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Integral of an exponential that has a polynomial

  1. Apr 29, 2012 #1
    How would one evaluate $$\Phi = \int_{-\infty}^{+\infty} e^{-(ax+bx^2)} dx$$.

    I was trying to change it into a product of an error function and a gamma function, but I needed an extra dx. Any other ideas?
  2. jcsd
  3. Apr 29, 2012 #2
    Let X = ? such as : ax+bx² = A X² +B
  4. Apr 29, 2012 #3
    I don't really see how that would work.

    I tried setting u=ax, v=bx^2. Then i would use
    $$ \Gamma(1) = \int_{0}^{\infty} e^{-u}du $$
    $${\rm erf}(x) = \int_{-\infty}^{x} e^{-v^2}dv $$
    I'm not exactly sure how to use that in
    $$ \int_{-\infty}^{+\infty} e^{-u} e^{-v^2} du$$
    without another differential dv somewhere.
  5. Apr 29, 2012 #4
    You are not on the right way !
    ax+bx² = b(x+(a/2b))² -a²/4b
    X= x+(a/2b)
    exp(-(ax+bx²)) = exp(a²/4b)*exp(-b X²)
    Then integrate exp(-b X²)*dX
  6. Apr 29, 2012 #5

    [tex]\displaystyle{ \int_{-\infty}^{+\infty} e^{-(ax+bx^2)} dx= \int_{-\infty}^{+\infty} e^{-b\left(x+\frac{a}{2b}\right)^2}e^{\frac{a^2}{4b}} dx=\frac{e^{\frac{a^2}{4b}}}{\sqrt{b}} \int_{-\infty}^{+\infty} e^{-x^2} dx=e^{\frac{a^2}{4b}}\sqrt{\frac{\pi}{b}}\,,\,\,with}[/tex]

    (1) first equality: completing the square

    (2) second equality: substituting [itex]\displaystyle{\sqrt{b}\left(x+\frac{a}{2b}\right) \to y}[/itex]

    (3) third equality: using [itex]\displaystyle{\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}}[/itex]


    Ps. Of course, I assume [itex]b>0[/itex]
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