prove if $f$ is bounded in [a,b] and integrable in every [a+$epsilon$ , b] for every $epsilon$ > 0 then...
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1
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I've started with showing that the improper integral in [a,b] is the limit when t goes to a+ (I've said that t=a+ $epsilon$ )
but now to say that the improper integral is defined I need to show that the limit exist and final.
integration improper-integrals
asked Nov 17 at 20:28
Dor
83
add a comment |
up vote
1
down vote
favorite
I've started with showing that the improper integral in [a,b] is the limit when t goes to a+ (I've said that t=a+ $epsilon$ )
but now to say that the improper integral is defined I need to show that the limit exist and final.
integration improper-integrals
asked Nov 17 at 20:28
Dor
83
Use upper and lower sums
– rubikscube09
Nov 17 at 20:29
$f$ is bounded, $m >= f >= M$ so the limit is bounded by $(b-a)M$ and $(b-a)m$ is that enough?
– Dor
Nov 17 at 20:33
Almost. When you have epsilon really small, the integral over the epsilon region is going to be small too because the function is bounded. Hence if the a b - e integral converges, you add only a small amount of mass (say M*epsilon ) by integrating over the rest of the region.
– rubikscube09
Nov 17 at 20:40
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I've started with showing that the improper integral in [a,b] is the limit when t goes to a+ (I've said that t=a+ $epsilon$ )
but now to say that the improper integral is defined I need to show that the limit exist and final.
integration improper-integrals
asked Nov 17 at 20:28
Dor
83
I've started with showing that the improper integral in [a,b] is the limit when t goes to a+ (I've said that t=a+ $epsilon$ )
but now to say that the improper integral is defined I need to show that the limit exist and final.
integration improper-integrals
integration improper-integrals
asked Nov 17 at 20:28
Dor
83
asked Nov 17 at 20:28
Dor
83
edited Nov 17 at 20:30
asked Nov 17 at 20:28
Dor
83
asked Nov 17 at 20:28
Dor
83
asked Nov 17 at 20:28
Dor
83
83
Use upper and lower sums
– rubikscube09
Nov 17 at 20:29
$f$ is bounded, $m >= f >= M$ so the limit is bounded by $(b-a)M$ and $(b-a)m$ is that enough?
– Dor
Nov 17 at 20:33
Almost. When you have epsilon really small, the integral over the epsilon region is going to be small too because the function is bounded. Hence if the a b - e integral converges, you add only a small amount of mass (say M*epsilon ) by integrating over the rest of the region.
– rubikscube09
Nov 17 at 20:40
add a comment |
Use upper and lower sums
– rubikscube09
Nov 17 at 20:29
$f$ is bounded, $m >= f >= M$ so the limit is bounded by $(b-a)M$ and $(b-a)m$ is that enough?
– Dor
Nov 17 at 20:33
Almost. When you have epsilon really small, the integral over the epsilon region is going to be small too because the function is bounded. Hence if the a b - e integral converges, you add only a small amount of mass (say M*epsilon ) by integrating over the rest of the region.
– rubikscube09
Nov 17 at 20:40
Use upper and lower sums
– rubikscube09
Nov 17 at 20:29
Use upper and lower sums
– rubikscube09
Nov 17 at 20:29
$f$ is bounded, $m >= f >= M$ so the limit is bounded by $(b-a)M$ and $(b-a)m$ is that enough?
– Dor
Nov 17 at 20:33
$f$ is bounded, $m >= f >= M$ so the limit is bounded by $(b-a)M$ and $(b-a)m$ is that enough?
– Dor
Nov 17 at 20:33
Almost. When you have epsilon really small, the integral over the epsilon region is going to be small too because the function is bounded. Hence if the a b - e integral converges, you add only a small amount of mass (say M*epsilon ) by integrating over the rest of the region.
– rubikscube09
Nov 17 at 20:40
Almost. When you have epsilon really small, the integral over the epsilon region is going to be small too because the function is bounded. Hence if the a b - e integral converges, you add only a small amount of mass (say M*epsilon ) by integrating over the rest of the region.
– rubikscube09
Nov 17 at 20:40
add a comment |
1 Answer
1
active
oldest
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up vote
0
down vote
accepted
Let $M=sup_{[a,b]}| f|$.
We can assume $M>0$.
Given $epsilon>0$ small enough.
$f$ is integrable at $[a+frac{epsilon}{4M},b]$ thus by Cauchy criterion,
there exist a partition $sigma$ of $[a+frac{epsilon}{4M},b]$ such that
$$U(f,sigma)-L(f,sigma)<frac{epsilon}{2}$$
Put $$Sigma=sigma cup{a}$$
then
$$U(f,Sigma)-L(f,Sigma)le$$
$$2Mfrac{epsilon}{4M}+U(f,sigma)-L(f,sigma)$$
$$<epsilon$$
done.
answered Nov 17 at 20:38
hamam_Abdallah
37k21534
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 $M=sup_{[a,b]}| f|$.
We can assume $M>0$.
Given $epsilon>0$ small enough.
$f$ is integrable at $[a+frac{epsilon}{4M},b]$ thus by Cauchy criterion,
there exist a partition $sigma$ of $[a+frac{epsilon}{4M},b]$ such that
$$U(f,sigma)-L(f,sigma)<frac{epsilon}{2}$$
Put $$Sigma=sigma cup{a}$$
then
$$U(f,Sigma)-L(f,Sigma)le$$
$$2Mfrac{epsilon}{4M}+U(f,sigma)-L(f,sigma)$$
$$<epsilon$$
done.
answered Nov 17 at 20:38
hamam_Abdallah
37k21534
add a comment |
up vote
0
down vote
accepted
Let $M=sup_{[a,b]}| f|$.
We can assume $M>0$.
Given $epsilon>0$ small enough.
$f$ is integrable at $[a+frac{epsilon}{4M},b]$ thus by Cauchy criterion,
there exist a partition $sigma$ of $[a+frac{epsilon}{4M},b]$ such that
$$U(f,sigma)-L(f,sigma)<frac{epsilon}{2}$$
Put $$Sigma=sigma cup{a}$$
then
$$U(f,Sigma)-L(f,Sigma)le$$
$$2Mfrac{epsilon}{4M}+U(f,sigma)-L(f,sigma)$$
$$<epsilon$$
done.
answered Nov 17 at 20:38
hamam_Abdallah
37k21534
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let $M=sup_{[a,b]}| f|$.
We can assume $M>0$.
Given $epsilon>0$ small enough.
$f$ is integrable at $[a+frac{epsilon}{4M},b]$ thus by Cauchy criterion,
there exist a partition $sigma$ of $[a+frac{epsilon}{4M},b]$ such that
$$U(f,sigma)-L(f,sigma)<frac{epsilon}{2}$$
Put $$Sigma=sigma cup{a}$$
then
$$U(f,Sigma)-L(f,Sigma)le$$
$$2Mfrac{epsilon}{4M}+U(f,sigma)-L(f,sigma)$$
$$<epsilon$$
done.
answered Nov 17 at 20:38
hamam_Abdallah
37k21534
Let $M=sup_{[a,b]}| f|$.
We can assume $M>0$.
Given $epsilon>0$ small enough.
$f$ is integrable at $[a+frac{epsilon}{4M},b]$ thus by Cauchy criterion,
there exist a partition $sigma$ of $[a+frac{epsilon}{4M},b]$ such that
$$U(f,sigma)-L(f,sigma)<frac{epsilon}{2}$$
Put $$Sigma=sigma cup{a}$$
then
$$U(f,Sigma)-L(f,Sigma)le$$
$$2Mfrac{epsilon}{4M}+U(f,sigma)-L(f,sigma)$$
$$<epsilon$$
done.
answered Nov 17 at 20:38
hamam_Abdallah
37k21534
edited Nov 17 at 20:44
answered Nov 17 at 20:38
hamam_Abdallah
37k21534
answered Nov 17 at 20:38
hamam_Abdallah
37k21534
answered Nov 17 at 20:38
hamam_Abdallah
37k21534
37k21534
add a comment |
add a comment |
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Use upper and lower sums
– rubikscube09
Nov 17 at 20:29
$f$ is bounded, $m >= f >= M$ so the limit is bounded by $(b-a)M$ and $(b-a)m$ is that enough?
– Dor
Nov 17 at 20:33
Almost. When you have epsilon really small, the integral over the epsilon region is going to be small too because the function is bounded. Hence if the a b - e integral converges, you add only a small amount of mass (say M*epsilon ) by integrating over the rest of the region.
– rubikscube09
Nov 17 at 20:40