Mixing
Conceptual understanding of convergence of improper integrals
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Statement
If $f$ is monotonic decreasing for all $xleq1$ and $f(x)to0$ as $xtoinfty$, then the integral $int_1^inftyf(x),dx$ and the series $sum_1^inftyf(n)$ both converge or diverge together.
This can be proved easily (proof similar to the proof of integral test).
Doubts
Give an example of nonmonotonic $f$ for which $sum_1^inftyf(n)$ converges but $int_1^inftyf(x),dx$ diverges.
If $f$ is monotonic, and if $lim_xtoinftyint_1^nf(x),dx$ exists, then the integral $int_1^inftyf(x),dx$ converges. How to prove this?
In the previous statements, is monotonicity necessary? What if $f$ is nonmonotonic but $lim_xtoinftyint_1^nf(x),dx$ exists?
If $lim_xtoinftyf(x)=0$ and $lim_xtoinftyint_1^nf(x),dx=A$, then $int_1^inftyf(x),dx$ converges to $A$. How to prove this?
If the sequence $leftlim_xtoinftyint_1^nf(x),dxright$ converges, then $int_1^inftyf(x),dx$ converges. True or false and why?
If $f$ is positive and if $lim_xtoinftyint_1^nf(x),dx=A$, then $int_1^inftyf(x),dx$ converges to A. True or false and why?
If $int_1^inftyf(x),dx$ converges, then $lim_xtoinftyint_1^nf(x),dx =0$.
Suppose $|f'(x)|<M$ for all $x>0$. If $lim_xtoinftyint_1^nf(x),dx=A$,then $int_1^inftyf(x),dx$ converges to $A$. Why?
calculus real-analysis
edited Aug 2 at 18:31
Robert Howard
1,263620
asked Aug 2 at 17:24
blue boy
528211
add a comment |Â
up vote
0
down vote
favorite
Statement
If $f$ is monotonic decreasing for all $xleq1$ and $f(x)to0$ as $xtoinfty$, then the integral $int_1^inftyf(x),dx$ and the series $sum_1^inftyf(n)$ both converge or diverge together.
This can be proved easily (proof similar to the proof of integral test).
Doubts
Give an example of nonmonotonic $f$ for which $sum_1^inftyf(n)$ converges but $int_1^inftyf(x),dx$ diverges.
If $f$ is monotonic, and if $lim_xtoinftyint_1^nf(x),dx$ exists, then the integral $int_1^inftyf(x),dx$ converges. How to prove this?
In the previous statements, is monotonicity necessary? What if $f$ is nonmonotonic but $lim_xtoinftyint_1^nf(x),dx$ exists?
If $lim_xtoinftyf(x)=0$ and $lim_xtoinftyint_1^nf(x),dx=A$, then $int_1^inftyf(x),dx$ converges to $A$. How to prove this?
If the sequence $leftlim_xtoinftyint_1^nf(x),dxright$ converges, then $int_1^inftyf(x),dx$ converges. True or false and why?
If $f$ is positive and if $lim_xtoinftyint_1^nf(x),dx=A$, then $int_1^inftyf(x),dx$ converges to A. True or false and why?
If $int_1^inftyf(x),dx$ converges, then $lim_xtoinftyint_1^nf(x),dx =0$.
Suppose $|f'(x)|<M$ for all $x>0$. If $lim_xtoinftyint_1^nf(x),dx=A$,then $int_1^inftyf(x),dx$ converges to $A$. Why?
calculus real-analysis
edited Aug 2 at 18:31
Robert Howard
1,263620
asked Aug 2 at 17:24
blue boy
528211
$f(x) = sin(2pi x)$ is an example where the series converges, but the integral does not.
– Doug M
Aug 2 at 17:31
" monotonic decreasing for all $xleq1$ "?
– zhw.
Aug 2 at 20:50
This is way too many questions posed as one question.
– zhw.
Aug 2 at 20:54
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Statement
If $f$ is monotonic decreasing for all $xleq1$ and $f(x)to0$ as $xtoinfty$, then the integral $int_1^inftyf(x),dx$ and the series $sum_1^inftyf(n)$ both converge or diverge together.
This can be proved easily (proof similar to the proof of integral test).
Doubts
Give an example of nonmonotonic $f$ for which $sum_1^inftyf(n)$ converges but $int_1^inftyf(x),dx$ diverges.
If $f$ is monotonic, and if $lim_xtoinftyint_1^nf(x),dx$ exists, then the integral $int_1^inftyf(x),dx$ converges. How to prove this?
In the previous statements, is monotonicity necessary? What if $f$ is nonmonotonic but $lim_xtoinftyint_1^nf(x),dx$ exists?
If $lim_xtoinftyf(x)=0$ and $lim_xtoinftyint_1^nf(x),dx=A$, then $int_1^inftyf(x),dx$ converges to $A$. How to prove this?
If the sequence $leftlim_xtoinftyint_1^nf(x),dxright$ converges, then $int_1^inftyf(x),dx$ converges. True or false and why?
If $f$ is positive and if $lim_xtoinftyint_1^nf(x),dx=A$, then $int_1^inftyf(x),dx$ converges to A. True or false and why?
If $int_1^inftyf(x),dx$ converges, then $lim_xtoinftyint_1^nf(x),dx =0$.
Suppose $|f'(x)|<M$ for all $x>0$. If $lim_xtoinftyint_1^nf(x),dx=A$,then $int_1^inftyf(x),dx$ converges to $A$. Why?
calculus real-analysis
edited Aug 2 at 18:31
Robert Howard
1,263620
asked Aug 2 at 17:24
blue boy
528211
Statement
If $f$ is monotonic decreasing for all $xleq1$ and $f(x)to0$ as $xtoinfty$, then the integral $int_1^inftyf(x),dx$ and the series $sum_1^inftyf(n)$ both converge or diverge together.
This can be proved easily (proof similar to the proof of integral test).
Doubts
Give an example of nonmonotonic $f$ for which $sum_1^inftyf(n)$ converges but $int_1^inftyf(x),dx$ diverges.
If $f$ is monotonic, and if $lim_xtoinftyint_1^nf(x),dx$ exists, then the integral $int_1^inftyf(x),dx$ converges. How to prove this?
In the previous statements, is monotonicity necessary? What if $f$ is nonmonotonic but $lim_xtoinftyint_1^nf(x),dx$ exists?
If $lim_xtoinftyf(x)=0$ and $lim_xtoinftyint_1^nf(x),dx=A$, then $int_1^inftyf(x),dx$ converges to $A$. How to prove this?
If the sequence $leftlim_xtoinftyint_1^nf(x),dxright$ converges, then $int_1^inftyf(x),dx$ converges. True or false and why?
If $f$ is positive and if $lim_xtoinftyint_1^nf(x),dx=A$, then $int_1^inftyf(x),dx$ converges to A. True or false and why?
If $int_1^inftyf(x),dx$ converges, then $lim_xtoinftyint_1^nf(x),dx =0$.
Suppose $|f'(x)|<M$ for all $x>0$. If $lim_xtoinftyint_1^nf(x),dx=A$,then $int_1^inftyf(x),dx$ converges to $A$. Why?
calculus real-analysis
edited Aug 2 at 18:31
Robert Howard
1,263620
asked Aug 2 at 17:24
blue boy
528211
edited Aug 2 at 18:31
Robert Howard
1,263620
edited Aug 2 at 18:31
Robert Howard
1,263620
edited Aug 2 at 18:31
Robert Howard
1,263620
1,263620
asked Aug 2 at 17:24
blue boy
528211
asked Aug 2 at 17:24
blue boy
528211
asked Aug 2 at 17:24
blue boy
528211
528211
$f(x) = sin(2pi x)$ is an example where the series converges, but the integral does not.
– Doug M
Aug 2 at 17:31
" monotonic decreasing for all $xleq1$ "?
– zhw.
Aug 2 at 20:50
This is way too many questions posed as one question.
– zhw.
Aug 2 at 20:54
add a comment |Â
$f(x) = sin(2pi x)$ is an example where the series converges, but the integral does not.
– Doug M
Aug 2 at 17:31
" monotonic decreasing for all $xleq1$ "?
– zhw.
Aug 2 at 20:50
This is way too many questions posed as one question.
– zhw.
Aug 2 at 20:54
$f(x) = sin(2pi x)$ is an example where the series converges, but the integral does not.
– Doug M
Aug 2 at 17:31
$f(x) = sin(2pi x)$ is an example where the series converges, but the integral does not.
– Doug M
Aug 2 at 17:31
" monotonic decreasing for all $xleq1$ "?
– zhw.
Aug 2 at 20:50
" monotonic decreasing for all $xleq1$ "?
– zhw.
Aug 2 at 20:50
This is way too many questions posed as one question.
– zhw.
Aug 2 at 20:54
This is way too many questions posed as one question.
– zhw.
Aug 2 at 20:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
For the first part, think of a function where $f(n)=0$ for all $n$.
The second part looks to be incorrectly stated, as the $x$ inside the integral is a dummy variable that I would not expect to be in the limit.
If I switch the $x$ to an $n$ in the limit, it's not true. Consider $f(x)=1$.
Essentially, we need to know which conditions apply to $f$ for part 2.
answered Aug 2 at 17:29
Dean C Wills
33919
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For the first part, think of a function where $f(n)=0$ for all $n$.
The second part looks to be incorrectly stated, as the $x$ inside the integral is a dummy variable that I would not expect to be in the limit.
If I switch the $x$ to an $n$ in the limit, it's not true. Consider $f(x)=1$.
Essentially, we need to know which conditions apply to $f$ for part 2.
answered Aug 2 at 17:29
Dean C Wills
33919
add a comment |Â
up vote
0
down vote
For the first part, think of a function where $f(n)=0$ for all $n$.
The second part looks to be incorrectly stated, as the $x$ inside the integral is a dummy variable that I would not expect to be in the limit.
If I switch the $x$ to an $n$ in the limit, it's not true. Consider $f(x)=1$.
Essentially, we need to know which conditions apply to $f$ for part 2.
answered Aug 2 at 17:29
Dean C Wills
33919
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For the first part, think of a function where $f(n)=0$ for all $n$.
The second part looks to be incorrectly stated, as the $x$ inside the integral is a dummy variable that I would not expect to be in the limit.
If I switch the $x$ to an $n$ in the limit, it's not true. Consider $f(x)=1$.
Essentially, we need to know which conditions apply to $f$ for part 2.
answered Aug 2 at 17:29
Dean C Wills
33919
For the first part, think of a function where $f(n)=0$ for all $n$.
The second part looks to be incorrectly stated, as the $x$ inside the integral is a dummy variable that I would not expect to be in the limit.
If I switch the $x$ to an $n$ in the limit, it's not true. Consider $f(x)=1$.
Essentially, we need to know which conditions apply to $f$ for part 2.
answered Aug 2 at 17:29
Dean C Wills
33919
answered Aug 2 at 17:29
Dean C Wills
33919
answered Aug 2 at 17:29
Dean C Wills
33919
answered Aug 2 at 17:29
Dean C Wills
33919
33919
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870308%2fconceptual-understanding-of-convergence-of-improper-integrals%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
$f(x) = sin(2pi x)$ is an example where the series converges, but the integral does not.
– Doug M
Aug 2 at 17:31
" monotonic decreasing for all $xleq1$ "?
– zhw.
Aug 2 at 20:50
This is way too many questions posed as one question.
– zhw.
Aug 2 at 20:54