# Rambling Thoughts

## A problem from the SMT

The 13th annual Stanford Mathematics Tournament was held on February 19, 2011. My favorite problem on the contest was the last problem on the calculus subject test.

Theorem 1 Compute the integral

$\displaystyle \int_0^\pi \ln (1 - 2a \cos x + a^2) \, dx$

for ${a > 1}$.

Solution 1:

This integral can be computed using a Riemann sum. Divide the interval of integration ${[0, \pi]}$ into ${n}$ parts to get the Riemann sum

$\displaystyle \frac{\pi}{n} \left[ \ln \left( a^2 - 2a \cos \frac{\pi}{n} + 1 \right) + \ln \left( a^2 - 2a \cos \frac{2 \pi}{n} + 1 \right) + \cdots + \ln \left( a^2 - 2a \cos \frac{(n-1)\pi}{n} + 1 \right) \right].$

Recall that

$\displaystyle \cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}.$

We can rewrite this sum of logs as a product and factor the inside to get

$\displaystyle \frac{\pi}{n} \ln \left[ \prod_{k = 1}^{n-1} \left( a^2 - 2a \cos \frac{k\pi}{n} + 1 \right) \right] = \frac{\pi}{n} \ln \left[ \prod_{k = 1}^{n-1} \left( a - e^{k\pi i / n} \right) \left( a - e^{-k\pi i / n} \right) \right].$

The terms ${e^{\pm k \pi i / n}}$ are all of the ${2n}$-th roots of unity except for ${\pm 1}$, so the inside product contains all of the factors of ${a^{2n} - 1}$ except for ${a -1}$ and ${a+1}$. The Riemann sum is therefore equal to

$\displaystyle \frac{\pi}{n} \ln \frac{a^{2n} - 1}{a^2 - 1}$

To compute the value of the desired integral, we compute the limit of the Riemann sum as ${n \rightarrow \infty}$; this is

$\displaystyle \lim_{n \rightarrow \infty} \frac{\pi}{n} \ln \frac{a^{2n} - 1}{a^2 - 1} = \lim_{n \rightarrow \infty} \pi \ln \sqrt[n]{\frac{a^{2n} - 1}{a^2 - 1}} = \lim_{n \rightarrow \infty} \pi \ln a^2 = 2 \pi \ln a.$

(This is problem 471 of Razvan Gelca and Titu Andreescu’s book Putnam and Beyond. The solution is due to Siméon Poisson.)

Solution 2:

Let the desired integral be ${I(a)}$, where we think of this integral as a function of the parameter ${a}$. In this solution, we differentiate by ${a}$ to convert the desired integral to an integral of a rational function in ${\cos x}$:

$\displaystyle \frac{d}{da} I(a) = \frac{d}{da} \int_0^\pi \ln(1 - 2a \cos x + a^2) \, dx = \int_0^\pi \frac{2a - 2\cos x}{1 - 2a \cos x + a^2} \, dx.$

All integrals of this form can be computed using the substitution ${t = \tan \frac x2}$. Then ${x = 2 \arctan t}$, so ${dx = \frac{2}{1 + t^2} \, dt}$ and

$\displaystyle \cos x = \cos (2 \arctan t) = 2 \cos (\arctan t)^2 - 1 = 2 \left( \frac{1}{1+t^2} \right) - 1 = \frac{1 - t^2}{1+t^2},$

so our integral becomes

$\displaystyle \frac{d}{da} I(a) = \int_0^\infty \frac{2a - 2 \frac{1-t^2}{1+t^2}}{1 - 2a \frac{1 - t^2}{1+t^2} + a^2} \frac{2}{1 + t^2} \, dt = 4 \int_0^\infty \frac{a (1 + t^2) - (1 - t^2)}{(1 + t^2) - 2a (1 - t^2) + a^2 (1 + t^2) } \frac{1}{1+t^2}\, dt$

$\displaystyle = 4 \int_0^\infty \frac{(a+1)t^2 + (a-1)}{((a+1)^2 t^2 + (a-1)^2)(1+t^2)} \, dt = \frac2a \int_0^\infty \frac{a^2 - 1}{(a+1)^2 t^2 + (a-1)^2} \, dt + \frac2a \int_0^\infty \frac{1}{1 + t^2} \, dt.$

In the first integral, we do the substitution ${t = \frac{a -1}{a+1} u}$. Then ${dt = \frac{a-1}{a+1} du}$ and we have

$\displaystyle = \frac2a \int_0^\infty \frac{1}{1+u^2} \, du + \frac2a \int_0^\infty \frac{1}{1+t^2} \, dt = \frac2a \left( \frac\pi2 + \frac\pi2 \right) = \frac{2\pi}a.$

Therefore, our desired integral is the integral of the previous quantity, or

$\displaystyle I = \int_0^\pi \ln (1 - 2a \cos x + a^2) \, dx = 2\pi \ln a.$

Solution 3:

We use Chebyshev polynomials. First, define the Chebyshev polynomial of the first kind to be ${T_n (x) = \cos (n \arccos x)}$. This is a polynomial in ${x}$, and note that ${T_n(\cos x) = \cos (nx)}$. Note that

$\displaystyle \cos((n+1)x) = \cos nx \cos x - \sin nx \sin x$

$\displaystyle \cos((n-1)x) = \cos nx \cos x + \sin nx \sin x$

so that ${\cos((n+1)x) = 2 \cos nx \cos x - \cos((n-1)x)}$ and hence the Chebyshev polynomials satisfy the recurrence ${T_{n+1}(x) = 2x T_n (x) - T_{n-1} (x)}$.

Therefore, the Chebyshev polynomials satisfy the generating function

$\displaystyle \sum_{n=0}^\infty T_n(x) t^n = \frac{1-tx}{1 - 2tx + t^2}.$

Now, substituting ${x \mapsto \cos x}$ and ${t \mapsto a^{-1}}$, we have

$\displaystyle \sum_{n=0}^\infty \cos (nx) a^{-n} = a\frac{a - \cos x}{a^2 - 2a \cos x + 1}.$

So

$\displaystyle 2 \sum_{n=0}^\infty \cos (nx) a^{-n-1} = \frac{2a - 2\cos x}{1 - 2a \cos x + a^2}.$

Then

$\displaystyle \int_0^\pi \frac{2a - 2\cos x}{1 - 2a \cos x + a^2} \, dx = 2 \int_0^\pi \sum_{n=0}^\infty \cos (nx) a^{-n-1} \, dx = 2 \sum_{n=0}^\infty \left( a^{-n-1} \int_0^\pi \cos (nx) \, dx \right) = 2 \pi a^{-1}.$

Now, since

$\displaystyle \ln (1 - 2a \cos x + a^2) = \int \frac{2a - 2 \cos x}{1 - 2a \cos x + a^2} \, da,$

we see that

$\displaystyle \int_0^\pi \ln (1 - 2 a \cos x + a^2) \, dx = \int 2 \pi a^{-1} \, da = 2\pi \ln a.$

Solution 4:

We can also give a solution based on physics. By symmetry, we can evaluate the integral from ${0}$ to ${2\pi}$ and divide the answer by ${2}$, so

$\displaystyle \int_0^{\pi} \ln (1 - 2a\cos x + a^2) \, dx = \int_0^{2\pi} \ln \sqrt{1 - 2a\cos x + a^2} \, dx.$

Now let’s calculate the 2D gravitational potential of a point mass falling along the ${x}$ axis towards a unit circle mass centered around the origin. We set the potential at infinity to ${0}$. We also note that, since the 2D gravitational force between two masses is proportional to ${\frac{1}{r}}$, the potential between two masses is proportional to ${-\ln r}$. So to calculate the gravitational potential, we integrate ${-\ln r}$ over the unit circle. But if the point mass is at ${(a,0)}$, then the distance between the point mass and the section of the circle at angle ${x}$ is ${\sqrt{1 - 2a\cos x + a^2}}$. So we get the integral

$\displaystyle -\int_0^{2\pi} \ln \sqrt{1 - 2a\cos x + a^2} \, dx$

This is exactly the integral we want to calculate! We can also calculate this potential by concentrating the mass of the circle at its center. The circle has mass ${2\pi}$ and its center is distance ${a}$ from the point mass. So the potential is simply ${-2\pi \ln a}$. Thus, the final answer is ${2\pi \ln(a)}$.

Solution 5:

This problem also has a solution which uses the Residue Theorem from complex analysis. It is easy to show that

$\displaystyle 2 \int_0^{\pi} \ln(1 - 2a \cos(x) + a^2) \, dx = \int_0^{2\pi} \ln(1 - 2a \cos(x) + a^2) \, dx.$

Furthermore, observe that ${1 - 2a \cos x + a^2 = (a - e^{ix})(a - e^{-ix})}$. Thus, our integral is

$\displaystyle I = \frac{1}{2} \left( \int_0^{2 \pi} \ln [ (a - e^{ix})(a - e^{-ix}) ] dx \right) = \frac{1}{2} \left( \int_0^{2\pi} \ln(a - e^{ix}) dx + \int_0^{2\pi} \ln(a - e^{-ix}) dx \right),$

where the integrals are performed on the real parts of the logarithms in the second expression. In the first integral, substitute ${z = e^{ix}}$, ${dz = ie^{ix} \, dx = iz \, dx}$; the resulting contour integral is

$\displaystyle \oint_{\| z \| = 1} \frac{\ln(a - z)}{iz} dz.$

By the Residue Theorem, this is equal to ${2 \pi i Res_{z = 0} \frac{\ln(a - z)}{iz} = 2\pi \ln(a)}$. The second integral is identical. Thus, the final answer is ${\frac{1}{2} (4\pi \ln(a)) = 2\pi \ln(a)}$.