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Please I need help with this Improper Integration between 0 and infinity
[tex]\int\sqrt{x}/(x^2 + 1)dx[/tex]
Thank you very much
[tex]\int\sqrt{x}/(x^2 + 1)dx[/tex]
Thank you very much
You can't really 'integrate' using the comparison method, but you can show the integral exists and find a bound on it. If that's all you want to do, it's a lot easier than actually doing the integration. To deal with the interval [0,1] you had the right idea when you looked at the graph and observed it's approximately between 0 and 1/2. You could just compute the upper bound by finding the derivative, looking for critical points etc. If you do you'll find it's actually a bit larger than 1/2.It doesn't work my friend. The only way to integrate this is via the comparison method.
Unfortunately I get stuck in the interval [0,1].
I am not yet familiar with contour integration. But thank you for the heads-up. I am going to run a google search for it.Computing the integral is a piece of cake using contour integration. In fact, it is so easy you can do it in your head. It is:
[tex]\frac{\pi}{\sqrt{2}}[/tex]
Thank you Dick...Your answer helped a lot in directing me towards the correct answer.You can't really 'integrate' using the comparison method, but you can show the integral exists and find a bound on it. If that's all you want to do, it's a lot easier than actually doing the integration. To deal with the interval [0,1] you had the right idea when you looked at the graph and observed it's approximately between 0 and 1/2. You could just compute the upper bound by finding the derivative, looking for critical points etc. If you do you'll find it's actually a bit larger than 1/2.
Indeed..Thanks for the editing. :)The maximum on [0,1] is attained at x=1/sqrt(3) making the value of the integral over [0,1] less than f(1/sqrt(3)) where f(x)=sqrt(x)/(x^2+1). That's what you meant to say, right? You could also do a rougher comparison by noting sqrt(x)/(1+x^2)<=1/1, the max of the numerator over the min of the denominator.
Dick, thank you for coaching me through this problem. It allowed me to review and practice some concepts. I hope we can repeat this in the future with more problems.Oh, great. So Wolfram Alpha will not only tell you the answer to your homework, it will tell you what intermediate steps to write down. That's nasty.