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    Integration by parts is well know to be

    INT f'(x) g(x) dx = f(x) g(x) - INT f(x) g'(x) dx

    I'm sorry if I don't know how to do the syntax/Latex stuff yet.

    If we have g(x)=x^n and f(x)=e^(x^2), we get

    INT 2x^(n+1) e^(x^2) dx = x^n e^(x^2) - INT nx^(n-1) e^(x^2) dx

    This recursive formula along with the easily derived INT 2x e^(x^2) dx = e^(x^2) can be easily used to find the solution for any even 'n' in INT 2x^(n+1) dx.

    For odd 'n', we get a power series, but in particular, I want to observe the power series for n = -1:

    INT 2e^(x^2) dx = x^(-1) e^(x^2) + INT x^(-2) e^(x^2) dx

    = x^(-1) e^(x^2) + 1/2 x^(-3)e^(x^2) + INT 3x^(-4) e^(x^2) dx

    =....

    What is the closed form solution? I know it will be of the form P(x)*e^(x^2), where P(x) is a power series, but what is the closed form?

    Also, I wish to know if this 'solution' can be used to take definite integrals from a to b on the function e^(x^2).
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    I'm 'naught but an A-Level student, but:

    \displaystyle \int \frac{e^{x^2}}{x} \, \mathrm{d}x = \frac{\text{Ei}(x^2)}{2}

    And a quick google should turn up the power series (I think) for the fairly well known special function \text{Ei}(x^2)
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    (Original post by HollyHSBLL)
    Integration by parts is well know to be

    INT f'(x) g(x) dx = f(x) g(x) - INT f(x) g'(x) dx

    I'm sorry if I don't know how to do the syntax/Latex stuff yet.

    If we have g(x)=x^n and f(x)=e^(x^2), we get

    INT 2x^(n+1) e^(x^2) dx = x^n e^(x^2) - INT nx^(n-1) e^(x^2) dx

    This recursive formula along with the easily derived INT 2x e^(x^2) dx = e^(x^2) can be easily used to find the solution for any even 'n' in INT 2x^(n+1) dx.

    For odd 'n', we get a power series, but in particular, I want to observe the power series for n = -1:

    INT 2e^(x^2) dx = x^(-1) e^(x^2) + INT x^(-2) e^(x^2) dx

    = x^(-1) e^(x^2) + 1/2 x^(-3)e^(x^2) + INT 3x^(-4) e^(x^2) dx

    =....

    What is the closed form solution? I know it will be of the form P(x)*e^(x^2), where P(x) is a power series, but what is the closed form?

    Also, I wish to know if this 'solution' can be used to take definite integrals from a to b on the function e^(x^2).
    Your last remark is the the important one. In order to make something useful of what you've got so far, you should turn these into definite integrals - this then allows you to evaluate the terms that you generate from the partial integrations.

    In particular, you are dealing with the "error function" and the "complementary error function" from special function theory. These are, respectively:

     \displaystyle \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} dt

     \displaystyle \frac{2}{\sqrt{\pi}} \int_{x}^{\infty} e^{-t^2} dt

    The partial integration that you have started off is the standard way of arriving at the asymptotic expansion of the complementary error function.
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    (Original post by Zacken)
    I'm 'naught but an A-Level student, but:

    \displaystyle \int \frac{e^{x^2}}{x} \, \mathrm{d}x = \frac{\text{Ei}(x^2)}{2}

    And a quick google should turn up the power series (I think) for the fairly well known special function \text{Ei}(x^2)
    Good way to start, but it's the asymptotic series that OP is chasing after here - not the ordinary power series expansion.
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    (Original post by Gregorius)
    Good way to start, but it's the asymptotic series that OP is chasing after here - not the ordinary power series expansion.
    In above my head once again. Your post looks very interesting, I've heard very roughly about asymptotic series but don't really know any of it myself.
 
 
 
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