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Math Help - Integrating a function that includes a series expansion

  1. #1
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    Integrating a function that includes a series expansion

    Hello everyone. First of all, greetings, this is my first time here. I have just begun a chemistry master's program in Germany.

    For my question, I am not looking for the solution for free, but more assistance on where I should even begin. The problem and the final equation are both given in the lecture notes; unfortunately, the gap between of how to get to the solution was not provided.

    Without further ado:

    problem

    Integrating a function that includes a series expansion-einfuehrung-die-theoretische-chemie.png

    I could really use a tip in the right direction. I have been refreshing my mind of geometric series, power series, and Taylor series. My first inclination is to try to rewrite the series w(k) as the function it represents. I stumble here, because it does not quite resemble a Taylor series since 1/n! is not present, and if I attempt to correlate it with a geometric or power series of the form a+ax+ax^2... (origin at zero for simplicity), I cannot match the derivative coefficient in w(k) to the coefficient "a" from a power series.

    In other words, I don't know how to plug in w(k) into the given equation in a way that I can then solve the integral. I've tried working backwards from the final equation. I can follow it mostly, except that I don't know where the sin function comes from. The Taylor series of a sin function has alternating positive and negative terms, which is not a feature of w(k).

    It's driving me crazy. I'd appreciate any links, starting points, words of advice that can be given. Thank you very much!
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  2. #2
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    Re: Integrating a function that includes a series expansion

    Hey blaisem.

    In this question, are you only using the approximation of w(k) given in the question or are you using the whole series expansion?

    In the case of the former, consider the real part of Integral e^(-iw(k)t)dk across the region.
    Thanks from blaisem
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  3. #3
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    Re: Integrating a function that includes a series expansion

    I AM SO CLOSE. Thanks a lot for your response, Chiro! I kept trying to plug in the series expansion to infinity, and I had no idea how to solve the subsequent integral. But if I just take the first two terms, my answer comes out very close to what was shown in the handout to be the answer.

    Original problem:

    Integrating a function that includes a series expansion-einfuehrung-die-theoretische-chemie.png

    I am stuck, though, and maybe I am missing something very obvious. I have reached the final equation, only the denominator term is reversed ie.

    instead of: w't-x
    I have: x-w't

    The same switch has occurred in my sin function as well.

    I backtracked my work, and the denominator term comes from the chain rule when I integrate the original function. My original function is simply the first two terms of the Taylor expansion substituted in for w(k).

    In other words, it looks like this:

    Integrating a function that includes a series expansion-einfuehrung-die-theoretische-chemie-solving-integral-.png

    I think that this is where the error comes from, namely the denominator is already reversed in sign.

    The rest of my work is as follows

    Integrating a function that includes a series expansion-einfuehrung-die-theoretische-chemie-simplifying-product-.png

    And the final answer is: (the last equation from the above image was reproduced as the first equation of the following image)

    Integrating a function that includes a series expansion-einfuehrung-die-theoretische-chemie-final-answer-.png

    I am not seeing where my mistake is, but maybe I have been looking at this problem too long now. Typical of me in such a situation, I am, of course, beginning to suspect the handout is wrong If anyone has time or interest to check the work, could you tell me if you see where I might be messing up? Thank you!

    EDIT: Ok, well, someone has advised me that the sin function is an odd function, so I can pull out a negative from it and reverse the sign of the denominator as well. This leads to the correct answer!

    Thanks chiro for setting me on the right path! Problem solved.
    Last edited by blaisem; October 28th 2012 at 08:29 AM.
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