Science Talk Science Talk    FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Forums Science Forums Biology Math Astronomy Physics Technology Chemistry Social Sciences History Psychology Philosophy Sociology Linguistics Religious Studies Economics Man Woman Ethno Ask an Expert World Records Society Issues Education People Alternative Science in L1 but not in L2          Science Talk Forum Index -> Mathematics View previous topic :: View next topic   Author Message Die stimme Guest Posted: Tue Jul 08, 2008 6:54 am    Post subject: in L1 but not in L2 Prove that: 1/(|x|(1+(log(x))^2)) is in L1(R) but not in L2(R). Guess I have to show that its module is integrable on R but not so for the square of its module. Clues? Thanks Back to top   Ads Advertising Sponsor Joubert Guest Posted: Tue Jul 08, 2008 6:55 am    Post subject: Re: in L1 but not in L2 On Tue, 08 Jul 2008 01:54:07 +0000, Die stimme wrote: Quote: Prove that: 1/(|x|(1+(log(x))^2)) is in L1(R) but not in L2(R). Guess I have to show that its module is integrable on R but not so for the square of its module. Clues? Thanks Correction: the function is 1/(|x|(1+(log(|x|))^2)) Back to top   Ads Advertising Sponsor Die stimme Guest Posted: Tue Jul 08, 2008 6:56 am    Post subject: Re: in L1 but not in L2 On Tue, 08 Jul 2008 01:54:07 +0000, Die stimme wrote: Quote: Prove that: 1/(|x|(1+(log(x))^2)) is in L1(R) but not in L2(R). Guess I have to show that its module is integrable on R but not so for the square of its module. Clues? Thanks Correction: it's 1/(|x|(1+(log(|x|))^2)) Back to top   Ads Advertising Sponsor The World Wide Wade Guest Posted: Tue Jul 08, 2008 7:47 am    Post subject: Re: in L1 but not in L2 In article <4872c949$0$35957$4fafbaef@reader2.news.tin.it>, Die stimme <massagein2stimmen765@hotmail.com> wrote: Quote: On Tue, 08 Jul 2008 01:54:07 +0000, Die stimme wrote: Prove that: 1/(|x|(1+(log(x))^2)) is in L1(R) but not in L2(R). Guess I have to show that its module is integrable on R but not so for the square of its module. Clues? Thanks Correction: it's 1/(|x|(1+(log(|x|))^2)) It's even so think about (0,oo). For L^1, try the substitution u = log(x). For L^2 the problem is near 0, not oo. You could make the same substitution here, but you'll learn more if you don't. Back to top   Ads Advertising Sponsor Die stimme Guest Posted: Tue Jul 08, 2008 11:14 am    Post subject: Re: in L1 but not in L2 On Mon, 07 Jul 2008 19:47:28 -0700, The World Wide Wade wrote: Quote: It's even so think about (0,oo). For L^1, try the substitution u = log(x). For L^2 the problem is near 0, not oo. You could make the same substitution here, but you'll learn more if you don't. I calculated the first integral through the substitution you suggested but can't get to grips with the divergence of the second. I mean I find it almost evident somehow but can't prove it formally. Back to top   Ads Advertising Sponsor Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First         Science Talk Forum Index -> Mathematics All times are GMT Page 1 of 1   Jump to: Select a forum Science Talk----------------Science TalkAstrologyNanotechnologySETISociology-- Elections-- LibertarianCreationismMathematicsPolitics  You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum Australian Debt Consolidation Experts medical insurance Wedding Website Escort e Accompagnatrici a Milano, Roma, Firenze, Bologna, Padova, Verona, Venezia Sex Pals Link Exchange Secured Loans Make Your Own Website Cheap International Calls Long island Cleaning service Mold UK Swingers Genuine Contacts Site floor machines Hoover Vacuum Parts 192 Attacks blocked Powered by phpBB © 2001, 2005 phpBB Group