When is $int_{-infty}^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?
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When is $int_{-infty}^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?
Normally, we would need to take two different limits, but I am wondering if there is necessary and/or sufficient condition that the integral can be evaluated by taking the limit simultaneously. If so, is $$limlimits_{n to infty} int_{-n}^n f(x) dx = limlimits_{n to infty} int_{-kn}^n f(x) dx$$ for any $k > 0$?
More specifically, if we know that the integral converges, why can we conclude that $int_infty^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?
limits improper-integrals
asked Nov 14 at 23:25
Cute Brownie
952316
add a comment |
up vote
0
down vote
favorite
When is $int_{-infty}^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?
Normally, we would need to take two different limits, but I am wondering if there is necessary and/or sufficient condition that the integral can be evaluated by taking the limit simultaneously. If so, is $$limlimits_{n to infty} int_{-n}^n f(x) dx = limlimits_{n to infty} int_{-kn}^n f(x) dx$$ for any $k > 0$?
More specifically, if we know that the integral converges, why can we conclude that $int_infty^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?
limits improper-integrals
asked Nov 14 at 23:25
Cute Brownie
952316
3
If $int_{-infty}^infty f(x),dx$ exists, then it is equal to its Cauchy Principal Value.
– Mark Viola
Nov 14 at 23:27
@MarkViola thank you, could you elaborate more with justification?
– Cute Brownie
Nov 14 at 23:33
1
What is the definition of $int_{-infty}^infty f?.$
– zhw.
Nov 14 at 23:49
1
If $f in L^1(mathbb{R})$, i.e. $int_{-infty}^infty |f(x)|,dx < infty$, then it's true. Also, the $L^1$ condition is equivalent to $int_{-n}^n |f(x)|,dx$ being bounded as a function of $n$.
– Daniel Schepler
Nov 14 at 23:52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When is $int_{-infty}^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?
Normally, we would need to take two different limits, but I am wondering if there is necessary and/or sufficient condition that the integral can be evaluated by taking the limit simultaneously. If so, is $$limlimits_{n to infty} int_{-n}^n f(x) dx = limlimits_{n to infty} int_{-kn}^n f(x) dx$$ for any $k > 0$?
More specifically, if we know that the integral converges, why can we conclude that $int_infty^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?
limits improper-integrals
asked Nov 14 at 23:25
Cute Brownie
952316
When is $int_{-infty}^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?
Normally, we would need to take two different limits, but I am wondering if there is necessary and/or sufficient condition that the integral can be evaluated by taking the limit simultaneously. If so, is $$limlimits_{n to infty} int_{-n}^n f(x) dx = limlimits_{n to infty} int_{-kn}^n f(x) dx$$ for any $k > 0$?
More specifically, if we know that the integral converges, why can we conclude that $int_infty^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?
limits improper-integrals
limits improper-integrals
asked Nov 14 at 23:25
Cute Brownie
952316
asked Nov 14 at 23:25
Cute Brownie
952316
edited Nov 14 at 23:44
asked Nov 14 at 23:25
Cute Brownie
952316
asked Nov 14 at 23:25
Cute Brownie
952316
asked Nov 14 at 23:25
Cute Brownie
952316
952316
3
If $int_{-infty}^infty f(x),dx$ exists, then it is equal to its Cauchy Principal Value.
– Mark Viola
Nov 14 at 23:27
@MarkViola thank you, could you elaborate more with justification?
– Cute Brownie
Nov 14 at 23:33
1
What is the definition of $int_{-infty}^infty f?.$
– zhw.
Nov 14 at 23:49
1
If $f in L^1(mathbb{R})$, i.e. $int_{-infty}^infty |f(x)|,dx < infty$, then it's true. Also, the $L^1$ condition is equivalent to $int_{-n}^n |f(x)|,dx$ being bounded as a function of $n$.
– Daniel Schepler
Nov 14 at 23:52
add a comment |
3
If $int_{-infty}^infty f(x),dx$ exists, then it is equal to its Cauchy Principal Value.
– Mark Viola
Nov 14 at 23:27
@MarkViola thank you, could you elaborate more with justification?
– Cute Brownie
Nov 14 at 23:33
1
What is the definition of $int_{-infty}^infty f?.$
– zhw.
Nov 14 at 23:49
1
If $f in L^1(mathbb{R})$, i.e. $int_{-infty}^infty |f(x)|,dx < infty$, then it's true. Also, the $L^1$ condition is equivalent to $int_{-n}^n |f(x)|,dx$ being bounded as a function of $n$.
– Daniel Schepler
Nov 14 at 23:52
3
3
If $int_{-infty}^infty f(x),dx$ exists, then it is equal to its Cauchy Principal Value.
– Mark Viola
Nov 14 at 23:27
If $int_{-infty}^infty f(x),dx$ exists, then it is equal to its Cauchy Principal Value.
– Mark Viola
Nov 14 at 23:27
@MarkViola thank you, could you elaborate more with justification?
– Cute Brownie
Nov 14 at 23:33
@MarkViola thank you, could you elaborate more with justification?
– Cute Brownie
Nov 14 at 23:33
1
1
What is the definition of $int_{-infty}^infty f?.$
– zhw.
Nov 14 at 23:49
What is the definition of $int_{-infty}^infty f?.$
– zhw.
Nov 14 at 23:49
1
1
If $f in L^1(mathbb{R})$, i.e. $int_{-infty}^infty |f(x)|,dx < infty$, then it's true. Also, the $L^1$ condition is equivalent to $int_{-n}^n |f(x)|,dx$ being bounded as a function of $n$.
– Daniel Schepler
Nov 14 at 23:52
If $f in L^1(mathbb{R})$, i.e. $int_{-infty}^infty |f(x)|,dx < infty$, then it's true. Also, the $L^1$ condition is equivalent to $int_{-n}^n |f(x)|,dx$ being bounded as a function of $n$.
– Daniel Schepler
Nov 14 at 23:52
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Let $F'(x)=f(x).$ Then using definitions, and the FTC we can see
$$int_{-infty}^infty f(x)dx= lim_{stoinfty}int_a^s f(x)dx +lim_{tto-infty}int_t^a f(x)dx$$
$$=lim_{stoinfty}F(s)-F(a) +lim_{tto-infty}F(a)-F(t)$$
$$=lim_{stoinfty}F(s)-lim_{tto-infty}F(t).$$
If these two limits exist, then it means that the result is finite, and in particular, it is well defined, and we don't need to evaluate these limits separately. Thus
$$lim_{stoinfty}F(s)-lim_{tto-infty}F(t)=lim_{stoinfty}F(s)-F(-s).$$
It is only in this situation that the principle value is equal to the result of the improper integral (equality can only be discussed because it converges, after all!).
answered Nov 14 at 23:55
Chickenmancer
3,209723
I'm sorry if this sounds basic, but why does two limits existing imply we can take both limits altogether?
– Cute Brownie
Nov 15 at 0:08
1
The reason is that these two limits do not influence each other. One limit applies to large positive values, the other to large negative values. Since there is no interaction, the two effects simply add up. That is what Chickenmancer demonstrates in his answer.
– M. Wind
Nov 15 at 0:52
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let $F'(x)=f(x).$ Then using definitions, and the FTC we can see
$$int_{-infty}^infty f(x)dx= lim_{stoinfty}int_a^s f(x)dx +lim_{tto-infty}int_t^a f(x)dx$$
$$=lim_{stoinfty}F(s)-F(a) +lim_{tto-infty}F(a)-F(t)$$
$$=lim_{stoinfty}F(s)-lim_{tto-infty}F(t).$$
If these two limits exist, then it means that the result is finite, and in particular, it is well defined, and we don't need to evaluate these limits separately. Thus
$$lim_{stoinfty}F(s)-lim_{tto-infty}F(t)=lim_{stoinfty}F(s)-F(-s).$$
It is only in this situation that the principle value is equal to the result of the improper integral (equality can only be discussed because it converges, after all!).
answered Nov 14 at 23:55
Chickenmancer
3,209723
I'm sorry if this sounds basic, but why does two limits existing imply we can take both limits altogether?
– Cute Brownie
Nov 15 at 0:08
1
The reason is that these two limits do not influence each other. One limit applies to large positive values, the other to large negative values. Since there is no interaction, the two effects simply add up. That is what Chickenmancer demonstrates in his answer.
– M. Wind
Nov 15 at 0:52
add a comment |
up vote
0
down vote
accepted
Let $F'(x)=f(x).$ Then using definitions, and the FTC we can see
$$int_{-infty}^infty f(x)dx= lim_{stoinfty}int_a^s f(x)dx +lim_{tto-infty}int_t^a f(x)dx$$
$$=lim_{stoinfty}F(s)-F(a) +lim_{tto-infty}F(a)-F(t)$$
$$=lim_{stoinfty}F(s)-lim_{tto-infty}F(t).$$
If these two limits exist, then it means that the result is finite, and in particular, it is well defined, and we don't need to evaluate these limits separately. Thus
$$lim_{stoinfty}F(s)-lim_{tto-infty}F(t)=lim_{stoinfty}F(s)-F(-s).$$
It is only in this situation that the principle value is equal to the result of the improper integral (equality can only be discussed because it converges, after all!).
answered Nov 14 at 23:55
Chickenmancer
3,209723
I'm sorry if this sounds basic, but why does two limits existing imply we can take both limits altogether?
– Cute Brownie
Nov 15 at 0:08
1
The reason is that these two limits do not influence each other. One limit applies to large positive values, the other to large negative values. Since there is no interaction, the two effects simply add up. That is what Chickenmancer demonstrates in his answer.
– M. Wind
Nov 15 at 0:52
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let $F'(x)=f(x).$ Then using definitions, and the FTC we can see
$$int_{-infty}^infty f(x)dx= lim_{stoinfty}int_a^s f(x)dx +lim_{tto-infty}int_t^a f(x)dx$$
$$=lim_{stoinfty}F(s)-F(a) +lim_{tto-infty}F(a)-F(t)$$
$$=lim_{stoinfty}F(s)-lim_{tto-infty}F(t).$$
If these two limits exist, then it means that the result is finite, and in particular, it is well defined, and we don't need to evaluate these limits separately. Thus
$$lim_{stoinfty}F(s)-lim_{tto-infty}F(t)=lim_{stoinfty}F(s)-F(-s).$$
It is only in this situation that the principle value is equal to the result of the improper integral (equality can only be discussed because it converges, after all!).
answered Nov 14 at 23:55
Chickenmancer
3,209723
Let $F'(x)=f(x).$ Then using definitions, and the FTC we can see
$$int_{-infty}^infty f(x)dx= lim_{stoinfty}int_a^s f(x)dx +lim_{tto-infty}int_t^a f(x)dx$$
$$=lim_{stoinfty}F(s)-F(a) +lim_{tto-infty}F(a)-F(t)$$
$$=lim_{stoinfty}F(s)-lim_{tto-infty}F(t).$$
If these two limits exist, then it means that the result is finite, and in particular, it is well defined, and we don't need to evaluate these limits separately. Thus
$$lim_{stoinfty}F(s)-lim_{tto-infty}F(t)=lim_{stoinfty}F(s)-F(-s).$$
It is only in this situation that the principle value is equal to the result of the improper integral (equality can only be discussed because it converges, after all!).
answered Nov 14 at 23:55
Chickenmancer
3,209723
answered Nov 14 at 23:55
Chickenmancer
3,209723
answered Nov 14 at 23:55
Chickenmancer
3,209723
answered Nov 14 at 23:55
Chickenmancer
3,209723
3,209723
I'm sorry if this sounds basic, but why does two limits existing imply we can take both limits altogether?
– Cute Brownie
Nov 15 at 0:08
1
The reason is that these two limits do not influence each other. One limit applies to large positive values, the other to large negative values. Since there is no interaction, the two effects simply add up. That is what Chickenmancer demonstrates in his answer.
– M. Wind
Nov 15 at 0:52
add a comment |
I'm sorry if this sounds basic, but why does two limits existing imply we can take both limits altogether?
– Cute Brownie
Nov 15 at 0:08
1
The reason is that these two limits do not influence each other. One limit applies to large positive values, the other to large negative values. Since there is no interaction, the two effects simply add up. That is what Chickenmancer demonstrates in his answer.
– M. Wind
Nov 15 at 0:52
I'm sorry if this sounds basic, but why does two limits existing imply we can take both limits altogether?
– Cute Brownie
Nov 15 at 0:08
I'm sorry if this sounds basic, but why does two limits existing imply we can take both limits altogether?
– Cute Brownie
Nov 15 at 0:08
1
1
The reason is that these two limits do not influence each other. One limit applies to large positive values, the other to large negative values. Since there is no interaction, the two effects simply add up. That is what Chickenmancer demonstrates in his answer.
– M. Wind
Nov 15 at 0:52
The reason is that these two limits do not influence each other. One limit applies to large positive values, the other to large negative values. Since there is no interaction, the two effects simply add up. That is what Chickenmancer demonstrates in his answer.
– M. Wind
Nov 15 at 0:52
add a comment |
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3
If $int_{-infty}^infty f(x),dx$ exists, then it is equal to its Cauchy Principal Value.
– Mark Viola
Nov 14 at 23:27
@MarkViola thank you, could you elaborate more with justification?
– Cute Brownie
Nov 14 at 23:33
1
What is the definition of $int_{-infty}^infty f?.$
– zhw.
Nov 14 at 23:49
1
If $f in L^1(mathbb{R})$, i.e. $int_{-infty}^infty |f(x)|,dx < infty$, then it's true. Also, the $L^1$ condition is equivalent to $int_{-n}^n |f(x)|,dx$ being bounded as a function of $n$.
– Daniel Schepler
Nov 14 at 23:52