Definite integral of gaussian

1. Mar 22, 2013

jashua

Is there a closed form expression for the following definite integral?

$\int_{-∞}^{∞} exp(\frac{-|z|^2}{2{\sigma}^2}-\alpha |\mu + z|)dz$

where $z$ is complex, and $\alpha, \sigma, \mu$ are real constants.

I couldn't obtain an expression similar to Gaussian integral, so I couldn't take the integral. So, how can we obtain a closed form expression for such an integral?

2. Mar 22, 2013

HallsofIvy

Staff Emeritus
If z is complex, what is the meaning of the limits of integration- what path are you taking from $-\infty$ to $\infty$?

3. Mar 22, 2013

jashua

Thank you for your reply, and I'm sorry for the incorrect statement of my question. Let me try to explain as follows:

Let z = x + jy. Then, the limit of x is from -∞ to ∞, and the limit of y is also from -∞ to ∞. In this case, I'm not sure about the limit of z. However, we can now restate the integral as follows:

$\int_{-∞}^{∞}\int_{-∞}^{∞} exp(\frac{-(x^2+y^2)}{2{\sigma}^2}-\alpha ((x+\mu)^2 + y^2)^{1/2})dxdy$

Last edited: Mar 22, 2013
4. Mar 22, 2013

Ray Vickson

There is something very wrong here. Integrals with respect to a complex z are defined, but they are 1-dimensional, like "line integrals" in the 2-dimensional plane. Your first message suggested that is what you want, but your later message implies that is NOT the case, but that, instead, you want a double integral over the plane. Which is it? What do you *really* want?

5. Mar 22, 2013

jashua

Actually, the second integral is what I want to ask.

However, if I'm not wrong, the second integral can be expressed as a contour integral in the complex plane, which I have tried to write in my first post. If I'm wrong please correct ( or what should be the path of z? )

Last edited: Mar 22, 2013
6. Mar 23, 2013

Ray Vickson

I don't think you can do the integral exactly, but you can reduce it to a one-dimensional integration involving a Bessel function, and that can be done numerically. In your function, change variables to
$$x = - \mu + r \cos(\theta), \: y = r \sin(\theta).$$
Your integrand f(x,y) becomes
$$f = \exp \left(-\frac{r^2 + \mu^2 + 2 \alpha \sigma^2 r}{2 \sigma^2} \right) e^{b \cos(\theta)}, \;\; b = \mu r \sigma^2.$$
The integral you want is
$$\int_0^{\infty} \int_0^{2 \pi} r f \, dr \, d\theta.$$
We can do the $\theta$ integration first, to get
$$\text{Answer} = \int_0^{\infty} r \exp \left(-\frac{r^2 + \mu^2 + 2 \alpha \sigma^2 r}{2 \sigma^2} \right) \text{BessellI}(0,\mu r/\sigma^2),$$
where $\text{BesselI}(0,v)$ is a modified Bessel function of the first kind. This Bessel function is a solution of the Bessel differential equation
$$x^2 y'' + xy' +x^2 y = 0,$$ and with intitial terms of its series given by
$$y = 1 + \frac{1}{4} x^2 + \frac{1}{64} x^4 + \cdots .$$ (Knowing a few terms of this series helps to relate this definition of BesselI(0,v) to possible other definitions you may know or encounter.)

Last edited: Mar 23, 2013
7. Mar 25, 2013

jashua

Ray, thank you very much for your detailed answer.