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Complex analysis

  • Thread starter KNUCK7ES
  • Start date
  • #1
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Homework Statement


compute

I=∫_2^∞ (1/(x(x-2)^.5)) dx

using the calculus of residues. be sure to choose an appropriate contour and to explain what happens on each part of that contour.

Homework Equations



transform to a complex integral

I= ∮ (1/(z(z-2)^.5)) dz

The Attempt at a Solution



i don't even know where to really begin. i keep ending up dividing by zero
 

Answers and Replies

  • #2
1,796
53

Homework Statement


compute

I=∫_2^∞ (1/(x(x-2)^.5)) dx

using the calculus of residues. be sure to choose an appropriate contour and to explain what happens on each part of that contour.

Homework Equations



transform to a complex integral

I= ∮ (1/(z(z-2)^.5)) dz

The Attempt at a Solution




i don't even know where to really begin. i keep ending up dividing by zero
First make sure you have it clearly expressed as in:

[tex]\int_2^{\infty}\frac{dx}{x\sqrt{x-2}}[/tex]

So how about use a key-hole contour that begins above the real-axis at x=2, travels down it to infinity, loops all the way round to right below the real axis at infinity, travles down the axis to 2 again, then loops around 2. That's a closed contour and I'll write the contour integral around that as:

[tex]\mathop\oint\limits_{K}\frac{dz}{z\sqrt{z-2}}[/tex]

Now, can you analyze the integral over each leg of that contour (I count 4 distinct legs, two horizontal ones, that big circular one, and that real small one around 2). Keep in mind that the horizontal legs are over a branch-cut of the square root function so on top, it's one value of the branch say [itex]\sqrt{z-2}[/itex] and on bottom, it's the other value, [itex]-\sqrt{z-2}[/itex].
 
Last edited:
  • #3
13
0
First make sure you have it clearly expressed as in:

[tex]\int_2^{\infty}\frac{dx}{x\sqrt{x-2}}[/tex]

So how about use a key-hole contour that begins above the real-axis at x=2, travels down it to infinity, loops all the way round to right below the real axis at infinity, travles down the axis to 2 again, then loops around 2. That's a closed contour and I'll write the contour integral around that as:

[tex]\mathop\oint\limits_{K}\frac{dz}{z\sqrt{z-2}}[/tex]

Now, can you analyze the integral over each leg of that contour (I count 4 distinct legs, two horizontal ones, that big circular one, and that real small one around 2). Keep in mind that the horizontal legs are over a branch-cut of the square root function so on top, it's one value of the branch say [itex]\sqrt{z-2}[/itex] and on bottom, it's the other value, [itex]-\sqrt{z-2}[/itex].
ok i did so, so it the contour breaks up into for parts, Jo + JI + Jtop+ Jbottom where Jo means the outer part of the contour and JI is the inner part to the contour. Jo=JI=0. now i just do the calculus of residues on the Jtop and Jbottom, whic are

[itex]\int\frac{1}{x\sqrt{x-2}} dx[/itex]

and

[itex]\int\frac{-1}{x\sqrt{x-2}} dx[/itex]

this since the bounds are flipped, the negative sign can be cancelled and the bounds are the same allowing me to add them,

2[itex]\int\frac{1}{x\sqrt{x-2}} dx[/itex]

i know that there is only one sing. in my contour which is z= 0. so

2[itex]\int\frac{1}{x\sqrt{x-2}} dx[/itex]=2*pi*i Ʃ residues

residue= lim[itex]z\rightarrow 0[/itex] [itex]\frac{2}{\sqrt{z^{2}-2}}[/itex]
 
  • #4
13
0
thanks btw
 

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