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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
Bachelier said: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].
Count Iblis said: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]
Dick said: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.
Dick said: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 said: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.