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Thanks. I'm not really sure that I ''needed'' to use some sort of comparison to show it didn't converge, as it did integrate nicely, but its good to have a second opinion, be that mathematical or otherwise. --[[User:Jmason|John Mason]] | Thanks. I'm not really sure that I ''needed'' to use some sort of comparison to show it didn't converge, as it did integrate nicely, but its good to have a second opinion, be that mathematical or otherwise. --[[User:Jmason|John Mason]] | ||
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| + | It does not converge for me either, but Josh, be careful with your comparison. It's not really valid to use <math>1/\sqrt{x^4}</math> as the function for comparison, because <math>x/\sqrt{x^4} < x/\sqrt{x^4}</math>, albeit infinitesimally less. I used <math>(g(x) = 1/x^{0.99}) > (f(x) = x/\sqrt{x^4}) </math>, in which case P < 1 and diverges. --[[User:Reckman|Randy Eckman]] 15:44, 26 October 2008 (UTC) | ||
Revision as of 11:44, 26 October 2008
8.8 #54
Does this indefinite integral converge for anyone? Also, if you are having trouble with the integral, take a look at the derivatives of inverse hyperbolic functions. --John Mason
It does not converge for me. I used direct comparison to test whether it converges or not. I started by comparing $ \sqrt{x^4-1} $ and $ \sqrt{x^4} $ It was easy from there. --Josh Visigothsandwich
Thanks. I'm not really sure that I needed to use some sort of comparison to show it didn't converge, as it did integrate nicely, but its good to have a second opinion, be that mathematical or otherwise. --John Mason
It does not converge for me either, but Josh, be careful with your comparison. It's not really valid to use $ 1/\sqrt{x^4} $ as the function for comparison, because $ x/\sqrt{x^4} < x/\sqrt{x^4} $, albeit infinitesimally less. I used $ (g(x) = 1/x^{0.99}) > (f(x) = x/\sqrt{x^4}) $, in which case P < 1 and diverges. --Randy Eckman 15:44, 26 October 2008 (UTC)
