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Thread: How can I solve the integral with singularity at endpoint?

  1. #1 How can I solve the integral with singularity at endpoint? 
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    I have a question about a integral that have singularity at the endpoint. The singularity will bring oscillation to the solution. I am wondering if there is a way to approximate this integrand with a recurrence relation so that we can smoothe out the oscillation? Here is the integral.
    integral exp(-k*L) /((Ro*k)^2 + Ro^2*(U*k-1)^2-1) *(exp(-Ro*k*z/(1-Ro^2*(U*k-1)^2)^0.5 ) *exp(i(kx-t)) dk from (Ro-1)/Ro/U to (Ro+1)/Ro/U. Here Ro(=0.707), U(=0.625), t(=0.25) are const; we can set x, z as arrays from -10 to 10 and from 0 to 10. As we notice that the singularity exists at endpoint (Ro+1)/Ro/U. My question is if it is possible to modify the integral to a suitable form which can smoothe out the singularity? For example, if we can simplify it to the form as integral exp(t)/t dt from const to infinity, it will be solvable.
    Thanks.


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  3. #2  
    Forum Bachelors Degree Demen Tolden's Avatar
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    I'm not sure if this is something I could help you with or not, but the notation you used isn't very clear to me.

    If you want to make an integral sign, you can do it like this:
    &#8747 ;
    without the space in between the 7 and ;

    If you want to make a superscript, as you would if you would raise something to a power or show the upper limit to an integrand, you can type
    < sup > 5 < / sup >
    without any spaces

    If you want to make a subscript, as you would perhaps to show the lower limit of an integrand
    < sub > 5 < / sub >
    without any spaces.

    To me, your "exp" I think could either refer to e or a power so its confusing to me.


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  4. #3  
    Forum Bachelors Degree Demen Tolden's Avatar
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    Am I right in thinking that this is your integrand?

    First, for the sake of clarity, I will temporarily make Ro = r

    Your lower limit looks to be
    (r-1)/r/u
    u(r-1)/r
    u - ( 1 / r )

    Your upper limit looks to be using the same steps as above:
    u + ( 1 / r )

    <sub>u-(1/r)</sub>∫<sup>u+(1/r)</sup> (-k L)e(r k)<sup>-2</sup> + ( r<sup>2</sup>(u k - 1)<sup>2</sup> - 1)e(-r k z)( (1 -r<sup>2</sup>)( (u k - 1)<sup>2</sup>)<sup>1/2</sup>)<sup>-1</sup>e( i ( kx - t ) ) dk

    If this is the case (as it most likely isn't), this integrand could be greatly simplified.
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