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Math Help - Integral of this exponential function: analytical solution?

  1. #1
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    Integral of this exponential function: analytical solution?

    Hi all,

    I'm trying to solve the definite integral between 0 and inf of:

    exp(a*x^2 + b*x + c)
    --------------------- dx
    1 + exp(m*x + n)

    with a,b,c,m,n real numbers and a < 0, m < 0 (negative number so it converges).

    I have tried to transform the denominator to cosh and integrate by parts,
    among many others alternatives but I didn't suceed.

    A way to obtain the exact solution would be perfect but an approximate result, even an upper/lowerbound would be fine as well.

    Any idea or help, please?

    Thanks in advance,
    FC.
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  2. #2
    MHF Contributor

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    Quote Originally Posted by fryderykchopin View Post
    Hi all,

    I'm trying to solve the definite integral between 0 and inf of:

    exp(a*x^2 + b*x + c)
    --------------------- dx
    1 + exp(m*x + n)

    with a,b,c,m,n real numbers and a < 0, m < 0 (negative number so it converges).

    I have tried to transform the denominator to cosh and integrate by parts,
    among many others alternatives but I didn't suceed.

    A way to obtain the exact solution would be perfect but an approximate result, even an upper/lowerbound would be fine as well.

    Any idea or help, please?

    Thanks in advance,
    FC.
    if we also assume that n \leq 0, then \frac{1}{1+e^{mx+n}} \approx 1 - e^{mx + n} + e^{2mx+2n} and thus: \text{your integral} \approx \int_0^{\infty}e^{ax^2+bx+c} \ dx \ - \ \int_0^{\infty} e^{ax^2+(b+m)x + c+n} \ dx \ + \ \int_0^{\infty} e^{ax^2+(b+2m)x + c+2n} \ dx.

    each of the three integrals in the right hand side can be found in terms of the error function. finally, there are several good approximations of the error function. just google it!
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  3. #3
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    Thanks a lot for your answer. It's very interesting, but unfortunately n>0 in my problem (a<0, b>0, c<0, m<0 but n>0). I don't know if your approach still holds for n>0, since I don't know where the approximation of the denominator comes from (may be Taylor series?). Anyway, thank you very much for your answer.
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