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Multidim. Gaussian integral with linear term

  1. Apr 29, 2010 #1
    Hey everyone,

    I know, lots of threads and online information about Gaussian integrals. But still, I couldn't find what I am looking for: Is there a general formula for the integral

    [tex] \int_{\mathbb{R}^d} d^d y \left|\vec{y}\right| \exp(-\alpha \vec{y}^2)[/tex]

    where y is a vector of arbitrary dimension d and alpha is a complex number? It could be done in hyperspherical coordinates but its cumbersome. So I wanted to look up a general expression but couldn't find one. Can anyone help?

    Thanks a lot
     
    Last edited: Apr 30, 2010
  2. jcsd
  3. May 5, 2010 #2
    Hi Orbb,

    I think the formula you are looking for is
    [tex]\frac{\pi^{\frac{d}{2}}}{\alpha}\frac{1}{\Gamma(\frac{d}{2})}[/tex]

    It is straightforward to get this result in spherical co-ordinates and I don’t think you can find any easier way to do it.
    Assume
    [tex] |\vec{y}| = r = \sqrt{x_{1}^2+x_{2}^2+\cdot\cdot\cdot x_{d}^2},[/tex]

    and
    [tex] y^2 = x_{1}^2+x_{2}^2+\cdot\cdot\cdot x_{d}^2.[/tex]

    Converting to spherical co-ords gives

    [tex]\int_{0}^{\infty}\int_{\phi_1=0}^{\pi}\int_{\phi_2=0}^{\pi}\cdot\cdot\cdot\int_{\phi_{d-2}=0}^{\pi}\int_{\phi_{d-1}}^{2\pi}\sin^{d-2}(\phi_1) \sin^{d-3}(\phi_2 )\cdot\cdot\cdot\sin(\phi_{d-2})r\exp(-\alpha r^2)d\phi_{1}d\phi_2\cdot\cdot\cdotd\phi_{d-1}dr.[/tex]

    Now
    [tex] \int_{0}^{\infty} r\exp(-\alpha r^2)dr = \frac{1}{2\alpha},
    [/tex]
    and
    [tex]
    \int_{0}^{\pi}\sin^{n}(\theta)d\theta=\frac{\sqrt{\pi}\Gamma(\frac{1+n}{2})}{\Gamma(1+\frac{n}{2})}.
    [/tex]

    Combining the above results (don’t forget [tex]\int_{\phi_{d-1}=0}^{2\pi}d\phi_{d-1}=2\pi[/tex]) and simplifying (must of the [tex]\Gamma[/tex] terms will cancel) results in
    [tex]\int_{\mathbb{R}^d} d^d y \left|\vec{y}\right| \exp(-\alpha \vec{y}^2)
    =\frac{\pi^{\frac{d}{2}}}{\alpha}\frac{1}{\Gamma(\frac{d}{2})}
    [/tex]
     
  4. May 6, 2010 #3
    Thank you for taking the time, appelberry. I tried the same by now. Only when converting to spherical coordinates, shouldn't the integral over r be:

    [tex] \int_0^\infty r^{d-1} \left[r \exp(-\alpha r^2)\right] dr [/tex] ?

    this should give

    [tex] \sqrt{\pi} \frac{(d-1)!!}{2^{d/2+1}} \alpha^{-\frac{d+1}{2}} [/tex]

    for odd d and

    [tex] \frac{[1/2(d-1)]!}{2} \alpha^{-\frac{d+1}{2}} [/tex]

    for even d. Is that correct? It is quite important wether the power of alpha depends on d in the result.
     
  5. May 6, 2010 #4
    Woops! yea, of course you are right. I forgot the [tex]r^{d-1}[/tex] term! I think your answer looks correct now.
     
  6. May 6, 2010 #5
    okay, thank you!
     
  7. Jun 1, 2010 #6
    Alright, again I'm troubled by a nasty integral. I guess it does not deserve it's own thread since it's a similar problem and the above has been solved. So I have an integral of the form:

    [tex] \int_{\mathbb{R}^d}d^d x \exp\left(i\vec{a}\cdot\vec{x}-i\sqrt{b\vec{x}^2+c}-d\vec{x}^2\right) [/tex]

    Ok I think there's no hope of solving it analytically without approximations. If there is a way, please let me know of course :wink: I think an expansion of the square root around x=0 isn't an option either because the taylor series doesn't converge for large x, right? So what I'm left with is the following: the square root actually comes from a relativistic expression [tex] \sqrt{p^2+m^2}[/tex], so the 'ultrarelativistic' or 'massless' limit p >> m (i.e. bx >> c) would give the integral:

    [tex] \int_{\mathbb{R}^d}d^d x \exp\left(i\vec{a}\cdot\vec{x}-ib|\vec{x}|-d\vec{x}^2\right) [/tex]

    This looks better. Written in polar coordinates, the scalar product gives an angular cos-dependence. Can I simply assume that my coordinates are oriented s.t. this dependence corresponds only to one of the d-1 angles I have to integrate over?

    Anyways, any suggestions or solutions greatly appreciated!
     
  8. Jun 8, 2010 #7
    Yes, I think you can assume that the angle is dependent on only one of the angles in a spherical polar co-ordinate system by assuming that the vector [tex]\vec{a}[/tex] lies along one of the axes in the system similar to the approach in the following integral:

    https://www.physicsforums.com/showthread.php?t=376233

    You can then integrate out the angular portion using the identity:

    [tex]
    \int^{2\pi}_{0}\exp(x\cos\theta)d\theta = 2\pi I_{0}(x)
    [/tex]

    which will leave you with a single integral over the magnitude of the vector [tex]\vec{x}[/tex].
     
  9. Jun 15, 2010 #8
    Thanks for your answer! This is what I did first. However, I was struggling because it seems to me that the dependence on all the other angles drops only if I assume [tex]\vec{a}[/tex] to lie along the polar axis. In this case, the angular integration would be

    [tex]\int_0^{\pi} sin^{d-2}\theta\exp(i|\vec{a}|rcos\theta)d\theta.[/tex]

    It would be beautiful if I could do the integration the way you suggested, but from making a sketch for the d=3 sphere it seems to me an integration over the azimuthal angle doesn't work. But maybe I'm mistaken.

    Also, I could live with a restriction to d=3 to make things simpler. For this case again I tried to manipulate the above integral to arrive at an expression in terms of bessel functions, but failed so far :wink:

    Edit: For the case d=3, [tex]\int_0^{\pi} sin\theta\exp(i|\vec{a}|rcos\theta)d\theta =\frac{2sin(r|\vec{a}|)}{r|\vec{a}|}.[/tex]
     
    Last edited: Jun 15, 2010
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