Solving Complex Analysis Integral Using Residues

[KNUCK7ES]

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

 

[jackmell]

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:

\int_2^{\infty}\frac{dx}{x\sqrt{x-2}}

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:

\mathop\oint\limits_{K}\frac{dz}{z\sqrt{z-2}}

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 \sqrt{z-2} and on bottom, it's the other value, -\sqrt{z-2}.

 

[KNUCK7ES]

First make sure you have it clearly expressed as in:

\int_2^{\infty}\frac{dx}{x\sqrt{x-2}}

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:

\mathop\oint\limits_{K}\frac{dz}{z\sqrt{z-2}}

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 \sqrt{z-2} and on bottom, it's the other value, -\sqrt{z-2}.

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

\int\frac{1}{x\sqrt{x-2}} dx

and

\int\frac{-1}{x\sqrt{x-2}} dx

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

2\int\frac{1}{x\sqrt{x-2}} dx

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

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

residue= limz\rightarrow 0 \frac{2}{\sqrt{z^{2}-2}}

 

[KNUCK7ES]

thanks btw