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Integral form of the unit step function

  • Thread starter ryanwilk
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  • #1
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Homework Statement



I need to show that the unit step function ([tex]\Theta(s) = 0 [/tex] for [tex] s<0, 1 [/tex] for [tex] s>0[/tex]) can be written as [tex]\Theta(s)=\frac{1}{2\pi i} \int_{-\infty}^{\infty} dx \frac{e^{ixs}}{x-i0}.[/tex]

Homework Equations



-

The Attempt at a Solution



Firstly, I'm unsure about what "x-i0" actually means. I've looked online and couldn't find anything but if it means "x minus an infinitessimal multiple of i", it kinda works.

There will be a pole in the upper half of the complex plane.

-For s>0, the pole will be contained, with residue [tex]e^0 = 1[/tex]. Then calculating the integral and dividing by [tex]2\pi i[/tex] will give [tex]\Theta(s) = 1 [/tex] for [tex] s>0.[/tex]

-For s<0, the pole won't be contained so the integral will be zero and [tex]\Theta(s) = 0 [/tex] for [tex] s<0.[/tex]

However, if "x-i0" just means "x", the pole is on the axis and it won't make a difference whether s is less or greater than 0...
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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You are right, you should actually read

[tex]
\Theta(s)= \lim_{\epsilon \downarrow 0} \frac{1}{2\pi i} \int_{-\infty}^{\infty} dx \frac{e^{ixs}}{x-i\epsilon}
[/tex]
(note how the limit is taken from above).

(As a side remark: In theoretical physics this is a very important equation, relating to causality of events).
 
  • #3
57
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You are right, you should actually read

[tex]
\Theta(s)= \lim_{\epsilon \downarrow 0} \frac{1}{2\pi i} \int_{-\infty}^{\infty} dx \frac{e^{ixs}}{x-i\epsilon}
[/tex]
(note how the limit is taken from above).

(As a side remark: In theoretical physics this is a very important equation, relating to causality of events).
Oh right, thanks a lot! :smile:
 
  • #4
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Any one can explain why the pole is not contained in the upper complex plane if s < 0? Thank you!
 
  • #5
CompuChip
Science Advisor
Homework Helper
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47
Well, the pole is the solution of [itex]x_0 - i \epsilon = 0[/itex], i.e it is at [itex]x_0 = i \epsilon[/itex].
So if you take the limit from above, where [itex]\epsilon > 0[/itex], then it is clearly in the top half of the complex plane (i.e. {z in C | Im(z) > 0 }).

You close the curve in counterclockwise direction, so [itex]\operatorname{Im}(x) \to +\infty[/itex], therefore it encircles the pole.
 
  • #6
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Thank you!

Well, the pole is the solution of [itex]x_0 - i \epsilon = 0[/itex], i.e it is at [itex]x_0 = i \epsilon[/itex].
So if you take the limit from above, where [itex]\epsilon > 0[/itex], then it is clearly in the top half of the complex plane (i.e. {z in C | Im(z) > 0 }).

You close the curve in counterclockwise direction, so [itex]\operatorname{Im}(x) \to +\infty[/itex], therefore it encircles the pole.
 

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