1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is this analytically solvable?

  1. Feb 25, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm trying to solve integrals of the form:
    ## \int_{-\infty}^{\infty} \frac{k}{k-1} e^{itk^2}e^{ikx}e^{\frac{-(k-k_0)^2}{\alpha^2}} dk \\
    \int_{-\infty}^{\infty} \frac{1}{k-1} e^{itk^2}e^{ikx}e^{\frac{-(k-k_0)^2}{\alpha^2}} dk ##

    A bit of background:
    I'm trying to set up a Mathematica notebook to show a wave packet going through a delta function and these integrals come from the transmission and reflection coefficients and i'm told i need to compute the integrals and plot then use "Animate" to display the propagation.
    2. Relevant equations


    3. The attempt at a solution
    I have tried using Mathematica but get an error saying the integral doesn't converge on the interval so I thought I'd try doing this analytically. I am told it should be possible but I'm not really sure how.
    I solve ones similar to this relatively easily but the fraction at the beginning is causing many issues.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 2, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Mar 4, 2017 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Seems to me the second integral can be rewritten as ##\int_{-\infty}^{\infty}\frac 1ke^{Q(k)}.dk##, where Q is a quadratic function with complex coefficients.
    The integral from 0 to K does not converge, but arguably you can cancel the infinities near 0+ with those near 0-.
    ##\int_{0}^{\infty}\frac 1k(e^{Q(k)}-e^{Q(-k)}).dk=\int_0^{\infty}\frac 1ke^{ak^2+c}(e^{bk}-e^{-bk}).dk##
    Since sinc(x)=sin(x)/x is much better behaved, that might work.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Is this analytically solvable?
  1. Is this solvable? (Replies: 1)

  2. Solvable groups (Replies: 1)

  3. Solvable Group (Replies: 1)

Loading...