Convergence of $int_0^1 (-ln x)^p ,dx$ [duplicate]

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  • For what real values of a does the integral $intlimits_0^1(-ln x)^a,dx$ converge?

    1 answer




For what values of $p$ does the following integral converge: $$int_0^1(-ln x)^p , dx$$




Any help would be appreciated




calculus integration




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marked as duplicate by Parcly Taxel, Robert Z calculus
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  • @gt6989b A direction would be appreciated
    – Um Shmum
    Jul 25 at 18:27










  • @gt6989b $-ln(x)>0$ for $xin(0,1)$.
    – Robert Z
    Jul 25 at 18:32











  • @RobertZ right, sorry
    – gt6989b
    Jul 25 at 18:32










  • $-ln x = ln frac 1x$ substitute $u = frac 1x$ this will change your limits to $[1, infty)$ and do a comparison test to functions that you know converge (and diverge).
    – Doug M
    Jul 25 at 18:34















up vote
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This question already has an answer here:



  • For what real values of a does the integral $intlimits_0^1(-ln x)^a,dx$ converge?

    1 answer




For what values of $p$ does the following integral converge: $$int_0^1(-ln x)^p , dx$$




Any help would be appreciated




calculus integration




share|cite|improve this question













marked as duplicate by Parcly Taxel, Robert Z calculus
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  • @gt6989b A direction would be appreciated
    – Um Shmum
    Jul 25 at 18:27










  • @gt6989b $-ln(x)>0$ for $xin(0,1)$.
    – Robert Z
    Jul 25 at 18:32











  • @RobertZ right, sorry
    – gt6989b
    Jul 25 at 18:32










  • $-ln x = ln frac 1x$ substitute $u = frac 1x$ this will change your limits to $[1, infty)$ and do a comparison test to functions that you know converge (and diverge).
    – Doug M
    Jul 25 at 18:34













up vote
-1
down vote

favorite
1









up vote
-1
down vote

favorite
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1






This question already has an answer here:



  • For what real values of a does the integral $intlimits_0^1(-ln x)^a,dx$ converge?

    1 answer




For what values of $p$ does the following integral converge: $$int_0^1(-ln x)^p , dx$$




Any help would be appreciated




calculus integration




share|cite|improve this question














This question already has an answer here:



  • For what real values of a does the integral $intlimits_0^1(-ln x)^a,dx$ converge?

    1 answer




For what values of $p$ does the following integral converge: $$int_0^1(-ln x)^p , dx$$




Any help would be appreciated





This question already has an answer here:



  • For what real values of a does the integral $intlimits_0^1(-ln x)^a,dx$ converge?

    1 answer





calculus integration





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 25 at 19:14









Michael Hardy

204k23186461




204k23186461









asked Jul 25 at 18:24









Um Shmum

1118




1118




marked as duplicate by Parcly Taxel, Robert Z calculus
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 Jul 25 at 18:36


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • @gt6989b A direction would be appreciated
    – Um Shmum
    Jul 25 at 18:27










  • @gt6989b $-ln(x)>0$ for $xin(0,1)$.
    – Robert Z
    Jul 25 at 18:32











  • @RobertZ right, sorry
    – gt6989b
    Jul 25 at 18:32










  • $-ln x = ln frac 1x$ substitute $u = frac 1x$ this will change your limits to $[1, infty)$ and do a comparison test to functions that you know converge (and diverge).
    – Doug M
    Jul 25 at 18:34

















  • @gt6989b A direction would be appreciated
    – Um Shmum
    Jul 25 at 18:27










  • @gt6989b $-ln(x)>0$ for $xin(0,1)$.
    – Robert Z
    Jul 25 at 18:32











  • @RobertZ right, sorry
    – gt6989b
    Jul 25 at 18:32










  • $-ln x = ln frac 1x$ substitute $u = frac 1x$ this will change your limits to $[1, infty)$ and do a comparison test to functions that you know converge (and diverge).
    – Doug M
    Jul 25 at 18:34
















@gt6989b A direction would be appreciated
– Um Shmum
Jul 25 at 18:27




@gt6989b A direction would be appreciated
– Um Shmum
Jul 25 at 18:27












@gt6989b $-ln(x)>0$ for $xin(0,1)$.
– Robert Z
Jul 25 at 18:32





@gt6989b $-ln(x)>0$ for $xin(0,1)$.
– Robert Z
Jul 25 at 18:32













@RobertZ right, sorry
– gt6989b
Jul 25 at 18:32




@RobertZ right, sorry
– gt6989b
Jul 25 at 18:32












$-ln x = ln frac 1x$ substitute $u = frac 1x$ this will change your limits to $[1, infty)$ and do a comparison test to functions that you know converge (and diverge).
– Doug M
Jul 25 at 18:34





$-ln x = ln frac 1x$ substitute $u = frac 1x$ this will change your limits to $[1, infty)$ and do a comparison test to functions that you know converge (and diverge).
– Doug M
Jul 25 at 18:34











1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










We have that for $y=1/x$



$$int_0^1(-ln x)^pdx=int_1^inftyfrac(ln y)^py^2dy=int_1^2frac(ln y)^py^2dy+int_2^inftyfrac(ln y)^py^2dy$$



and



$$int_2^inftyfrac(ln y)^py^2dy$$



converges for any $p$ by limit comparison test with $int frac1y^frac32 dy$ but



$$int_1^2frac(ln y)^py^2dy$$



diverges for $ple -1$, indeed for $p<0$, let $q=-p$ and $z=y-1$



$$int_1^2frac(ln y)^py^2dy=int_0^1frac1ln^q(1+z)(z+1)^2dz$$



and for $zto 0^+$



$$frac1ln^q(1+z) sim frac1z^q$$






share|cite|improve this answer























  • Can you explain how to complete the proof? I get that it converges for all $p$ which doesn't make sense to me.
    – Um Shmum
    Jul 25 at 18:55










  • To complete we can refer to limit comparison test for example with $int frac1y^frac32 dy$ and show that the given integral converge for any p.
    – gimusi
    Jul 25 at 19:01










  • @UmShmum For $p leq -1$ the integral diverges though.
    – ComplexYetTrivial
    Jul 25 at 19:31










  • @ComplexYetTrivial Ops...yes of course we have problem at 1 in taht case. Thanks I fix that.
    – gimusi
    Jul 25 at 19:36

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










We have that for $y=1/x$



$$int_0^1(-ln x)^pdx=int_1^inftyfrac(ln y)^py^2dy=int_1^2frac(ln y)^py^2dy+int_2^inftyfrac(ln y)^py^2dy$$



and



$$int_2^inftyfrac(ln y)^py^2dy$$



converges for any $p$ by limit comparison test with $int frac1y^frac32 dy$ but



$$int_1^2frac(ln y)^py^2dy$$



diverges for $ple -1$, indeed for $p<0$, let $q=-p$ and $z=y-1$



$$int_1^2frac(ln y)^py^2dy=int_0^1frac1ln^q(1+z)(z+1)^2dz$$



and for $zto 0^+$



$$frac1ln^q(1+z) sim frac1z^q$$






share|cite|improve this answer























  • Can you explain how to complete the proof? I get that it converges for all $p$ which doesn't make sense to me.
    – Um Shmum
    Jul 25 at 18:55










  • To complete we can refer to limit comparison test for example with $int frac1y^frac32 dy$ and show that the given integral converge for any p.
    – gimusi
    Jul 25 at 19:01










  • @UmShmum For $p leq -1$ the integral diverges though.
    – ComplexYetTrivial
    Jul 25 at 19:31










  • @ComplexYetTrivial Ops...yes of course we have problem at 1 in taht case. Thanks I fix that.
    – gimusi
    Jul 25 at 19:36














up vote
2
down vote



accepted










We have that for $y=1/x$



$$int_0^1(-ln x)^pdx=int_1^inftyfrac(ln y)^py^2dy=int_1^2frac(ln y)^py^2dy+int_2^inftyfrac(ln y)^py^2dy$$



and



$$int_2^inftyfrac(ln y)^py^2dy$$



converges for any $p$ by limit comparison test with $int frac1y^frac32 dy$ but



$$int_1^2frac(ln y)^py^2dy$$



diverges for $ple -1$, indeed for $p<0$, let $q=-p$ and $z=y-1$



$$int_1^2frac(ln y)^py^2dy=int_0^1frac1ln^q(1+z)(z+1)^2dz$$



and for $zto 0^+$



$$frac1ln^q(1+z) sim frac1z^q$$






share|cite|improve this answer























  • Can you explain how to complete the proof? I get that it converges for all $p$ which doesn't make sense to me.
    – Um Shmum
    Jul 25 at 18:55










  • To complete we can refer to limit comparison test for example with $int frac1y^frac32 dy$ and show that the given integral converge for any p.
    – gimusi
    Jul 25 at 19:01










  • @UmShmum For $p leq -1$ the integral diverges though.
    – ComplexYetTrivial
    Jul 25 at 19:31










  • @ComplexYetTrivial Ops...yes of course we have problem at 1 in taht case. Thanks I fix that.
    – gimusi
    Jul 25 at 19:36












up vote
2
down vote



accepted







up vote
2
down vote



accepted






We have that for $y=1/x$



$$int_0^1(-ln x)^pdx=int_1^inftyfrac(ln y)^py^2dy=int_1^2frac(ln y)^py^2dy+int_2^inftyfrac(ln y)^py^2dy$$



and



$$int_2^inftyfrac(ln y)^py^2dy$$



converges for any $p$ by limit comparison test with $int frac1y^frac32 dy$ but



$$int_1^2frac(ln y)^py^2dy$$



diverges for $ple -1$, indeed for $p<0$, let $q=-p$ and $z=y-1$



$$int_1^2frac(ln y)^py^2dy=int_0^1frac1ln^q(1+z)(z+1)^2dz$$



and for $zto 0^+$



$$frac1ln^q(1+z) sim frac1z^q$$






share|cite|improve this answer















We have that for $y=1/x$



$$int_0^1(-ln x)^pdx=int_1^inftyfrac(ln y)^py^2dy=int_1^2frac(ln y)^py^2dy+int_2^inftyfrac(ln y)^py^2dy$$



and



$$int_2^inftyfrac(ln y)^py^2dy$$



converges for any $p$ by limit comparison test with $int frac1y^frac32 dy$ but



$$int_1^2frac(ln y)^py^2dy$$



diverges for $ple -1$, indeed for $p<0$, let $q=-p$ and $z=y-1$



$$int_1^2frac(ln y)^py^2dy=int_0^1frac1ln^q(1+z)(z+1)^2dz$$



and for $zto 0^+$



$$frac1ln^q(1+z) sim frac1z^q$$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 25 at 19:47


























answered Jul 25 at 18:39









gimusi

65k73583




65k73583











  • Can you explain how to complete the proof? I get that it converges for all $p$ which doesn't make sense to me.
    – Um Shmum
    Jul 25 at 18:55










  • To complete we can refer to limit comparison test for example with $int frac1y^frac32 dy$ and show that the given integral converge for any p.
    – gimusi
    Jul 25 at 19:01










  • @UmShmum For $p leq -1$ the integral diverges though.
    – ComplexYetTrivial
    Jul 25 at 19:31










  • @ComplexYetTrivial Ops...yes of course we have problem at 1 in taht case. Thanks I fix that.
    – gimusi
    Jul 25 at 19:36
















  • Can you explain how to complete the proof? I get that it converges for all $p$ which doesn't make sense to me.
    – Um Shmum
    Jul 25 at 18:55










  • To complete we can refer to limit comparison test for example with $int frac1y^frac32 dy$ and show that the given integral converge for any p.
    – gimusi
    Jul 25 at 19:01










  • @UmShmum For $p leq -1$ the integral diverges though.
    – ComplexYetTrivial
    Jul 25 at 19:31










  • @ComplexYetTrivial Ops...yes of course we have problem at 1 in taht case. Thanks I fix that.
    – gimusi
    Jul 25 at 19:36















Can you explain how to complete the proof? I get that it converges for all $p$ which doesn't make sense to me.
– Um Shmum
Jul 25 at 18:55




Can you explain how to complete the proof? I get that it converges for all $p$ which doesn't make sense to me.
– Um Shmum
Jul 25 at 18:55












To complete we can refer to limit comparison test for example with $int frac1y^frac32 dy$ and show that the given integral converge for any p.
– gimusi
Jul 25 at 19:01




To complete we can refer to limit comparison test for example with $int frac1y^frac32 dy$ and show that the given integral converge for any p.
– gimusi
Jul 25 at 19:01












@UmShmum For $p leq -1$ the integral diverges though.
– ComplexYetTrivial
Jul 25 at 19:31




@UmShmum For $p leq -1$ the integral diverges though.
– ComplexYetTrivial
Jul 25 at 19:31












@ComplexYetTrivial Ops...yes of course we have problem at 1 in taht case. Thanks I fix that.
– gimusi
Jul 25 at 19:36




@ComplexYetTrivial Ops...yes of course we have problem at 1 in taht case. Thanks I fix that.
– gimusi
Jul 25 at 19:36


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