Kinetic integrals

It is useful to list some common integrals that occur frequently in statistical physics. Consider:

$$I_{2n}(a) = \int_0^\infty x^{2n} e^{-a x^2} dx = (-1)^n \frac{d^n I_0(a)}{d a^n}$$ (16.1)

and

$$I_{2n+1}(a) = \int_0^\infty x^{2n+1} e^{-a x^2} dx = (-1)^n \frac{d^n I_1(a)}{d a^n},$$ (16.2)

with $n$ any positive integer. To compute $I_0(a)$ we first make a change of variable $y = \sqrt{a} x$, and then we define:

$$I = \frac{1}{\sqrt{a}} \int_{-\infty}^\infty e^{-y^2} dy = 2 I_0(a).$$ (16.3)

Now consider:

$$I^2 = \frac{1}{a} \int_{-\infty}^\infty e^{- z^2} dz \int_{-\inft... ...frac{1}{a} \int_{-\infty}^\infty \int_{-\infty}^\infty e^{- (y^2 + z^2)} dy dz.$$ (16.4)

This integral can be easily calculated in polar coordinates:

$$I^2 = \frac{1}{a} \int_{0}^{2\pi} d\phi \int_{0}^\infty r e^{- r^... ...rac{1}{a} 2\pi \frac{1}{2} \left [ e^{- r^2} \right ]_0^\infty = \frac{\pi}{a},$$ (16.5)

from which we obtain:

$$I_0(a) = \frac{1}{2}\left(\frac{\pi}{a}\right )^{1/2}.$$ (16.6)

The integral $I_1(a)$ is straightforwardly computed:

$$I_1(a) = \int_0^\infty x e^{-a x^2} dx = \frac{1}{a} \int_0^\inft... ... = -\frac{1}{a} \frac{1}{2} \left [ e^{- y^2} \right ]_0^\infty = \frac{1}{2a}.$$ (16.7)

Using Eqs. , we can summarise the first few integrals:

The integrals  and  can also be written as:

$$I_{n}(a) = \int_0^\infty x^n e^{-a x^2} dx = \frac{1}{2} \frac{\Gamma\left(\frac{n+1}{2}\right )}{a^{\frac{n+1}{2}}},$$ (16.9)

with the $\Gamma$ function defined by:

$$\Gamma(z) = \int_0^\infty x^{z-1} e^{-x} dx.$$ (16.10)

Integrating by parts Eq.  we obtain $\Gamma(z+1) = z \Gamma(z)$, which makes the connection with the factorial for integer numbers: $n! = \Gamma(n+1)$. Substituting $y = \sqrt{x}$ in Eq. , for $z = 1/2$ we obtain:

$$\Gamma(1/2) = 2\int_0^\infty e^{-y^2} dy = \sqrt{\pi}.$$ (16.11)