Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Exponential integral over removeable singularity gives wrong result

  1. Aug 14, 2014 #1
    Hi,

    I am struggling for some time to solve the following integral:

    $$
    \int_{-n}^{N-n} \left( \frac{e^{-j\pi(\alpha-1)\tau}}{\tau} - \frac{e^{-j\pi(\alpha+1)\tau}}{\tau} \right) d\tau
    $$

    [itex]N[/itex] is a positive integer, [itex]n[/itex] is an integer, [itex]\alpha[/itex] can be a negative or positive rational number.

    I want to express it analytically using exponential integrals [itex]E_1(z)[/itex]:

    $$
    \mathrm{E}_1(z) = \int_z^\infty \frac{e^{-t}}{t}\, dt,\qquad|{\rm Arg}(z)|<\pi
    $$

    In my opinion, the result should just be

    $$
    E_1(i\pi(\alpha-1)n) - E_1(-i\pi(\alpha-1)(N-n)) - E_1(-i\pi(\alpha+1)n) + E_1(-i\pi(\alpha+1)(N-n))
    $$

    I compare my analytical results against a numerical evaluation in MATLAB.
    I calculate my analytical results by using [itex]E_1(z)[/itex] (expint in MATLAB) and the numerical version by creating a vector and using trapezoidal rule for integration.

    Both match for approximately half of my values but the rest is different by absolute values of 1 or 0.5. Sometimes I even get NaN (undefined) values because [itex]\alpha[/itex] will be -1 and 1 at some point and [itex]n[/itex] will be zero at some point. Still, practically this makes no sense and should not happen (it does not happen with the numerical integration too and the result of the numerical integrations is exactly what I expect).

    In order to match the results with my numerical integration, I split the integration and leave out the problematic point around zero ([itex]\epsilon\rightarrow 0[/itex]):

    $$
    E_1(i\pi(\alpha-1)n) - E_1(-i\pi(\alpha-1)\epsilon) + E_1(i\pi(\alpha-1)\epsilon) - E_1(-i\pi(\alpha-1)(N-n))
    -E_1(-i\pi(\alpha+1)n)+E_1(-i\pi(\alpha+1)\epsilon)-E_1(i\pi(\alpha+1)\epsilon)+E_1(-i\pi(\alpha+1)(N-n))
    $$

    I would expect [itex]-E_1(-i\pi(\alpha-1)\epsilon) + E_1(i\pi(\alpha-1)\epsilon)[/itex] and [itex]E_1(-i\pi(\alpha+1)\epsilon)-E_1(i\pi(\alpha+1)\epsilon)[/itex] to be zero. If they would be, numerical and analytical results would match.


    However, depending on [itex]n[/itex], they are [itex]\pm i\pi[/itex] (which ultimately results in the wrong result).

    Despite the fact the it is a removeable singularity, do I need to care about something special here?
     
  2. jcsd
  3. Aug 27, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
     
  4. Sep 4, 2014 #3
    Didn’t get into details, but… you apparently forgot to say that “ j ” is the imaginary unit. If my guess is correct, then the expression under the integral certainly contains sinc function as a factor.
     
  5. Mar 4, 2015 #4
    Hi Incnis,
    Yes, j is the imaginary unit. And yes, it contains some sort of the sinc function. To be precise, I want to integrate parts of the sinc function (this is where my formula comes from in the first place).

    I still have no idea why and where I get my mismatch from :( :(

    divB
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Exponential integral over removeable singularity gives wrong result
  1. Removable Singularity (Replies: 4)

Loading...