# Integrate 0 to ∞: Improper √x/(x^2+1)

• Bachelier

#### Bachelier

Please I need help with this Improper Integration between 0 and infinity

$$\int\sqrt{x}/(x^2 + 1)dx$$

Thank you very much

Try Integration By Parts

$$\int$$ u dv = uv - $$\int$$ v du

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].

substitute u = sqr(x), then use partial fractions

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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.

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].

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.

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:

$$\frac{\pi}{\sqrt{2}}$$

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:

$$\frac{\pi}{\sqrt{2}}$$

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.

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.

in [0,1], I get a maxima of $$1/\sqrt{3}$$. The graph interestingly goes beyond y=1/2 but never above y= 0.6, making the area less than or equal to $$1/\sqrt{3}$$ in the said interval.

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:
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.

Indeed..Thanks for the editing. :)
regarding your quote: "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.

You are absolutely correct, but if we wanted to narrow down the area, we can use the graph to calculate the maximum value of y, here y= f(1/sqrt(3))=0.59 and since x=1, then the area under the graph is smaller or equal to x*y = 0.59...

What do you think?

please see attached pic for the graph. :)

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Sure you could. If you wanted to narrow it more you could cut [0,1] into subintervals, and also split [1,infinity) into subintervals. Depends on what you want to do. You could also with some work integrate it exactly using either substitution and partial fraction as waht suggested or a contour integration as Count Iblis suggested. If you haven't done contour integrals before I wouldn't suggest starting with this one. And waht's technique is kind of a long haul too, but doesn't require complex analysis. Like I say, depends on what kind of answer you are happy with.

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.

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.

On another note, the greatest discovery for me was the engine called Wolfram Alpha mentioned by Waht. This discovery by itself was worth the thread. Although Waht decided to delete the link, I am still thankful. I think this is a great tool for someone who wants to learn and understand mathematics, and not just do homework or get an A or B in a class.

I have Maple and Mathematica 7 installed on my computer, yet I found alpha a lot smoother and more fun.

You're welcome. Wolfram Alpha seems pretty dangerous to me. I can imagine some people might have a hard time resisting the urge to peek at the answer before they even try to solve the problem. If the contents of WA's 'Show steps' were posted on the Forum as the response to a question the moderators would delete it.