How can the difficult Gaussian integral be solved using standard tricks?

[Ben D.]

Hi everyone,

in the course of trying to solve a rather complicated statistics problem, I stumbled upon a few difficult integrals. The most difficult looks like:

$$I(k,a,b,c) = \int_{-\infty}^{\infty} dx\, \frac{e^{i k x} e^{-\frac{x^2}{2}} x}{(a + 2 i x)(b+2 i x)(c+2 i x)}$$

where $$a,b,c$$ are real positive numbers and $$k$$ is a real number. This integral cannot be done by simple contour integration because of the Gaussian factor. From the context, I expect error functions to appear in the solution.

Any clever ideas?

B.D.

 

[fzero]

Hi everyone,

in the course of trying to solve a rather complicated statistics problem, I stumbled upon a few difficult integrals. The most difficult looks like:

$$I(k,a,b,c) = \int_{-\infty}^{\infty} dx\, \frac{e^{i k x} e^{-\frac{x^2}{2}} x}{(a + 2 i x)(b+2 i x)(c+2 i x)}$$

I think you can do this with a few standard tricks. First, assume that we can do partial fraction decomposition, so that we can consider three integrals of the form
$$I_{k,a} = \int_{-\infty}^{\infty} dx\, \frac{e^{i k x} e^{-\frac{x^2}{2}} x}{ x - ia/2 }.$$
Next, use the Schwinger parametrization
$$ \frac{1}{x - ia/2 } = -i \int_0^\infty du\, e^{i u(x - i a/2)},$$
so
$$I_{k,a} = -i e^{a/2} \int_0^\infty du\, \int_{-\infty}^{\infty} dx\, e^{i (u+k) x} e^{-\frac{x^2}{2}} x.$$
Next, the factor of $x$ can be handled by replacing it with a derivative $-i d/dk$, so
$$I_{k,a} = -e^{a/2} \frac{d}{dk} \int_0^\infty du\, \int_{-\infty}^{\infty} dx\, e^{i (u+k) x} e^{-\frac{x^2}{2}} .$$
The Gaussian integral is done by completing the square:
$$I_{k,a} = - e^{a/2} \frac{d}{dk} \int_0^\infty du\, \sqrt{2\pi} e^{-(u+k)^2/2 } .$$
Finally, we redefine variable to $u'=u+k$ and use the fundamental theorem of calculus to find
$$I_{k,a} = \sqrt{2\pi} e^{(-k^2 + a)/2} .$$
This has the right behavior in the limit as $k,a\rightarrow 0$.

Note that you can skip the partial fractions and just do a more complicated Schwinger parameterization in your original integral, but I'm not sure that would save too much effort.