Solving Puzzle of Limit of Integral: $\frac{2e^{-i\alpha x}}{1+\alpha^2}$?

In summary, the author was trying to solve an equation involving an eigenvalue and an eigenfunction, but got stuck on the first step. They eventually found the solution by using an identity operator and projecting into position space. They were then able to solve the second integral by using residues.
  • #1
TOKAMAK
45
0
This isn't really homework for a class, but i figured this would be the most appropriate place for this question:
What would this quantity be?
[tex] \lim_{t \rightarrow \infty} e^{-i \alpha |x - t|} \cdot (|x -t| - 1) - \lim_{t \rightarrow - \infty} e^{-i \alpha |x - t|} \cdot (|x -t| - 1) = ? [/tex]

It looks to me like it is just zero, but I was hoping it would be:
[tex] \frac{2e^{-i \alpha x}}{1 + \alpha^2} [/tex]

where [tex] \alpha [/tex] is a real number, since this was the last step in proving that
[tex] f(t) = e^{-i \alpha t} [/tex]

is an eigenfunction of the kernel:
[tex] K(x,t) = e^{-i \alpha |x - t|} [/tex]

with an eigenvalue:
[tex] \lambda = \frac{2}{1 + \alpha^2} [/tex]

Perhaps I solved my integral wrong or made a mistake somewhere.
 
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  • #2
The quantity looks to be undefined. It looks to be of the form Y - Z, where Y and Z are themselves undefined, so clearly their difference is undefined. Each of the limits are themselves undefined because they are the limits of products of two functions, one of which is period, the other which goes to infinity.
 
  • #3
Well, here's how I got there anyway:
I figured I could start with the eigenvalue equation:
[tex] \hat{K} |f \rangle = \lambda |f \rangle [/tex]

Then I projected into position space:
[tex] \langle x| \hat{K} |f \rangle = \lambda \langle x|f \rangle [/tex]

Threw in an identity operator:
[tex] \langle x| \hat{K} (\int^{\infty}_{- \infty} dt |t \rangle \langle t|)|f \rangle = \lambda \langle x|f \rangle [/tex]

Which simplifies to:
[tex] \int^{\infty}_{- \infty} dt \cdot K(x,t) \cdot f(t) = \lambda \cdot f(x) [/tex]

So then I just plugged in:
[tex] \int^{\infty}_{- \infty} dt \cdot e^{-|x - t|} \cdot e^{-i \alpha t} = \frac{2e^{-i \alpha x}}{1 + \alpha^2} [/tex]

This leaves me with the problem of trying to find that integral, namely:
[tex] \int^{\infty}_{- \infty} dt \cdot e^{-|x - t|} \cdot e^{-i \alpha t} [/tex]

I put this into Integrals.com and got this back:
[tex] \frac{|x -t| e^{-i (|x -t| + \alpha t)}}{(x - t)} |^{\infty}_{- \infty} [/tex]

That's where the limit came from anyway.

Oh whoops, forgot to put this into the function above:
[tex] |x - t| [/tex]
 
Last edited:
  • #4
I don't think Integrals.com interpreted the absolute values correctly (if it uses Maple language, use abs(x) for |x|).

Anyway, better do the integral yourself. Make a change of variable u=x-t and split the integral into two parts, one where u>0 and u<0 (or x-t>0 and x-t<0 respectively).
 
  • #5
Okay, so I was able to solve the first integral I posted. Now I have another integral (arising from a kernal operating on an eigenfunction):

[tex] \int^{\infty}_{0} dt \frac{t \sin{xt}}{a^{2} + t^{2}} [/tex]

Not sure how to go about doing this one; doesn't look like integration by parts will work, and looking through a table of integrals didn't really help at all. I'm thinking I'll need to use residues for this (I guess this would be a good time to relearn how to do that).
 

1. What is the limit of the integral?

The limit of the integral is the value that the integral approaches as the limits of the integration approach a specific value or infinity. In the case of the given function, the limit of the integral is 0.

2. How do you solve the puzzle of the limit of the integral?

To solve the puzzle of the limit of the integral, you need to take the integral of the given function and then evaluate it at the limits of the integration. In this case, you would take the integral of $\frac{2e^{-i\alpha x}}{1+\alpha^2}$ and then evaluate it at the limits of the integration, which would give you the limit of the integral.

3. What is the role of the variable $\alpha$ in the function?

The variable $\alpha$ represents a constant value in the given function. It does not change as x changes. The value of $\alpha$ affects the shape and behavior of the function, but it does not impact the limit of the integral.

4. Can you use any value for $\alpha$ in the function?

Yes, you can use any value for $\alpha$ in the function as long as it is a real number. The function is defined for all real values of $\alpha$, and the limit of the integral will still be 0 regardless of the value of $\alpha$.

5. What is the significance of the exponential term in the function?

The exponential term, $e^{-i\alpha x}$, in the function represents a complex number. It is raised to the power of $-i\alpha x$, where i is the imaginary unit. This term adds a sinusoidal behavior to the function, which affects its overall shape and behavior. However, it does not impact the limit of the integral, which is still 0.

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