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February 19th, 2008, 09:35 AM   #1
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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?
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February 19th, 2008, 10:41 AM   #2
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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/x-1)
u'=1

this gives you:

I=x sqrt(z/x-1)+z/2 int[1/(x sqrt(z/x-1))dx]

now we have another integral that can be solved using substitution:

t=sqrt(z/x-1)

This gives you:

I=x sqrt(z/x-1)-z arctan(sqrt(z/x-1))+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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