- #1
Bachelier
- 376
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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.
An improper integral is an integral where either the upper or lower limit of integration is infinite, or the integrand has a discontinuity within the interval of integration. It is called "improper" because it does not meet the standard conditions for a definite integral.
The square root in the integrand indicates that the function is not continuous for all values of x. This means that the integral is improper and must be evaluated using special techniques.
To solve an improper integral, you can use a mathematical technique called "limit of integration". This involves taking the limit of the integral as the upper or lower bound approaches infinity (or a point of discontinuity). If the limit exists, then it is the value of the improper integral.
The limit of integration for this integral is infinity, since the upper bound is ∞. This means that the upper bound is not a specific number, but rather a concept of approaching infinity.
No, the Fundamental Theorem of Calculus only applies to definite integrals where the upper and lower bounds are finite. Improper integrals require special techniques to evaluate, such as limit of integration or other methods.