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Integral of x^2*exp(x^2) using erfc tables

  1. Mar 13, 2009 #1
    Can anybody show me how to integrate x^2*exp(-x^2) between the limits 0 to infinity(symbol=00) and 1.5 to infinity with detail steps. I want this by using error function table. I know the 'multiplying integral method'. Here is what I did so far
    int(0,00) x^2*exp(-x^2)
    =x^2*int(0,00)exp(-x^2)-int(0,00)[2*x*int(0,00)exp(-x^2)]
    =x^2(pi^0.5/2)-int(0,00)[2*x*(pi^0.5/2)] ...first term also between (0,00)
    This seems to be giving 00! Where am I missing?
    Thanks.

    Also please show me how to do this between limits (1.5,00) with error function tables.
     
    Last edited: Mar 13, 2009
  2. jcsd
  3. Mar 13, 2009 #2

    alxm

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    Did you get that right? [tex]x^2e^{x^2}[/tex] is divergent, those integrals will be [tex]\infty[/tex].
     
  4. Mar 13, 2009 #3
    Re: integral of x^2*exp(-x^2) using erfc tables

    Oh, I am terribly sorry. I meant x^2*exp(-x^2). Forgot the negative sign. I corrected the original post (could not correct the Sub line). Thanks.
     
  5. Mar 13, 2009 #4

    alxm

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    Hmm, okay. Well what I'd do is first do a substitution, e.g.
    [tex]y = x^2[/tex]
    Resulting in:
    [tex]\frac{1}{2}\int\sqrt ye^{-y}dy[/tex]
    Then use partial integration which will land you with a infinite series for the indefinite integral.

    For the case of:
    [tex]\frac{1}{2}\int^\infty_0\sqrt ye^{-y}dy [/tex] you can 'cheat' and note that it's [tex]= \frac{1}{2}\Gamma(1.5) [/tex]
     
  6. Mar 13, 2009 #5
    Re: integral of x^2*exp(-x^2) using erfc tables

    alxm, I don't know whether your approach is correct. But its pretty straightforward with 'multiplying integration'. You can find a method somewhere on the web.

    My problem is trying to use error function approach.
     
  7. Mar 18, 2009 #6
    Hello ronphysics,

    You need to use the following four equations in order to solve it the way you want. These are:

    [tex]\int_0^{\infty}e^{-ax^2}dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}[/tex]
    [tex]erf(x)=\frac{2}{\sqrt{\pi}}\int_0^{x} e^{-t^2}dt[/tex]
    [tex]erfc(x)=\frac{2}{\sqrt{\pi}}\int_{x}^{\infty} e^{-t^2}dt[/tex]
    [tex]erf(x)=1-erfc(x)[/tex]

    Now I will show how to obtain the general question as:

    [tex]I=\int_{\beta}^{\infty}x^2e^{-x^2}dx[/tex]

    Afterwards you need to find for yourself the solution to your integrals, and additionally check them by using the definite numbers for [itex]\beta[/itex]. If any problems occur, just post. OK, so the first step is to use partial integration. This gives:

    [tex]I=\int_{\beta}^{\infty}x^2e^{-x^2}dx= -\int_{\beta}^{\infty}\frac{x}{2}d\left(e^{-x^2}\right)= -\left[\frac{x}{2}e^{-x^2}\right]_{\beta}^{\infty}+\frac{1}{2}\int_{\beta}^{\infty}e^{-x^2}dx= \frac{\beta}{2}e^{-{\beta}^2}+\frac{1}{2}erfc(\beta)\frac{\sqrt{\pi}}{2}[/tex]

    Therefore, after rewriting:

    [tex]I=\frac{\beta}{2}e^{-{\beta}^2}+\frac{\sqrt{\pi}}{4}erfc({\beta})= \frac{\beta}{2}e^{-{\beta}^2}+\frac{\sqrt{\pi}}{4}\left(1-erf({\beta})\right)[/tex]

    Is this the method you were looking for?

    coomast
     
  8. Mar 18, 2009 #7
    coomast,

    Thanks. Thats exactly what I was looking for. My mind was stuck in separating variables as x^2 and exp(-x^2). However, the useful way to seprate is x and x*exp(-x^2). Thanks again.
     
  9. Dec 1, 2011 #8
    Hi all,

    I am trying for the following integration

    ∫x2 exp(a0+a1x+a2x2) dx

    Please help me to find a solution.

    Sivaji
     
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