This week’s column is relatively short but sweet. It covers the general operations that can be carried out on the Gaussian family of integrals. Since these integrals are relatively easy to compute and they find applications in numerous fields they are a cornerstone of many physical disciplines and appear in many models of basic systems.
The prototype integral upon which the whole family is built is the traditional
\[ I = \int_{-\infty}^{\infty} dx \; e^{-a x^2} \; .\]
Although the derivation of this integral is well known, I’ll cover the computation for the sake of completeness.
Solving for the value of $$I$$ is much easier when one squares both sides to get
\[ I^2 = \int_{-\infty}^{\infty} dx \, e^{-a x^2} \int_{-\infty}^{\infty} dy \, e^{-a y^2} \; . \]
The next step is to put the repeated integral together as a double integral and then to change coordinates from Cartesian to plane polar. Following this program, one soon arrives at
\[ I^2 = \int_{0}^{\infty} \int_{0}^{2\pi}\, dr \, d\theta \, r e^{-ar^2} \; .\]
The $$\theta$$ integral is easily done leaving
\[ I^2 = 2 \pi \int_{0}^{\infty} \, dr \, r e^{-a r^2} \; .\]
Letting $$s = a r^2$$ gives $$ds = 2\, a\, r\, dr$$ with limits $$s(r=0) = 0$$ and $$s(r=\infty) = \infty$$. Plugging these in yields
\[ I^2 = \frac{\pi}{a} \int_{0}^{\infty} ds \, e^{-s} \; ,\]
which is a standard integral that can be immediate solved to give
\[ I^2 = \frac{\pi}{a} \left. e^{-s} \right|_{0}^{\infty} = \frac{\pi}{a} \]
or, more simply
\[ I = \int_{-\infty}^{\infty} \, dx \, e^{-a x^2} = \sqrt{ \frac{\pi}{a} } \; .\]
This can be generalized without much more work to the more fearsome-looking (but actually not much harder) integral family
\[ I_n = \int_{-\infty}^{\infty} \, dx \, x^n e^{-a x^2 + b x } \; n = 0, 1, 2, \ldots \; .\]
The first step is to evaluate $$I_0$$, which is the only true integral that will have to be done. The integral
\[ I_0 = \int_{-\infty}^{\infty} \, dx \, e^{-a x^2 + b x} \]
is tackled by completing the square:
\[ \int_{-\infty}^{\infty}\, dx e^{-a x^2 + b x} = \int_{-\infty}^{\infty} dx \; e^{-(\sqrt{a} x – \frac{b}{2\sqrt{a}})^2} e^{b^2/4a} \; . \]
The last term is a constant that can be brought out of the integral. The remaining piece is best computed by first defining
\[ y = \sqrt{a} x – \frac{b}{2\sqrt{a}} \; \]
and
\[ dy = \sqrt{a} dx \; .\]
The limits of integration stay the same the new form of the integral is
\[ I_0 = e^{b^2/4a} \frac{1}{\sqrt{a}} \int_{-\infty}^{\infty} \, dy e^{-y^2} = \sqrt{\frac{\pi}{a}} e^{b^2/4a} \; . \]
This result is so important it deserves its own box
All the other integrals come from the clever trick that says that powers of $$x$$ can be obtained from $$I_0$$ by differentiating with respect to either $$b$$ (all powers) or $$a$$ (even powers only).
For example, by definition
\[ I_1 = \int_{-\infty}^{\infty} \, dx \, x e^{-a x^2 + bx} \; ,\]
but, by construction
\[ \frac{\partial}{\partial b} I_0 = \frac{\partial}{\partial b} \int_{-\infty}^{\infty} \, dx \, e^{-a x^2 + bx} = \int_{-\infty}^{\infty} \, dx \, \frac{\partial}{\partial b} e^{-a x^2 + bx} \\ = \int_{-\infty}^{\infty} \, dx \, x e^{-a x^2 + bx} \; .\]
So
\[ I_1 = \frac{\partial}{\partial b} I_0 = \frac{\partial}{\partial b} \sqrt{\frac{\pi}{a}} e^{b^2/4a} \]
and the computation of this integral is done by performing a differentiation – a much easier task – to get
\[ I_1 = \frac{b}{4a} \sqrt{\frac{\pi}{a}} e^{b^2/4a} \; . \]
This generalizes immediately to
\[ I_n = \frac{\partial^n}{\partial b^n} I_0 \; .\]
Sometimes the integrals we wish to compute have $$b=0$$. In this case, which is fairly common, we can get the desired results almost as easily.
For $$n$$ odd, the integrand will be a product of an even function $$\exp({-ax^2})$$ and an odd function $$x^n$$ over an even domain. The result is exactly zero.
The $$n$$ even case comes from a similar differentiation with respect to $$a$$. For example,
\[ I_2 = -\frac{\partial}{\partial a} I_0 = \int_{-\infty}^{\infty} \, dx \, x^2 e^{-a x^2} \; .\]
Evaluating the derivative gives
\[ I_2 = -\frac{\partial}{\partial a} \sqrt{\frac{\pi}{a}} = \frac{1}{2} \sqrt{\frac{\pi}{a^3}} \; .\]
There are two consistency checks that are worth noting.
First, the recipe involving $$b$$ can still be used in the $$b = 0$$ case by taking the limit as $$b$$ approaches zero as the last step. In this limit,
\[ \lim_{b\rightarrow 0} I_1 = \lim_{b\rightarrow 0} \frac{b}{4a} \sqrt{\pi}{a} e^{b^2/4a} = 0 \; , \]
which checks with the partity argument given above.
Second, one would like to see that
\[ -\frac{\partial}{\partial a} I_0 = \frac{\partial^2}{\partial b^2} I_0 \]
for the computation of $$I_2$$. The left-hand side gives
\[ -\frac{\partial}{\partial a} I_0 = \frac{1}{2} \sqrt{\frac{\pi}{a^3}} e^{b^2/4a} + \frac{b^2}{4 a^2} \sqrt{\frac{\pi}{a}} e^{b^2/4a} \; .\]
Happily, the right-hand side gives the same:
\[ \frac{\partial^2}{\partial b^2} I_0 = \frac{\partial}{\partial b} \frac{b}{4a} \sqrt{\frac{\pi}{a}} e^{b^2/4a} = \frac{1}{2} \sqrt{\frac{\pi}{a^3}} e^{b^2/4a} + \frac{b^2}{4 a^2} \sqrt{\frac{\pi}{a}} e^{b^2/4a} \; .\]
In closing, a note on strategy. It seems that even in the case where $$b \neq 0$$ it is easier to differentiate with respect to $$a$$ for even powers.