Definite integral of infinite product

Question

I have been struggling for a while on evaluating this definite infinite product integral:

$$\int_{-\frac{\pi}{4}}^0(1+\tan{x})(1+\tan^2x)(1+\tan^4x)(1+\tan^8x)(1+\tan^{16}x)...dx$$

This is a question given by my maths teacher a while back and I have been struggling with it ever since. I have tried so many different substitutions and I have even tried integrating by parts (do NOT do this), but nothing has led me even close to an answer. I'm guessing there is trig identity I must be missing in order to simplify the inside of the integral? or some wonder substitution?

Any help would be greatly appreciated.

Answer 1

Here's a short method that uses a trigonometric substitution. Substituting $u = \tan \theta$, so that $d\theta = \frac{du}{1 + u^2}$, transforms the integral to $$\int_{-1}^0 \frac{(1 + u) (1 + u^2) (1 + u^4) \cdots}{1 + u^2} \,du .$$ Verify that (for $u \in (-1, 0]$), $$(1 + u) (1 + u^2) (1 + u^4) \cdots = 1 + u + u^2 + u^3 + \cdots = \frac{1}{1 - u} ,$$ and substituting realizes the integrand explicitly as a rational function.

The integral becomes $$\int_{-1}^0 \frac{du}{(1 - u) (1 + u^2)} = \boxed{\frac{1}{4} \log 2 + \frac{\pi}{8}}.$$

Answer 2

HINT

Multiply and divide the integrand by $1 - \tan(x)$.

Answer 3

Hint: Try to prove that for $x\in (-\pi/4,0)$, $$\prod_{n=0}^\infty (1+\tan^{2^n}x)=\frac{1}{1-\tan x}.$$ EDIT: Take $$\prod_{n=0}^N (1+\tan^{2^n}x)=(1+\tan x)(1+\tan^2x)\cdots (1+\tan^{2^N}x)=\frac{(1-\tan x)(1+\tan x)(1+\tan^2x)\cdots (1+\tan^{2^N}x)}{1-\tan x}=\frac{(1-\tan^2x)(1+\tan^2x)(1+\tan^4x)\cdots (1+\tan^{2^N}x)}{1-\tan x}=\frac{(1-\tan^4x)(1+\tan^4x)\cdots (1+\tan^{2^N}x)}{1-\tan x}=\cdots=\frac{1-\tan^{2^{N+1}}x}{1-\tan x}\to \frac{1}{1-\tan x},$$ as $N\to \infty$, since $x\in (-\pi/4,0)$. From here is straightforward to compute the integral. The answer is $\frac{\ln 2}{4}+\frac{\pi}{8}.$

Answer 4

$\int_{-\frac{\pi}{4}}^0(1+\tan{x})(1+\tan^2x)(1+\tan^4x)(1+\tan^8x)(1+\tan^{16}x)...dx$

$=\int_{-\frac{\pi}{4}}^0\color{red}{\frac{1}{1-\tan{x}}(1-\tan{x})}(1+\tan{x})(1+\tan^2x)(1+\tan^4x)(1+\tan^8x)(1+\tan^{16}x)...dx$

(then collapse the product)

$=\int_{-\frac{\pi}{4}}^0\frac{1}{1-\tan{x}}(1-(\tan{x})^\infty)dx$ (probably shouldn't write this step, but it might be helpful)

$=\int_{-\frac{\pi}{4}}^0\frac{1}{1-\tan{x}}dx$

$=\int_{-\frac{\pi}{4}}^0\frac12 \left(1+\tan{\left(x+\frac{\pi}{4}\right)}\right)dx$

$=\dots$

$=\frac18(\pi+\log 4)$