One of my favorite integrals is found in *Mechanics* by Landau and Lifshitz. In the context of figuring out to what extent the potential energy \( U(x) \) of a field can be determined if the period of oscillation \( T(E) \) is known as a function of the energy \( E \), this integral appears:

\[ \int_\alpha^\beta \frac{dx}{\sqrt{(\beta - x) (x - \alpha)}} . \]

I changed the notation from the book. The assumption is that \( \beta > \alpha > 0\). With characteristic Russian tersity, the authors write in reference to this integral: “The integral over \(E\) is elementary; its value is \(\pi\).”

Here is a plot of the integrand for \(\alpha = 1\) and \(\beta = 2\):

Mathematica can do this integral if you give it the assumption \( \beta > \alpha > 0\). But this doesn’t give any insight into why \(\pi\) is showing up.

One way to solve it by hand is to first complete the square for the term under the square root:

\[ - x^2 + x (\alpha + \beta) - \alpha \beta = - \left( x - \frac{\alpha + \beta}{2} \right)^2 + \frac{1}{4}(\beta - \alpha)^2 .\]

Next let

\[ u = x - \frac{\alpha + \beta}{2} \]

and

\[ z = \frac{\beta - \alpha}{2} > 0 .\]

Making these substitutions, the integral is

\[ \int_{-z}^z \frac{du}{\sqrt{z^2 - u^2}}, \]

which is a well-known trig integral:

\[ \int_{-z}^z \frac{du}{\sqrt{z^2 - u^2}} = \arcsin(1) - \arcsin(-1) = \frac{\pi}{2} - \left( - \frac{\pi}{2} \right) = \pi .\]

This solves the integral, but again doesn’t provide much intuition as to why \(\pi\) shows up. It feels like there should be a nice geometric argument explaining why the integral is \(\pi\). If you know of one, please let me know!