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

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Discussion Overview

The discussion centers around the evaluation of the integral \(\int \exp(-a \cdot \text{abs}(x)) \cdot \exp(i \cdot (k_0 - k) \cdot x) \, dx\) from \(-\infty\) to \(\infty\). Participants explore various techniques for solving this integral, including potential changes of variables and splitting the integral into parts.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses difficulty in finding a solution and suggests that a change of variables might help.
  • Another participant proposes splitting the integral into two parts to eliminate the absolute value function, leading to a specific solution involving complex numbers.
  • A later reply acknowledges the initial oversight and discusses using trigonometric identities and Euler's relation to further simplify the integral, noting that the sine component drops out due to symmetry.

Areas of Agreement / Disagreement

There is no consensus on the solution to the integral, as participants present different approaches and insights without resolving the overall question of whether a nice solution exists.

Contextual Notes

Participants mention potential changes of variables and the use of trigonometric identities, but the discussion does not clarify all assumptions or steps involved in the proposed solutions.

quasar_4
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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

\int 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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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 \frac{1}{a-bi} + \frac{1}{a+bi} = \frac{2a}{a^2+b^2}

BTW. b = k_0 - k in my solution.
 
Last edited:
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... :smile:
 

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