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February 19th, 2008, 09:35 AM  #1 
Newbie Joined: Feb 2008 Posts: 3 Thanks: 0  integrating squareroot[ (z/x)  1 ] dx
I'm having trouble integrating the form that's in the subject squareroot[ (z/x)  1 ] dx from 0 to infinity anyone willing to share some insight on where to start? is there some identity for this form? 
February 19th, 2008, 10:41 AM  #2 
Senior Member Joined: May 2007 Posts: 402 Thanks: 0 
First of all, the integral does not converge on 0 to infinity. This isn't so hard to see since the function it self isn't defined for x > z. Still, let's try to find the integral it self, not concerning about the boundaries (I'll do the calculation replacing infinity with z). First, use integration by parts. Using: v=sqrt(z/x1) u'=1 this gives you: I=x sqrt(z/x1)+z/2 int[1/(x sqrt(z/x1))dx] now we have another integral that can be solved using substitution: t=sqrt(z/x1) This gives you: I=x sqrt(z/x1)z arctan(sqrt(z/x1))+C Now, let us do the boundaries: lim(I(x),x>z) = 0 lim(I(x),x>0) = z arctan(infinity)=z pi/2 so: I form 0 to z = z pi/2 for all z in R 

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integrating, squareroot, z or x 
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