problem-solving

The integral ${I=\int_0^\pi \log \sin x dx}$ is normally computed (e.g. in Ahlfors’ book) to be ${-\pi \log 2}$ using complex integration over a suitable almost-rectangular contour. There is also a simple and direct way to get the value of this integral by a substitution and elementary calculus.

First, by the substitution ${x=2t}$ and the identity ${\sin(2x)=2\sin x \cos x}$,

$\displaystyle I = 2 \int_0^{\pi/2} \log \sin t dt + 2 \int_0^{\pi/2} \log \cos t dt + \pi \log 2;$

then using the symmetry of ${\sin}$ and ${\cos}$ gives:

$\displaystyle I = I + I + \pi \log 2,$

whence the result. There are slight technicalities regarding the improperness of these integrals, but they can be directly justified (or one may use the Lebesgue integral).

[Edit (7/25)- Todd Trimble posted solutions to similar integrals, which use the result of this post as a lemma, here.  AM]

So, in a break from the earlier series I was doing on Lie algebras, I want to discuss a very elementary question about polynomials. The answer is well-known but is interesting.   It would make a good competition type problem (indeed, it’s an exercise in Serge Lang’s Algebra).  Moreover, ironically, it’s useful in algebra: At some point one of us will probably discuss Hilbert polynomials, which take integer values, so this result tells us something about them.

We have a polynomial ${P(X) \in \mathbb{Q}[X]}$ which takes integer values at all sufficiently large ${n \in \mathbb{N}}$. What can we say about ${P}$?

Denote the set of such ${P}$ by ${\mathfrak{I}}$. Then clearly ${\mathbb{Z}[X] \subset \mathfrak{I}}$. But the converse is false.