The third most important integral in physics ?

$$\int_{-\infty}^{\infty} e^{-\alpha z^2} dz = \sqrt{\frac{\p... ...a}} \hspace*{1cm} \mbox{{\rm for any}} \hspace*{1cm} \alpha \gt$$

Right...I'll confess that I'm not quite sure of my ranking system here. I just wrote that heading to encourage you to pay more attention than you might otherwise to a dry-looking integral.

Since I've got your attention for a moment I'll explain:

• how you prove this

• why it is particularly useful.

First, the proof. Begin by noting that we can expose the way the integral depends on $\alpha$ by making the change of variables:

$$z \rightarrow x/\sqrt{\alpha}$$

yielding immediately

$$\int_{-\infty}^{\infty} e^{-\alpha z^2} dz = \frac{I}{\sqrt{\alpha}}$$

where

$$I=\int_{-\infty}^{\infty} e^{-x^2} dx$$

I've chosen to use the variable x again for a good reason --despite the clash with the x we are using for our random variable.

So, we are after the value of I. Let's do something that looks silly...but will turn into something rather cunning. We note that we could also write

$$I=\int_{-\infty}^{\infty} e^{-y^2} dy$$

And now we can multiply the two expressions for I together to give

$$I^2= \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} dx dy$$ Although this looks worse, it is easier to evaluate. We just change to polar coordinates, r and $\theta$ with $$x=r \cos{\theta} \hspace*{0.5cm} \mbox{{\rm and}} \hspace*{0.5cm} y=r \sin{\theta}$$

Then

$$I^2= \int_{0}^{\infty}dr \int_{0}^{2\pi} d\theta e^{-r^2} r = 2\pi\int_{0}^{\infty} e^{-r^2} r dr$$

where we have done the easy integral on $\theta$.

The remaining integral is also easy:

$$I^2 = 2\pi \int_{0}^{\infty} e^{-r^2} r dr = 2\pi \left [- \frac{1}{2} e^{-r^2}\right ]^{\infty}_{0} =\pi$$

So we identify $I=\sqrt{\pi}$ and our claimed result follows.

I did that for you, because it's a nice example of having to take a little step backwards in order to make a larger one forwards.

Now, why is the result useful? Well, let's take it and differentiate it (both sides!) with respect to $\alpha$:

Differentiating the RHS we find

$$\frac{d}{d \alpha} \sqrt{\frac{\pi}{\alpha}} =- \frac{\sqrt{\pi}}{2 \alpha ^{3/2}}$$

Differentiating the LHS

$$\frac{d}{d \alpha} \int_{-\infty}^{\infty} e^{-\alpha z^2} ... ...pha z^2} dz = -\int_{-\infty}^{\infty} z^2 e^{-\alpha z^2} dz$$

Equating the two expressions we conclude that

$$\int_{-\infty}^{\infty} z^2 e^{-\alpha z^2} dz = \frac{\sqrt{\pi}} {2\alpha ^{3/2}}$$

So now we have a result for a second integral. We can use it immediately to show that

$$(2 \pi \sigma ^2)^{-1/2}\int_{-\infty}^{\infty} [x- \mu] ^2 e^{-(x-\mu)^2/(2 \sigma ^2)} = \sigma ^2$$

confirming that $\sigma$ in KP4.4 does indeed give the standard deviation of the gaussian pdf.

There are other ways of getting this result (I suggest integration by parts in the main text). But the trick of generating the integral you want by differentiating the standard integral is well worth noting...if you have aspirations to travel far in physics.