showing an integral diverges












2












$begingroup$


I am working on the following problem:



problem



Edit: So after thinking a little bit more about the problem, here is my (hopefully correct) solution. If there is an error, I would appreciate if someone could point it out.



Suppose that $int_a^b f(x)g(x) ,dx $ exists. From the boundedness of $f$, we have $rhoint_a^b g(x) ,dx leq int_a^bf(x)g(x),dx $.
By Cauchy's Criterion, $exists r in (a,b)$ such that $x_1,x_2 in (r,b) implies |int_{x_1}^{x_2} f(x)g(x) ,dx| < epsilon$.



Now assume $g > 0$ (otherwise consider $-g$). Then $ rho M leqrhoint_{x_1}^{x_2} g(x) ,dx leqint_{x_1}^{x_2} f(x)g(x) ,dx =|int_{x_1}^{x_2} f(x)g(x) ,dx| < epsilon$



Or more simply, $rho M leq int_{x_1}^{x_2} f(x)g(x) ,dx < epsilon$. However epsilon is arbitrary, $implies$ $rho M leq 0$, a contradiction. Thus $int_a^b f(x)g(x) ,dx$ diverges.






real-analysis







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  • $begingroup$
    Note that the inclusion of border points in integration boundaries is optional, i.e. $$int_{[a,b]} = int_{(a,b)}$$
    $endgroup$
    – AlexR
    Oct 30 '13 at 18:16










  • $begingroup$
    I have made an attempt at a solution, any feedback would be appreciated.
    $endgroup$
    – user104352
    Oct 30 '13 at 23:18










  • $begingroup$
    @Glorfindel: Maybe go a little easy on the editing. In the past couple of hours, you've bumped almost three dozen years-old questions to the top of the queue. There's already so much current traffic on Math.SE that it's best to limit the influx of old stuff. (See, for instance, this recent meta post in the wake of rampant re-tagging of sangaku questions.)
    $endgroup$
    – Blue
    Dec 24 '18 at 9:49












  • $begingroup$
    @Blue heh, thanks for the warning. It's an automated script; I kept an eye on the front page but didn't notice any relevant activity there. Somehow the caching is off, I guess.
    $endgroup$
    – Glorfindel
    Dec 24 '18 at 9:51












  • $begingroup$
    It's good that the process is automated; fixing all those image links by hand would be tedious. :) Even so, there are over a dozen of your recent edits on the Top Questions/Interesting queue; your eye may not be noticing them, but mine certainly is. :) I see in your Revisions list that another couple-dozen edits were made 12 hours ago. Before then, maybe a dozen or so, then I got bored of counting. You might want to tweak your script to max-out at, say, five edits at a time.
    $endgroup$
    – Blue
    Dec 24 '18 at 10:02
















2












$begingroup$


I am working on the following problem:



problem



Edit: So after thinking a little bit more about the problem, here is my (hopefully correct) solution. If there is an error, I would appreciate if someone could point it out.



Suppose that $int_a^b f(x)g(x) ,dx $ exists. From the boundedness of $f$, we have $rhoint_a^b g(x) ,dx leq int_a^bf(x)g(x),dx $.
By Cauchy's Criterion, $exists r in (a,b)$ such that $x_1,x_2 in (r,b) implies |int_{x_1}^{x_2} f(x)g(x) ,dx| < epsilon$.



Now assume $g > 0$ (otherwise consider $-g$). Then $ rho M leqrhoint_{x_1}^{x_2} g(x) ,dx leqint_{x_1}^{x_2} f(x)g(x) ,dx =|int_{x_1}^{x_2} f(x)g(x) ,dx| < epsilon$



Or more simply, $rho M leq int_{x_1}^{x_2} f(x)g(x) ,dx < epsilon$. However epsilon is arbitrary, $implies$ $rho M leq 0$, a contradiction. Thus $int_a^b f(x)g(x) ,dx$ diverges.






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that the inclusion of border points in integration boundaries is optional, i.e. $$int_{[a,b]} = int_{(a,b)}$$
    $endgroup$
    – AlexR
    Oct 30 '13 at 18:16










  • $begingroup$
    I have made an attempt at a solution, any feedback would be appreciated.
    $endgroup$
    – user104352
    Oct 30 '13 at 23:18










  • $begingroup$
    @Glorfindel: Maybe go a little easy on the editing. In the past couple of hours, you've bumped almost three dozen years-old questions to the top of the queue. There's already so much current traffic on Math.SE that it's best to limit the influx of old stuff. (See, for instance, this recent meta post in the wake of rampant re-tagging of sangaku questions.)
    $endgroup$
    – Blue
    Dec 24 '18 at 9:49












  • $begingroup$
    @Blue heh, thanks for the warning. It's an automated script; I kept an eye on the front page but didn't notice any relevant activity there. Somehow the caching is off, I guess.
    $endgroup$
    – Glorfindel
    Dec 24 '18 at 9:51












  • $begingroup$
    It's good that the process is automated; fixing all those image links by hand would be tedious. :) Even so, there are over a dozen of your recent edits on the Top Questions/Interesting queue; your eye may not be noticing them, but mine certainly is. :) I see in your Revisions list that another couple-dozen edits were made 12 hours ago. Before then, maybe a dozen or so, then I got bored of counting. You might want to tweak your script to max-out at, say, five edits at a time.
    $endgroup$
    – Blue
    Dec 24 '18 at 10:02














2












2








2





$begingroup$


I am working on the following problem:



problem



Edit: So after thinking a little bit more about the problem, here is my (hopefully correct) solution. If there is an error, I would appreciate if someone could point it out.



Suppose that $int_a^b f(x)g(x) ,dx $ exists. From the boundedness of $f$, we have $rhoint_a^b g(x) ,dx leq int_a^bf(x)g(x),dx $.
By Cauchy's Criterion, $exists r in (a,b)$ such that $x_1,x_2 in (r,b) implies |int_{x_1}^{x_2} f(x)g(x) ,dx| < epsilon$.



Now assume $g > 0$ (otherwise consider $-g$). Then $ rho M leqrhoint_{x_1}^{x_2} g(x) ,dx leqint_{x_1}^{x_2} f(x)g(x) ,dx =|int_{x_1}^{x_2} f(x)g(x) ,dx| < epsilon$



Or more simply, $rho M leq int_{x_1}^{x_2} f(x)g(x) ,dx < epsilon$. However epsilon is arbitrary, $implies$ $rho M leq 0$, a contradiction. Thus $int_a^b f(x)g(x) ,dx$ diverges.






real-analysis







share|cite|improve this question











$endgroup$




I am working on the following problem:



problem



Edit: So after thinking a little bit more about the problem, here is my (hopefully correct) solution. If there is an error, I would appreciate if someone could point it out.



Suppose that $int_a^b f(x)g(x) ,dx $ exists. From the boundedness of $f$, we have $rhoint_a^b g(x) ,dx leq int_a^bf(x)g(x),dx $.
By Cauchy's Criterion, $exists r in (a,b)$ such that $x_1,x_2 in (r,b) implies |int_{x_1}^{x_2} f(x)g(x) ,dx| < epsilon$.



Now assume $g > 0$ (otherwise consider $-g$). Then $ rho M leqrhoint_{x_1}^{x_2} g(x) ,dx leqint_{x_1}^{x_2} f(x)g(x) ,dx =|int_{x_1}^{x_2} f(x)g(x) ,dx| < epsilon$



Or more simply, $rho M leq int_{x_1}^{x_2} f(x)g(x) ,dx < epsilon$. However epsilon is arbitrary, $implies$ $rho M leq 0$, a contradiction. Thus $int_a^b f(x)g(x) ,dx$ diverges.






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 21:18









Glorfindel

3,41981830




3,41981830










asked Oct 30 '13 at 18:11









user104352user104352

162




162












  • $begingroup$
    Note that the inclusion of border points in integration boundaries is optional, i.e. $$int_{[a,b]} = int_{(a,b)}$$
    $endgroup$
    – AlexR
    Oct 30 '13 at 18:16










  • $begingroup$
    I have made an attempt at a solution, any feedback would be appreciated.
    $endgroup$
    – user104352
    Oct 30 '13 at 23:18










  • $begingroup$
    @Glorfindel: Maybe go a little easy on the editing. In the past couple of hours, you've bumped almost three dozen years-old questions to the top of the queue. There's already so much current traffic on Math.SE that it's best to limit the influx of old stuff. (See, for instance, this recent meta post in the wake of rampant re-tagging of sangaku questions.)
    $endgroup$
    – Blue
    Dec 24 '18 at 9:49












  • $begingroup$
    @Blue heh, thanks for the warning. It's an automated script; I kept an eye on the front page but didn't notice any relevant activity there. Somehow the caching is off, I guess.
    $endgroup$
    – Glorfindel
    Dec 24 '18 at 9:51












  • $begingroup$
    It's good that the process is automated; fixing all those image links by hand would be tedious. :) Even so, there are over a dozen of your recent edits on the Top Questions/Interesting queue; your eye may not be noticing them, but mine certainly is. :) I see in your Revisions list that another couple-dozen edits were made 12 hours ago. Before then, maybe a dozen or so, then I got bored of counting. You might want to tweak your script to max-out at, say, five edits at a time.
    $endgroup$
    – Blue
    Dec 24 '18 at 10:02


















  • $begingroup$
    Note that the inclusion of border points in integration boundaries is optional, i.e. $$int_{[a,b]} = int_{(a,b)}$$
    $endgroup$
    – AlexR
    Oct 30 '13 at 18:16










  • $begingroup$
    I have made an attempt at a solution, any feedback would be appreciated.
    $endgroup$
    – user104352
    Oct 30 '13 at 23:18










  • $begingroup$
    @Glorfindel: Maybe go a little easy on the editing. In the past couple of hours, you've bumped almost three dozen years-old questions to the top of the queue. There's already so much current traffic on Math.SE that it's best to limit the influx of old stuff. (See, for instance, this recent meta post in the wake of rampant re-tagging of sangaku questions.)
    $endgroup$
    – Blue
    Dec 24 '18 at 9:49












  • $begingroup$
    @Blue heh, thanks for the warning. It's an automated script; I kept an eye on the front page but didn't notice any relevant activity there. Somehow the caching is off, I guess.
    $endgroup$
    – Glorfindel
    Dec 24 '18 at 9:51












  • $begingroup$
    It's good that the process is automated; fixing all those image links by hand would be tedious. :) Even so, there are over a dozen of your recent edits on the Top Questions/Interesting queue; your eye may not be noticing them, but mine certainly is. :) I see in your Revisions list that another couple-dozen edits were made 12 hours ago. Before then, maybe a dozen or so, then I got bored of counting. You might want to tweak your script to max-out at, say, five edits at a time.
    $endgroup$
    – Blue
    Dec 24 '18 at 10:02
















$begingroup$
Note that the inclusion of border points in integration boundaries is optional, i.e. $$int_{[a,b]} = int_{(a,b)}$$
$endgroup$
– AlexR
Oct 30 '13 at 18:16




$begingroup$
Note that the inclusion of border points in integration boundaries is optional, i.e. $$int_{[a,b]} = int_{(a,b)}$$
$endgroup$
– AlexR
Oct 30 '13 at 18:16












$begingroup$
I have made an attempt at a solution, any feedback would be appreciated.
$endgroup$
– user104352
Oct 30 '13 at 23:18




$begingroup$
I have made an attempt at a solution, any feedback would be appreciated.
$endgroup$
– user104352
Oct 30 '13 at 23:18












$begingroup$
@Glorfindel: Maybe go a little easy on the editing. In the past couple of hours, you've bumped almost three dozen years-old questions to the top of the queue. There's already so much current traffic on Math.SE that it's best to limit the influx of old stuff. (See, for instance, this recent meta post in the wake of rampant re-tagging of sangaku questions.)
$endgroup$
– Blue
Dec 24 '18 at 9:49






$begingroup$
@Glorfindel: Maybe go a little easy on the editing. In the past couple of hours, you've bumped almost three dozen years-old questions to the top of the queue. There's already so much current traffic on Math.SE that it's best to limit the influx of old stuff. (See, for instance, this recent meta post in the wake of rampant re-tagging of sangaku questions.)
$endgroup$
– Blue
Dec 24 '18 at 9:49














$begingroup$
@Blue heh, thanks for the warning. It's an automated script; I kept an eye on the front page but didn't notice any relevant activity there. Somehow the caching is off, I guess.
$endgroup$
– Glorfindel
Dec 24 '18 at 9:51






$begingroup$
@Blue heh, thanks for the warning. It's an automated script; I kept an eye on the front page but didn't notice any relevant activity there. Somehow the caching is off, I guess.
$endgroup$
– Glorfindel
Dec 24 '18 at 9:51














$begingroup$
It's good that the process is automated; fixing all those image links by hand would be tedious. :) Even so, there are over a dozen of your recent edits on the Top Questions/Interesting queue; your eye may not be noticing them, but mine certainly is. :) I see in your Revisions list that another couple-dozen edits were made 12 hours ago. Before then, maybe a dozen or so, then I got bored of counting. You might want to tweak your script to max-out at, say, five edits at a time.
$endgroup$
– Blue
Dec 24 '18 at 10:02




$begingroup$
It's good that the process is automated; fixing all those image links by hand would be tedious. :) Even so, there are over a dozen of your recent edits on the Top Questions/Interesting queue; your eye may not be noticing them, but mine certainly is. :) I see in your Revisions list that another couple-dozen edits were made 12 hours ago. Before then, maybe a dozen or so, then I got bored of counting. You might want to tweak your script to max-out at, say, five edits at a time.
$endgroup$
– Blue
Dec 24 '18 at 10:02










1 Answer
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Using the definition of an integral will also work here. Note that you can choose an infinite sequence of pairs $(a_1,b_1),(a_2,b_2) ldots$ such that $a < a_1 < b_1 < a_2 < b_2 ldots$ and all pairs are less than $b$, such that $|int_{a_j}^{b_j} g(x) dx| geq M$. Then in the limit as $j to infty$, you will eventually have $|int_{a_j}^{b_j} f(x)g(x) dx| > M rho / 2$ since $f$ is continuous. Since $M rho / 2$ is a fixed constant, this is enough to show the integral diverges.






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    $begingroup$

    Using the definition of an integral will also work here. Note that you can choose an infinite sequence of pairs $(a_1,b_1),(a_2,b_2) ldots$ such that $a < a_1 < b_1 < a_2 < b_2 ldots$ and all pairs are less than $b$, such that $|int_{a_j}^{b_j} g(x) dx| geq M$. Then in the limit as $j to infty$, you will eventually have $|int_{a_j}^{b_j} f(x)g(x) dx| > M rho / 2$ since $f$ is continuous. Since $M rho / 2$ is a fixed constant, this is enough to show the integral diverges.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Using the definition of an integral will also work here. Note that you can choose an infinite sequence of pairs $(a_1,b_1),(a_2,b_2) ldots$ such that $a < a_1 < b_1 < a_2 < b_2 ldots$ and all pairs are less than $b$, such that $|int_{a_j}^{b_j} g(x) dx| geq M$. Then in the limit as $j to infty$, you will eventually have $|int_{a_j}^{b_j} f(x)g(x) dx| > M rho / 2$ since $f$ is continuous. Since $M rho / 2$ is a fixed constant, this is enough to show the integral diverges.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Using the definition of an integral will also work here. Note that you can choose an infinite sequence of pairs $(a_1,b_1),(a_2,b_2) ldots$ such that $a < a_1 < b_1 < a_2 < b_2 ldots$ and all pairs are less than $b$, such that $|int_{a_j}^{b_j} g(x) dx| geq M$. Then in the limit as $j to infty$, you will eventually have $|int_{a_j}^{b_j} f(x)g(x) dx| > M rho / 2$ since $f$ is continuous. Since $M rho / 2$ is a fixed constant, this is enough to show the integral diverges.






        share|cite|improve this answer









        $endgroup$



        Using the definition of an integral will also work here. Note that you can choose an infinite sequence of pairs $(a_1,b_1),(a_2,b_2) ldots$ such that $a < a_1 < b_1 < a_2 < b_2 ldots$ and all pairs are less than $b$, such that $|int_{a_j}^{b_j} g(x) dx| geq M$. Then in the limit as $j to infty$, you will eventually have $|int_{a_j}^{b_j} f(x)g(x) dx| > M rho / 2$ since $f$ is continuous. Since $M rho / 2$ is a fixed constant, this is enough to show the integral diverges.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 30 '13 at 20:01









        user2566092user2566092

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