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Help for integral

  1. Feb 21, 2008 #1
    Integral of exp{-(a*x^2+b*x+c)} from 0 to infinite



    i know the answer of Integral of exp{-(a*x^2+b*x+c)} from -inf to inf
    = sqrt(pi/a)*exp((b^2-4ac)/4a)



    thanks!!
     
  2. jcsd
  3. Feb 21, 2008 #2
  4. Feb 22, 2008 #3

    ssd

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    Write the exponent in the form -g(kx+d)^2 + f. Put g(kx+d)=u. Then the integral
    (-inf to inf) becomes twice the same integral over (0 to inf), since the integrand is
    now of the form e^(-u^2) which is symmetric about 0.
     
  5. Feb 23, 2008 #4
    @ssd, if you mean that the end result of the integral in the original post is half the result of the other one given in there, you are making a calculation error.

    OK yf920, let's calculate the result given. The argument of the exponential function can be written as:

    [tex]-ax^2-bx-c=-a\left(x+\frac{b}{2a}\right)^2 +\frac{b^2-4ac}{4a}[/tex]

    Putting this into the integral gives then:

    [tex]I=e^{\frac{b^2-4ac}{4a}}\int_{-\infty}^{\infty} e^{-a\left(x+\frac{b}{2a}\right)^2}dx[/tex]

    Putting as ssd said:

    [tex]\sqrt{a}\left(x+\frac{b}{2a}\right)=t[/tex]

    gives now:

    [tex]I=\frac{e^{\frac{b^2-4ac}{4a}}}{\sqrt{a}}\int_{-\infty}^{\infty} e^{-t^2}dt[/tex]

    The following integral is known:

    [tex]\int_{-\infty}^{\infty} e^{-t^2}dt=\sqrt{\pi}[/tex]

    Giving the final result as:

    [tex]I=\sqrt{\frac{\pi}{a}} e^{\frac{b^2-4ac}{4a}}[/tex]

    Now, the original question was:

    [tex]I=\int_{0}^{\infty} e^{-\left(ax^2+bx+c\right)}dx[/tex]

    Which is not the double of the one just derived. Just go through all the steps with this new integral limits and use the result:

    [tex]erf(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{x}e^{-z^2}dz[/tex]

    Do these calculations, and come back with the result you found, it will be our pleasure to check it.
     
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