Closed form of $\prod_{n=0}^{\infty}\frac1{1+x^{2^n}}$

Question

In the qualifying exam for the MIT integration bee 2023, the following question was asked: $$\int_0^1\prod_{n=0}^{\infty}\frac1{1+x^{2^n}}dx$$I graphed the integrand (denoted as $f(x)$) and from what I could see it is equal to $-x+1$ when $|x|<1$ and $0$ elsewhere. All I was able to do was prove that $f(0)=1$ and $f(1)=0$. So how could I find a closed form for this product?

Answer 1

This is a telescoping product.

$$(1-x^{2^n})(1+x^{2^n}) = 1 - x^{2^{n+1}}.$$

Therefore, $$(1-x) \prod_{n=0}^N (1+x^{2^n}) = 1 - x^{2^{N+1}}.$$

The rest of the details I leave to you as an exercise.

Answer 2

Re-write the product: $$\prod_{k=0}^\infty\frac1{1+x^{2^k}}=\frac1{\prod_{k=0}^\infty(1+x^{2^k})}$$Expanding the first few partial products of the denominator, we get that it seems to be $$\prod_{k=0}^N(1+x^{2^k})=\frac{1-x^{2\epsilon(N)}}{1-x^2}(1+x)=\frac{1-x^{2\epsilon(N)}}{1-x}$$Where $\epsilon(N)=\sum_{k=1}^n2^k$. As $N$ approaches infinity, the numerator becomes $1$. Taking the reciprocal gives the answer.