Integral form of the unit step function

[ryanwilk]

Homework Statement

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

Homework Equations

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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 e^0 = 1. Then calculating the integral and dividing by 2\pi i will give \Theta(s) = 1 for s>0.

-For s<0, the pole won't be contained so the integral will be zero and \Theta(s) = 0 for s&lt;0.

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...

 

[CompuChip]

You are right, you should actually read

<br /> \Theta(s)= \lim_{\epsilon \downarrow 0} \frac{1}{2\pi i} \int_{-\infty}^{\infty} dx \frac{e^{ixs}}{x-i\epsilon}<br />
(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).

 

[ryanwilk]

You are right, you should actually read

<br /> \Theta(s)= \lim_{\epsilon \downarrow 0} \frac{1}{2\pi i} \int_{-\infty}^{\infty} dx \frac{e^{ixs}}{x-i\epsilon}<br />
(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!

 

[wzr9999]

Any one can explain why the pole is not contained in the upper complex plane if s < 0? Thank you!

 

[CompuChip]

Well, the pole is the solution of x_0 - i \epsilon = 0, i.e it is at x_0 = i \epsilon.
So if you take the limit from above, where \epsilon &gt; 0, 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 \operatorname{Im}(x) \to +\infty, therefore it encircles the pole.

 

[wzr9999]

Thank you!

Well, the pole is the solution of x_0 - i \epsilon = 0, i.e it is at x_0 = i \epsilon.
So if you take the limit from above, where \epsilon &gt; 0, 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 \operatorname{Im}(x) \to +\infty, therefore it encircles the pole.