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Computing a Gaussian integral

  1. Sep 19, 2014 #1
    1. The problem statement, all variables and given/known data
    Let [itex]a,b[/itex] be real with [itex]a > 0[/itex]. Compute the integral
    [tex]
    I = \int_{-\infty}^{\infty} e^{-ax^2 + ibx}\,dx.
    [/tex]


    2. Relevant equations
    Equation (1):
    [tex]\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}[/tex]

    Equation (2):
    [tex]-ax^2 + ibx = -a\Big(x - \frac{ib}{2a}\Big)^2 - \frac{b^2}{4a}[/tex]


    3. The attempt at a solution
    Completing the square in [itex]-ax^2 + ibx[/itex] gives me Equation (2), so that my integral is now
    [tex]
    I = e^{-b^2/4a}\int_{-\infty}^{\infty} e^{-a(x-ib/2a)^2}\,dx.
    [/tex]
    Making the substitution [itex]u = \sqrt{a}(x-ib/2a)[/itex] I get [itex]du = \sqrt{a}\,dx[/itex] so tht my integral becomes
    [tex]
    I = \frac{1}{\sqrt{a}}\,e^{-b^2/4a}\int_{-\infty - ib/2\sqrt{a}}^{\infty - ib/2\sqrt{a}} e^{-u^2}\,du.
    [/tex]

    But this doesn't seem right.
     
  2. jcsd
  3. Sep 19, 2014 #2

    ZetaOfThree

    User Avatar
    Gold Member

    You've got it right. Just note that ##-\infty - ib/2\sqrt{a}=-\infty## and ##\infty - ib/2\sqrt{a}=\infty##, and you're basically done.
     
  4. Sep 20, 2014 #3
    Thanks Zeta. Huge brain fart on my part in making it rigorous. The integral I was trying to compute is the limit of

    [tex]I(R_1, R_2) = \int_{-R_1}^{R_2} e^{-ax^2 + bx}\,dx[/tex]

    as [itex]R_1, R_2 \to \infty[/itex]. Then I can make a subsitution [itex]z = \sqrt{a}(x-ib/2a)[/itex] to get the integral

    [tex]I(R_1,R_2) = \frac{1}{\sqrt{a}}\,e^{-b^2/4a}\int_{-\sqrt{a}R_1-ib/2\sqrt{a}}^{\sqrt{a}R_2 - ib/2\sqrt{a}}e^{-z^2}\,dz.[/tex]

    The integrand [itex]e^{-z^2}[/itex] is analytic on the entire complex plane, so the integral is path independent. So in particular I can take it on a contour consisting of:

    (1) A straight line up from [itex]z = -\sqrt{a}R_1 - ib/\sqrt{a}[/itex] up to the real axis at point [itex]z = -\sqrt{a}R_1[/itex].

    (2) A straight line on the real axis from [itex]z = -\sqrt{a}R_1[/itex] to [itex]z = \sqrt{a}R_2[/itex].

    (3) A straight line down from the real axis at point [itex]z = \sqrt{a}R_2[/itex] to [itex]z = \sqrt{a}R_2 - ib/2\sqrt{a}[/itex].

    Taking the limit as [itex]R_1, R_2 \to \infty[/itex] the integrals on contour sections (1) and (3) vanish since [itex]\lvert z\rvert \to \infty[/itex] and thus [itex]e^{-z^2} \to 0[/itex] on these two vertical lines. The integral on contour (2) then becomes

    [tex]\lim_{R_1, R_2 \to \infty}\int_2 e^{-z^2}\,dz = \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}.[/tex]

    Thus we have

    \begin{align*}
    \int_{-\infty}^{\infty} e^{-ax^2 + bx}\,dx
    & = \lim_{R_1,R_2 \to \infty} I(R_1,R_2) \\
    & =
    \lim_{R_1,R_2 \to \infty} \frac{1}{\sqrt{a}}\,e^{-b^2/4a}\int_{-\sqrt{a}R_1-ib/2\sqrt{a}}^{\sqrt{a}R_2 - ib/2\sqrt{a}}e^{-z^2}\,dz \\
    & = \frac{1}{\sqrt{a}}\,e^{-b^2/4a}\sqrt{\pi}.
    \end{align*}
     
  5. Sep 20, 2014 #4
    Oops, contour (1) should be from [itex]z = -\sqrt{a}R_1 - ib/2\sqrt{a}[/itex] to [itex]z = -\sqrt{a}R_1[/itex].
     
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