Is there a solution to this integral? -exp(-a*abs(x))*exp(i*(k0-k)*x)*dx

In summary, the conversation is about solving the integral \int exp(-a*abs(x))*exp(i*(k0-k)*x)*dx from -infinity to infinity. The participants are discussing different techniques and solutions, including splitting the integral into two parts and using trig identities and Euler's relation. They also mention the variable b = k0 - k and the importance of considering the absolute value when using certain approaches.
  • #1
quasar_4
290
0
Hi everyone. Maple and I have collectively racked our brains and I've tried most of the integration techniques I know. Does anyone know the solution to the integral

[tex]\int[/tex] exp(-a*abs(x))*exp(i*(k0-k)*x)*dx

from -infinity to infinity (not sure how to get the limits over the integral sign here in this text box)?

There might be a good change of variables, but my brain is now too fried to think of it. Or does this baby just not have a nice solution? anyone?
 
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  • #2
Not me :O)
 
  • #3
What? That's trivial if you just split the integral into two parts, one part from -infinity to zero and the other from 0 to infinity as it let's you get rid of the annoying abs(x).

I got [tex] \frac{1}{a-bi} + \frac{1}{a+bi} = \frac{2a}{a^2+b^2}[/tex]

BTW. [tex]b = k_0 - k[/tex] in my solution.
 
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  • #4
Haha, yes, I realized that once I got home and felt REALLY dumb for posting the previous msg. I think in fact you can use trig identities and Euler's relation as well to split into a sin and cos part, then the sin drops out (since it's over a symmetric interval) and you can take twice the integral of the cos part from 0-infinity. That hadn't worked at the time, but it turns out I was being brain-dead and forgetting to drop my abs. Duh! That's what I get for doing homework on 2 hours of sleep... :rofl:
 

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