Contour Integration

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Homework Statement



[tex] lim_{\alpha\rightarrow 0} \int_{-\infty}^{\infty} dx. e^{ixy}/(2\Pi i (x-i \alpha)) = H(y)[/tex]
where H(y) is the step function ie. H(y) = 1 for y > 0, H(y) = 0 (otherwise)
Compute using an appropriate contour integral.


Homework Equations


-Laurent series
-Residue theorem
-Concepts from contour integration


The Attempt at a Solution


I found the residue and its [tex] e^{-y \alpha} [/tex] for a semi circular contour lying in the top half plane. Mathematica vouches for that.

To find the integral, I followed the usual procedure.
1. Define a semi circular contour
2. Separate the contour into two parts one along the x-axis and the other constituting the semicircular arc. The straight line becomes the desired integral.

Problems:
-How do I reduce the Integral for the semi-circular arc to zero.
-If it IS zero, how do I relate the residue which is an exponential to the step function.

I was thinking that this might be something like an inverse fourier transform. But the fourier transform of the Step function involves a Delta function: http://mathworld.wolfram.com/FourierTransformHeavisideStepFunction.html

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Dick
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You can only close that in the upper half plane if y>0. This is because you need to be able to ignore the contribution from the semicircle as r->infinity. In the upper half plane Im(x)>0 so you have a negative real part of the exponential. If y<0 then you need to close in the lower half plane. That's where the 'step function' quality comes from.
 
  • #3
Office_Shredder
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Maybe I'm just crazy, but as a goes to 0, you don't know if it's going to be in your contour, so you probably need to have an indentation around the origin (with the origin contained in your contour) in order to ensure that for small values of |a| the pole is in the contour (and then at the very end you take the limit of the radius of the indentation going to zero).
 
  • #4
Dick
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Maybe I'm just crazy, but as a goes to 0, you don't know if it's going to be in your contour, so you probably need to have an indentation around the origin (with the origin contained in your contour) in order to ensure that for small values of |a| the pole is in the contour (and then at the very end you take the limit of the radius of the indentation going to zero).

You can't really do that. It's important to the argument that the pole is in the upper half plane. If alpha goes to zero from the negative direction you get the wrong step function. The problem should have stated it goes to zero from the positive direction.
 
  • #5
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You can only close that in the upper half plane if y>0. This is because you need to be able to ignore the contribution from the semicircle as r->infinity. In the upper half plane Im(x)>0 so you have a negative real part of the exponential. If y<0 then you need to close in the lower half plane. That's where the 'step function' quality comes from.

That clears up a lot in my foggy brain but there are still some issues. The y in the integral. Is this the complex part of the variable in the contour integral. i.e. z = x+iy. Or is it just some arbitrary variable. I thought that it was not connected to the integration variable in the contour integral and simply substituted z for x
 
  • #6
Dick
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That clears up a lot in my foggy brain but there are still some issues. The y in the integral. Is this the complex part of the variable in the contour integral. i.e. z = x+iy. Or is it just some arbitrary variable. I thought that it was not connected to the integration variable in the contour integral and simply substituted z for x

It's not connected with the integration variable. The y in the integral is the y in H(y). It's a real number.
 
  • #7
Dick
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You have to look at the e^(i*z*y) part in the numerator. Since y is real the real part of i*z*y is -y*Im(z), right? To argue that you can ignore the contribution for the circular part of the contour you want to be able to say that is SMALL for large values of Im(z). If y>0 that means you want Im(z)>0 and large. That means upper half plane. Vice versa for y<0. Making the contour move to the lower half plane when y<0 is just what you want! No enclosed pole means integral=0, H(y)=0.
 
  • #8
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It's not connected with the integration variable. The y in the integral is the y in H(y). It's a real number.

Eureka.

Thanks a lot
 

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