorite
up vote
5
down vote
favorite
In integration, there is a property that says: If you're integrating from -a to a some odd function f(x), then the area under the curve between -a and a is zero.
I was listening to this in class , and then I thought about integrating some odd function, like x^3, from negative infinity to positive infinity.
But, if you integrate x^3 and then solve it from negative infinity to positive infinity, wouldn't you end up subtracting infinity from infinity, which is undefined?
Given this, which answer is the correct one: is the area 0 or is it undefined?
integration improper-integrals even-and-odd-functions
asked Aug 9 at 15:23
Skatinima
475
In integration, there is a property that says: If you're integrating from -a to a some odd function f(x), then the area under the curve between -a and a is zero.
I was listening to this in class , and then I thought about integrating some odd function, like x^3, from negative infinity to positive infinity.
But, if you integrate x^3 and then solve it from negative infinity to positive infinity, wouldn't you end up subtracting infinity from infinity, which is undefined?
Given this, which answer is the correct one: is the area 0 or is it undefined?
integration improper-integrals even-and-odd-functions
asked Aug 9 at 15:23
Skatinima
475
edited Aug 9 at 15:27
asked Aug 9 at 15:23
Skatinima
475
asked Aug 9 at 15:23
Skatinima
475
asked Aug 9 at 15:23
Skatinima
475
475
2
It is only zero if the integral is defined. The function $x mapsto x^3$ is not integrable on the real line, so the integral is not defined. You can define an improper integral as the limit, but the integral per so is not defined.
– copper.hat
Aug 9 at 15:31
In general when you have improper integrals you consider the bounds as limits (ie $int_0^infty$ actually represents $lim_n to infty int_0^n$)(same principle if both bounds are infinite)
– aidangallagher4
Aug 9 at 15:32
@copper.hat Not sure what you are implying about $x^3$ not being integratable.
– Rushabh Mehta
Aug 9 at 15:32
@RushabhMehta: Integrable usually means that the integral is finite.
– Clayton
Aug 9 at 15:32
@aidangallagher4 that's correct, which is why what TheSimpliFire did is quite incorrect.
– Rushabh Mehta
Aug 9 at 15:32
 |Â
show 1 more comment
2
It is only zero if the integral is defined. The function $x mapsto x^3$ is not integrable on the real line, so the integral is not defined. You can define an improper integral as the limit, but the integral per so is not defined.
– copper.hat
Aug 9 at 15:31
In general when you have improper integrals you consider the bounds as limits (ie $int_0^infty$ actually represents $lim_n to infty int_0^n$)(same principle if both bounds are infinite)
– aidangallagher4
Aug 9 at 15:32
@copper.hat Not sure what you are implying about $x^3$ not being integratable.
– Rushabh Mehta
Aug 9 at 15:32
@RushabhMehta: Integrable usually means that the integral is finite.
– Clayton
Aug 9 at 15:32
@aidangallagher4 that's correct, which is why what TheSimpliFire did is quite incorrect.
– Rushabh Mehta
Aug 9 at 15:32
2
2
It is only zero if the integral is defined. The function $x mapsto x^3$ is not integrable on the real line, so the integral is not defined. You can define an improper integral as the limit, but the integral per so is not defined.
– copper.hat
Aug 9 at 15:31
It is only zero if the integral is defined. The function $x mapsto x^3$ is not integrable on the real line, so the integral is not defined. You can define an improper integral as the limit, but the integral per so is not defined.
– copper.hat
Aug 9 at 15:31
In general when you have improper integrals you consider the bounds as limits (ie $int_0^infty$ actually represents $lim_n to infty int_0^n$)(same principle if both bounds are infinite)
– aidangallagher4
Aug 9 at 15:32
In general when you have improper integrals you consider the bounds as limits (ie $int_0^infty$ actually represents $lim_n to infty int_0^n$)(same principle if both bounds are infinite)
– aidangallagher4
Aug 9 at 15:32
@copper.hat Not sure what you are implying about $x^3$ not being integratable.
– Rushabh Mehta
Aug 9 at 15:32
@copper.hat Not sure what you are implying about $x^3$ not being integratable.
– Rushabh Mehta
Aug 9 at 15:32
@RushabhMehta: Integrable usually means that the integral is finite.
– Clayton
Aug 9 at 15:32
@RushabhMehta: Integrable usually means that the integral is finite.
– Clayton
Aug 9 at 15:32
@aidangallagher4 that's correct, which is why what TheSimpliFire did is quite incorrect.
– Rushabh Mehta
Aug 9 at 15:32
@aidangallagher4 that's correct, which is why what TheSimpliFire did is quite incorrect.
– Rushabh Mehta
Aug 9 at 15:32
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
17
down vote
accepted
For improper integrals, you're correct: you have to be careful. Both limits need to exist independently of each other. In your case, $int_-infty^0 x^3,dx$ is $-infty$, hence the integral "doesn't exist" (except in the extended real numbers case). There is something called a principal value, where you take the limits simultaneously, e.g., $$lim_Ntoinftyint_-N^N x^3,dx=lim_Ntoinfty0=0,$$ and in this sense, the limit will always give $0$.
answered Aug 9 at 15:30
Clayton
18.3k22883
3
Thank you, this is exactly what OP needs (+1)
– Rushabh Mehta
Aug 9 at 15:31
add a comment |Â
up vote
4
down vote
The improper integral is defined as $int_-infty^inftyx^3dx$=$lim_alphato-inftylim_betatoinftyint_alpha^betax^3 dx$, which as you said is undefined as the final calculation is $infty-infty$.
However, this integral can be given a value by something called the Cauchy Principal Value, which is a method of giving a value to double infinite integrals such as this. It is defined as $PVint_-infty^inftyx^3dx$$=lim_Rtoinftyint_-R^Rx^3dx$, which will converge to $0$ as your heuristic has you believe.
answered Aug 9 at 15:39
Tyler6
565210
3
There is a small defect in your notation. You have written that the two limits are taken sequentially. Correctly, they are taken simultaneously and independently, that is, as $lim_(alpha, beta) rightarrow (-infty, infty) dots$.
– Eric Towers
Aug 9 at 20:56
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
17
down vote
accepted
For improper integrals, you're correct: you have to be careful. Both limits need to exist independently of each other. In your case, $int_-infty^0 x^3,dx$ is $-infty$, hence the integral "doesn't exist" (except in the extended real numbers case). There is something called a principal value, where you take the limits simultaneously, e.g., $$lim_Ntoinftyint_-N^N x^3,dx=lim_Ntoinfty0=0,$$ and in this sense, the limit will always give $0$.
answered Aug 9 at 15:30
Clayton
18.3k22883
3
Thank you, this is exactly what OP needs (+1)
– Rushabh Mehta
Aug 9 at 15:31
add a comment |Â
up vote
17
down vote
accepted
For improper integrals, you're correct: you have to be careful. Both limits need to exist independently of each other. In your case, $int_-infty^0 x^3,dx$ is $-infty$, hence the integral "doesn't exist" (except in the extended real numbers case). There is something called a principal value, where you take the limits simultaneously, e.g., $$lim_Ntoinftyint_-N^N x^3,dx=lim_Ntoinfty0=0,$$ and in this sense, the limit will always give $0$.
answered Aug 9 at 15:30
Clayton
18.3k22883
3
Thank you, this is exactly what OP needs (+1)
– Rushabh Mehta
Aug 9 at 15:31
add a comment |Â
up vote
17
down vote
accepted
up vote
17
down vote
accepted
For improper integrals, you're correct: you have to be careful. Both limits need to exist independently of each other. In your case, $int_-infty^0 x^3,dx$ is $-infty$, hence the integral "doesn't exist" (except in the extended real numbers case). There is something called a principal value, where you take the limits simultaneously, e.g., $$lim_Ntoinftyint_-N^N x^3,dx=lim_Ntoinfty0=0,$$ and in this sense, the limit will always give $0$.
answered Aug 9 at 15:30
Clayton
18.3k22883
For improper integrals, you're correct: you have to be careful. Both limits need to exist independently of each other. In your case, $int_-infty^0 x^3,dx$ is $-infty$, hence the integral "doesn't exist" (except in the extended real numbers case). There is something called a principal value, where you take the limits simultaneously, e.g., $$lim_Ntoinftyint_-N^N x^3,dx=lim_Ntoinfty0=0,$$ and in this sense, the limit will always give $0$.
answered Aug 9 at 15:30
Clayton
18.3k22883
answered Aug 9 at 15:30
Clayton
18.3k22883
answered Aug 9 at 15:30
Clayton
18.3k22883
answered Aug 9 at 15:30
Clayton
18.3k22883
18.3k22883
3
Thank you, this is exactly what OP needs (+1)
– Rushabh Mehta
Aug 9 at 15:31
add a comment |Â
3
Thank you, this is exactly what OP needs (+1)
– Rushabh Mehta
Aug 9 at 15:31
3
3
Thank you, this is exactly what OP needs (+1)
– Rushabh Mehta
Aug 9 at 15:31
Thank you, this is exactly what OP needs (+1)
– Rushabh Mehta
Aug 9 at 15:31
add a comment |Â
up vote
4
down vote
The improper integral is defined as $int_-infty^inftyx^3dx$=$lim_alphato-inftylim_betatoinftyint_alpha^betax^3 dx$, which as you said is undefined as the final calculation is $infty-infty$.
However, this integral can be given a value by something called the Cauchy Principal Value, which is a method of giving a value to double infinite integrals such as this. It is defined as $PVint_-infty^inftyx^3dx$$=lim_Rtoinftyint_-R^Rx^3dx$, which will converge to $0$ as your heuristic has you believe.
answered Aug 9 at 15:39
Tyler6
565210
3
There is a small defect in your notation. You have written that the two limits are taken sequentially. Correctly, they are taken simultaneously and independently, that is, as $lim_(alpha, beta) rightarrow (-infty, infty) dots$.
– Eric Towers
Aug 9 at 20:56
add a comment |Â
up vote
4
down vote
The improper integral is defined as $int_-infty^inftyx^3dx$=$lim_alphato-inftylim_betatoinftyint_alpha^betax^3 dx$, which as you said is undefined as the final calculation is $infty-infty$.
However, this integral can be given a value by something called the Cauchy Principal Value, which is a method of giving a value to double infinite integrals such as this. It is defined as $PVint_-infty^inftyx^3dx$$=lim_Rtoinftyint_-R^Rx^3dx$, which will converge to $0$ as your heuristic has you believe.
answered Aug 9 at 15:39
Tyler6
565210
3
There is a small defect in your notation. You have written that the two limits are taken sequentially. Correctly, they are taken simultaneously and independently, that is, as $lim_(alpha, beta) rightarrow (-infty, infty) dots$.
– Eric Towers
Aug 9 at 20:56
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The improper integral is defined as $int_-infty^inftyx^3dx$=$lim_alphato-inftylim_betatoinftyint_alpha^betax^3 dx$, which as you said is undefined as the final calculation is $infty-infty$.
However, this integral can be given a value by something called the Cauchy Principal Value, which is a method of giving a value to double infinite integrals such as this. It is defined as $PVint_-infty^inftyx^3dx$$=lim_Rtoinftyint_-R^Rx^3dx$, which will converge to $0$ as your heuristic has you believe.
answered Aug 9 at 15:39
Tyler6
565210
The improper integral is defined as $int_-infty^inftyx^3dx$=$lim_alphato-inftylim_betatoinftyint_alpha^betax^3 dx$, which as you said is undefined as the final calculation is $infty-infty$.
However, this integral can be given a value by something called the Cauchy Principal Value, which is a method of giving a value to double infinite integrals such as this. It is defined as $PVint_-infty^inftyx^3dx$$=lim_Rtoinftyint_-R^Rx^3dx$, which will converge to $0$ as your heuristic has you believe.
answered Aug 9 at 15:39
Tyler6
565210
answered Aug 9 at 15:39
Tyler6
565210
answered Aug 9 at 15:39
Tyler6
565210
answered Aug 9 at 15:39
Tyler6
565210
565210
3
There is a small defect in your notation. You have written that the two limits are taken sequentially. Correctly, they are taken simultaneously and independently, that is, as $lim_(alpha, beta) rightarrow (-infty, infty) dots$.
– Eric Towers
Aug 9 at 20:56
add a comment |Â
3
There is a small defect in your notation. You have written that the two limits are taken sequentially. Correctly, they are taken simultaneously and independently, that is, as $lim_(alpha, beta) rightarrow (-infty, infty) dots$.
– Eric Towers
Aug 9 at 20:56
3
3
There is a small defect in your notation. You have written that the two limits are taken sequentially. Correctly, they are taken simultaneously and independently, that is, as $lim_(alpha, beta) rightarrow (-infty, infty) dots$.
– Eric Towers
Aug 9 at 20:56
There is a small defect in your notation. You have written that the two limits are taken sequentially. Correctly, they are taken simultaneously and independently, that is, as $lim_(alpha, beta) rightarrow (-infty, infty) dots$.
– Eric Towers
Aug 9 at 20:56
add a comment |Â
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2
It is only zero if the integral is defined. The function $x mapsto x^3$ is not integrable on the real line, so the integral is not defined. You can define an improper integral as the limit, but the integral per so is not defined.
– copper.hat
Aug 9 at 15:31
In general when you have improper integrals you consider the bounds as limits (ie $int_0^infty$ actually represents $lim_n to infty int_0^n$)(same principle if both bounds are infinite)
– aidangallagher4
Aug 9 at 15:32
@copper.hat Not sure what you are implying about $x^3$ not being integratable.
– Rushabh Mehta
Aug 9 at 15:32
@RushabhMehta: Integrable usually means that the integral is finite.
– Clayton
Aug 9 at 15:32
@aidangallagher4 that's correct, which is why what TheSimpliFire did is quite incorrect.
– Rushabh Mehta
Aug 9 at 15:32