Mixing
If the integral of a positive/negative function is convergent, then the limit of the function will be zero?
$begingroup$
Prove or disprove:
a) $f$ is Reimann integrable in the interval $[1,b]$ $forall$ $b>1$, $f(x) > 0$ $forall x$, and $int_1^infty f(x)dx $ is convergent. Then,
$$lim_{xrightarrowinfty}f(x)=0$$
b) Does your conclusion remain the same if $f(x)$ can take any value?
I am struggling with this question. Here is what I know but I can't bring the pieces together. Need a rigorous proof.
For part a, since $int_1^infty f(x)dx $ is convergent,
$int_1^infty f(x)dx = lim_{x to infty}int_1^x f(u)du $ exists.
Intuitively, I think I should show that since the function is positive and the integral is convergent, it must be that there exist a large enough $N$ such that $forall n >N$, $f(x_n)-f(x_{n-1})<epsilon$ for any arbitrary $epsilon>0$ which should imply that the limit of the function is zero. But I am not sure about this.
For part b, my guess is that the conclusion will not hold as we need the function to be positive to prove the statement in part a. But I don't know how can I show it.
Thanks for your help.
real-analysis integration convergence improper-integrals
asked Jan 22 at 22:45
Sher AfghanSher Afghan
1789
$endgroup$
|
show 5 more comments
$begingroup$
Prove or disprove:
a) $f$ is Reimann integrable in the interval $[1,b]$ $forall$ $b>1$, $f(x) > 0$ $forall x$, and $int_1^infty f(x)dx $ is convergent. Then,
$$lim_{xrightarrowinfty}f(x)=0$$
b) Does your conclusion remain the same if $f(x)$ can take any value?
I am struggling with this question. Here is what I know but I can't bring the pieces together. Need a rigorous proof.
For part a, since $int_1^infty f(x)dx $ is convergent,
$int_1^infty f(x)dx = lim_{x to infty}int_1^x f(u)du $ exists.
Intuitively, I think I should show that since the function is positive and the integral is convergent, it must be that there exist a large enough $N$ such that $forall n >N$, $f(x_n)-f(x_{n-1})<epsilon$ for any arbitrary $epsilon>0$ which should imply that the limit of the function is zero. But I am not sure about this.
For part b, my guess is that the conclusion will not hold as we need the function to be positive to prove the statement in part a. But I don't know how can I show it.
Thanks for your help.
real-analysis integration convergence improper-integrals
asked Jan 22 at 22:45
Sher AfghanSher Afghan
1789
$endgroup$
$begingroup$
Are you assuming $f(x)$ is increasing in part a)? But you wrote $f(x)>0$ then the limit cannot be zero.
$endgroup$
– Eclipse Sun
Jan 22 at 22:56
1
$begingroup$
I think even answer for a) is negative. Take $f(x) = 1 / x^2$ if $x notin mathbb{N}$ and $f(x) = 1$ if $x in mathbb{N}$. Maybe you also want assumption that $f$ is continuous?
$endgroup$
– prosinac
Jan 22 at 22:59
$begingroup$
@prosinac if $f$ is Reimann integrable, doesn't that imply that it is continuous as well? Also I didn't understand how your counter example disprove the statement :/
$endgroup$
– Sher Afghan
Jan 22 at 23:22
$begingroup$
@SherAfghan: Can you state all of the assumptions clearly for part (a)? Is $f$ positive? Is $f$ nonnegative? I now believe it is not monotone from the barrage of downvotes I received from my now deleted answer.
$endgroup$
– RRL
Jan 22 at 23:27
$begingroup$
@RRL I am sorry for the confusion that I created. The question is edited and have all the assumptions therein. For part a, $f$ is positive. For part b, it can take any value. We don't have any assumption about the monotonicity.
$endgroup$
– Sher Afghan
Jan 22 at 23:30
|
show 5 more comments
$begingroup$
Prove or disprove:
a) $f$ is Reimann integrable in the interval $[1,b]$ $forall$ $b>1$, $f(x) > 0$ $forall x$, and $int_1^infty f(x)dx $ is convergent. Then,
$$lim_{xrightarrowinfty}f(x)=0$$
b) Does your conclusion remain the same if $f(x)$ can take any value?
I am struggling with this question. Here is what I know but I can't bring the pieces together. Need a rigorous proof.
For part a, since $int_1^infty f(x)dx $ is convergent,
$int_1^infty f(x)dx = lim_{x to infty}int_1^x f(u)du $ exists.
Intuitively, I think I should show that since the function is positive and the integral is convergent, it must be that there exist a large enough $N$ such that $forall n >N$, $f(x_n)-f(x_{n-1})<epsilon$ for any arbitrary $epsilon>0$ which should imply that the limit of the function is zero. But I am not sure about this.
For part b, my guess is that the conclusion will not hold as we need the function to be positive to prove the statement in part a. But I don't know how can I show it.
Thanks for your help.
real-analysis integration convergence improper-integrals
asked Jan 22 at 22:45
Sher AfghanSher Afghan
1789
$endgroup$
Prove or disprove:
a) $f$ is Reimann integrable in the interval $[1,b]$ $forall$ $b>1$, $f(x) > 0$ $forall x$, and $int_1^infty f(x)dx $ is convergent. Then,
$$lim_{xrightarrowinfty}f(x)=0$$
b) Does your conclusion remain the same if $f(x)$ can take any value?
I am struggling with this question. Here is what I know but I can't bring the pieces together. Need a rigorous proof.
For part a, since $int_1^infty f(x)dx $ is convergent,
$int_1^infty f(x)dx = lim_{x to infty}int_1^x f(u)du $ exists.
Intuitively, I think I should show that since the function is positive and the integral is convergent, it must be that there exist a large enough $N$ such that $forall n >N$, $f(x_n)-f(x_{n-1})<epsilon$ for any arbitrary $epsilon>0$ which should imply that the limit of the function is zero. But I am not sure about this.
For part b, my guess is that the conclusion will not hold as we need the function to be positive to prove the statement in part a. But I don't know how can I show it.
Thanks for your help.
real-analysis integration convergence improper-integrals
real-analysis integration convergence improper-integrals
asked Jan 22 at 22:45
Sher AfghanSher Afghan
1789
asked Jan 22 at 22:45
Sher AfghanSher Afghan
1789
edited Jan 22 at 23:02
Sher Afghan
asked Jan 22 at 22:45
Sher AfghanSher Afghan
1789
asked Jan 22 at 22:45
Sher AfghanSher Afghan
1789
asked Jan 22 at 22:45
Sher AfghanSher Afghan
1789
1789
$begingroup$
Are you assuming $f(x)$ is increasing in part a)? But you wrote $f(x)>0$ then the limit cannot be zero.
$endgroup$
– Eclipse Sun
Jan 22 at 22:56
1
$begingroup$
I think even answer for a) is negative. Take $f(x) = 1 / x^2$ if $x notin mathbb{N}$ and $f(x) = 1$ if $x in mathbb{N}$. Maybe you also want assumption that $f$ is continuous?
$endgroup$
– prosinac
Jan 22 at 22:59
$begingroup$
@prosinac if $f$ is Reimann integrable, doesn't that imply that it is continuous as well? Also I didn't understand how your counter example disprove the statement :/
$endgroup$
– Sher Afghan
Jan 22 at 23:22
$begingroup$
@SherAfghan: Can you state all of the assumptions clearly for part (a)? Is $f$ positive? Is $f$ nonnegative? I now believe it is not monotone from the barrage of downvotes I received from my now deleted answer.
$endgroup$
– RRL
Jan 22 at 23:27
$begingroup$
@RRL I am sorry for the confusion that I created. The question is edited and have all the assumptions therein. For part a, $f$ is positive. For part b, it can take any value. We don't have any assumption about the monotonicity.
$endgroup$
– Sher Afghan
Jan 22 at 23:30
|
show 5 more comments
$begingroup$
Are you assuming $f(x)$ is increasing in part a)? But you wrote $f(x)>0$ then the limit cannot be zero.
$endgroup$
– Eclipse Sun
Jan 22 at 22:56
1
$begingroup$
I think even answer for a) is negative. Take $f(x) = 1 / x^2$ if $x notin mathbb{N}$ and $f(x) = 1$ if $x in mathbb{N}$. Maybe you also want assumption that $f$ is continuous?
$endgroup$
– prosinac
Jan 22 at 22:59
$begingroup$
@prosinac if $f$ is Reimann integrable, doesn't that imply that it is continuous as well? Also I didn't understand how your counter example disprove the statement :/
$endgroup$
– Sher Afghan
Jan 22 at 23:22
$begingroup$
@SherAfghan: Can you state all of the assumptions clearly for part (a)? Is $f$ positive? Is $f$ nonnegative? I now believe it is not monotone from the barrage of downvotes I received from my now deleted answer.
$endgroup$
– RRL
Jan 22 at 23:27
$begingroup$
@RRL I am sorry for the confusion that I created. The question is edited and have all the assumptions therein. For part a, $f$ is positive. For part b, it can take any value. We don't have any assumption about the monotonicity.
$endgroup$
– Sher Afghan
Jan 22 at 23:30
$begingroup$
Are you assuming $f(x)$ is increasing in part a)? But you wrote $f(x)>0$ then the limit cannot be zero.
$endgroup$
– Eclipse Sun
Jan 22 at 22:56
$begingroup$
Are you assuming $f(x)$ is increasing in part a)? But you wrote $f(x)>0$ then the limit cannot be zero.
$endgroup$
– Eclipse Sun
Jan 22 at 22:56
1
1
$begingroup$
I think even answer for a) is negative. Take $f(x) = 1 / x^2$ if $x notin mathbb{N}$ and $f(x) = 1$ if $x in mathbb{N}$. Maybe you also want assumption that $f$ is continuous?
$endgroup$
– prosinac
Jan 22 at 22:59
$begingroup$
I think even answer for a) is negative. Take $f(x) = 1 / x^2$ if $x notin mathbb{N}$ and $f(x) = 1$ if $x in mathbb{N}$. Maybe you also want assumption that $f$ is continuous?
$endgroup$
– prosinac
Jan 22 at 22:59
$begingroup$
@prosinac if $f$ is Reimann integrable, doesn't that imply that it is continuous as well? Also I didn't understand how your counter example disprove the statement :/
$endgroup$
– Sher Afghan
Jan 22 at 23:22
$begingroup$
@prosinac if $f$ is Reimann integrable, doesn't that imply that it is continuous as well? Also I didn't understand how your counter example disprove the statement :/
$endgroup$
– Sher Afghan
Jan 22 at 23:22
$begingroup$
@SherAfghan: Can you state all of the assumptions clearly for part (a)? Is $f$ positive? Is $f$ nonnegative? I now believe it is not monotone from the barrage of downvotes I received from my now deleted answer.
$endgroup$
– RRL
Jan 22 at 23:27
$begingroup$
@SherAfghan: Can you state all of the assumptions clearly for part (a)? Is $f$ positive? Is $f$ nonnegative? I now believe it is not monotone from the barrage of downvotes I received from my now deleted answer.
$endgroup$
– RRL
Jan 22 at 23:27
$begingroup$
@RRL I am sorry for the confusion that I created. The question is edited and have all the assumptions therein. For part a, $f$ is positive. For part b, it can take any value. We don't have any assumption about the monotonicity.
$endgroup$
– Sher Afghan
Jan 22 at 23:30
$begingroup$
@RRL I am sorry for the confusion that I created. The question is edited and have all the assumptions therein. For part a, $f$ is positive. For part b, it can take any value. We don't have any assumption about the monotonicity.
$endgroup$
– Sher Afghan
Jan 22 at 23:30
|
show 5 more comments
1 Answer
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oldest
votes
$begingroup$
If $f$ is positive (and even continuous) , then we can have $int_1^infty f(x) , dx$ converge and $f(x) notto 0$ as $x to infty$.
Take $f(x) = x^{-2} + sum_{k=2}^infty phi_k(x)$ where
$$phi_k(x) = begin{cases}k^2(x - k + k^{-2}) & k- k^{-2} leqslant x leqslant k \ k^2(k + k^{-2}-x) & k < x < k + k^{-2}\ 0& text{otherwise} end{cases}$$
In this case $limsup f(x) = 1$ and $liminf f(x) = 0$ so the limit does not exist, but
$$int_1^infty f(x) , dx = 1 + sum_{k=2}^infty frac{1}{k^2} < infty$$
answered Jan 22 at 23:40
RRLRRL
52.5k42573
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
If $f$ is positive (and even continuous) , then we can have $int_1^infty f(x) , dx$ converge and $f(x) notto 0$ as $x to infty$.
Take $f(x) = x^{-2} + sum_{k=2}^infty phi_k(x)$ where
$$phi_k(x) = begin{cases}k^2(x - k + k^{-2}) & k- k^{-2} leqslant x leqslant k \ k^2(k + k^{-2}-x) & k < x < k + k^{-2}\ 0& text{otherwise} end{cases}$$
In this case $limsup f(x) = 1$ and $liminf f(x) = 0$ so the limit does not exist, but
$$int_1^infty f(x) , dx = 1 + sum_{k=2}^infty frac{1}{k^2} < infty$$
answered Jan 22 at 23:40
RRLRRL
52.5k42573
$endgroup$
add a comment |
$begingroup$
If $f$ is positive (and even continuous) , then we can have $int_1^infty f(x) , dx$ converge and $f(x) notto 0$ as $x to infty$.
Take $f(x) = x^{-2} + sum_{k=2}^infty phi_k(x)$ where
$$phi_k(x) = begin{cases}k^2(x - k + k^{-2}) & k- k^{-2} leqslant x leqslant k \ k^2(k + k^{-2}-x) & k < x < k + k^{-2}\ 0& text{otherwise} end{cases}$$
In this case $limsup f(x) = 1$ and $liminf f(x) = 0$ so the limit does not exist, but
$$int_1^infty f(x) , dx = 1 + sum_{k=2}^infty frac{1}{k^2} < infty$$
answered Jan 22 at 23:40
RRLRRL
52.5k42573
$endgroup$
add a comment |
$begingroup$
If $f$ is positive (and even continuous) , then we can have $int_1^infty f(x) , dx$ converge and $f(x) notto 0$ as $x to infty$.
Take $f(x) = x^{-2} + sum_{k=2}^infty phi_k(x)$ where
$$phi_k(x) = begin{cases}k^2(x - k + k^{-2}) & k- k^{-2} leqslant x leqslant k \ k^2(k + k^{-2}-x) & k < x < k + k^{-2}\ 0& text{otherwise} end{cases}$$
In this case $limsup f(x) = 1$ and $liminf f(x) = 0$ so the limit does not exist, but
$$int_1^infty f(x) , dx = 1 + sum_{k=2}^infty frac{1}{k^2} < infty$$
answered Jan 22 at 23:40
RRLRRL
52.5k42573
$endgroup$
If $f$ is positive (and even continuous) , then we can have $int_1^infty f(x) , dx$ converge and $f(x) notto 0$ as $x to infty$.
Take $f(x) = x^{-2} + sum_{k=2}^infty phi_k(x)$ where
$$phi_k(x) = begin{cases}k^2(x - k + k^{-2}) & k- k^{-2} leqslant x leqslant k \ k^2(k + k^{-2}-x) & k < x < k + k^{-2}\ 0& text{otherwise} end{cases}$$
In this case $limsup f(x) = 1$ and $liminf f(x) = 0$ so the limit does not exist, but
$$int_1^infty f(x) , dx = 1 + sum_{k=2}^infty frac{1}{k^2} < infty$$
answered Jan 22 at 23:40
RRLRRL
52.5k42573
edited Jan 22 at 23:53
answered Jan 22 at 23:40
RRLRRL
52.5k42573
answered Jan 22 at 23:40
RRLRRL
52.5k42573
answered Jan 22 at 23:40
RRLRRL
52.5k42573
52.5k42573
add a comment |
add a comment |
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$begingroup$
Are you assuming $f(x)$ is increasing in part a)? But you wrote $f(x)>0$ then the limit cannot be zero.
$endgroup$
– Eclipse Sun
Jan 22 at 22:56
1
$begingroup$
I think even answer for a) is negative. Take $f(x) = 1 / x^2$ if $x notin mathbb{N}$ and $f(x) = 1$ if $x in mathbb{N}$. Maybe you also want assumption that $f$ is continuous?
$endgroup$
– prosinac
Jan 22 at 22:59
$begingroup$
@prosinac if $f$ is Reimann integrable, doesn't that imply that it is continuous as well? Also I didn't understand how your counter example disprove the statement :/
$endgroup$
– Sher Afghan
Jan 22 at 23:22
$begingroup$
@SherAfghan: Can you state all of the assumptions clearly for part (a)? Is $f$ positive? Is $f$ nonnegative? I now believe it is not monotone from the barrage of downvotes I received from my now deleted answer.
$endgroup$
– RRL
Jan 22 at 23:27
$begingroup$
@RRL I am sorry for the confusion that I created. The question is edited and have all the assumptions therein. For part a, $f$ is positive. For part b, it can take any value. We don't have any assumption about the monotonicity.
$endgroup$
– Sher Afghan
Jan 22 at 23:30