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

[TOKAMAK]

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?
\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) = ?

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

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

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

with an eigenvalue:
\lambda = \frac{2}{1 + \alpha^2}

Perhaps I solved my integral wrong or made a mistake somewhere.

 

[AKG]

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.

 

[TOKAMAK]

Well, here's how I got there anyway:
I figured I could start with the eigenvalue equation:
\hat{K} |f \rangle = \lambda |f \rangle

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

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

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

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

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

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

That's where the limit came from anyway.

Oh whoops, forgot to put this into the function above:
|x - t|

 

[Galileo]

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).

 

[TOKAMAK]

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):

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

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).