# Choosing convenient limits of integration on Basel problem

I have recently discovered $$\sum_{k=1}^{\infty}\frac{\cos\left(k\alpha\right)}{k^{2}}-\sum_{k=1}^{\infty}\frac{\cos\left(k\beta\right)}{k^{2}}=\int_{\alpha}^{\beta}\tan^{-1}\left(\cot\left(\theta/2\right)\right)\,\mathrm{d}\theta$$ which seems to be a nice way to attack Basel problem. The integral nicely reduces to $$\int_{0}^{\beta}\frac{\pi}{2}-\frac{\theta}{2}\,\mathrm{d}\theta=\frac{2\pi\beta-\beta^{2}-2\pi\alpha+\alpha^{2}}{4}=\frac{1}{4}\left(\alpha-\beta\right)\left(\alpha+\beta-2\pi\right)$$ when $\alpha,\,\beta\in\left[0,\,2\pi\right]$. The question: how do I choose $\alpha$ and $\beta$ such that $\cos\left(k\alpha\right)-\cos\left(k\beta\right)=1$ for all $k\in\mathbb{N}$? Does this $k$ even exist?

• I don't think this is possible. Intuitively, for any $\alpha\neq\beta$, there should be some large $k$ such that $k\beta\approx\beta\pmod{2\pi}$, but $k\alpha\not\approx\alpha\pmod{2\pi}$, and thus $\cos(k\alpha)-\cos(k\beta)\neq\cos(\alpha)-\cos(\beta).$ – Samuel Jan 13 '13 at 20:51

## 2 Answers

Sadly, there's no such $\alpha$ and $\beta$. We'll try and solve the more general equation - $\cos(k\alpha)-\cos(k\beta)$ is a non-zero constant, say $C$.

Assume there are such $\alpha, \beta$. Let $x=\cos \alpha, y = \cos \beta$.

Note that $\cos(2t) = 2\cos(t)^2 - 1, \cos(3t) = 4\cos(t)^{3} - 3\cos(t)$. Then the following 2 equalities hold: $$x-y = C$$ $$2x^2-2y^2 = C$$ Factoring $2x^2-2y^2$ as $2(x-y)(x+y)$ and plugging the first equation, we find that $x+y = 0.5$ This leads to $x = \frac{C+0.5}{2}, y=\frac{0.5 - C}{2}$. Now we choose $k=3$: $$4x^3 - 3x - (4y^3 - 3y) = C$$ The LHS factors as $(x-y)(4(x^2+xy+y^2)-3)=(x-y)(4(x+y)^2-4xy-3)$, which is $$C(4(0.5)^2-4(\frac{1}{16} - \frac{C^2}{4})-3)=C$$ Cancelling the $C$s, it becomes $$C^2 = 3.25$$ So $C\neq 1$!

Now we note that $\cos(4t) = 8(\cos(t)^4-\cos(t)^2)+1$, which gives the following equation: $$8(x^4-x^2-(y^4-y^2)) = 1$$ The LHS factors as $8(x^2-y^2)(x^2+y^2-1)=1$. We can plug $2x^2-2y^2 = C$ and this becomes $$x^2+y^2 = \frac{1}{4C}+1$$ But $x^2+y^2 = \frac{C^2}{2}+\frac{1}{8} = 1.75$, so it follows that $C=\frac{1}{3}$, a contradiction to $C^2 = 3.25$.

Although the $\alpha$ and $\beta$ you seek cannot exist, taking $\alpha=0$ and $\beta = \pi$ will allow you to attack the Basel problem. These values give: $$\sum_{k=1}^{\infty}\frac{2}{(2k-1)^2} = \frac{\pi^2}{4}$$

and then since $$\sum_{k=1}^{\infty}\frac{1}{k^2}= \sum_{m=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(2^m(2k-1))^2}=\sum_{n=0}^{\infty} \frac{1}{4^n}( \sum_{k=1}^{\infty}\frac{1}{(2k-1)^2})=\frac{4}{3}\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}$$ The first equality comes from the fact any integer can be written uniquely as an odd integer times a power of 2. The second is from bringing the powers of 2 outside the sum. The $\frac{4}{3}$ arises as the sum of the geometric series. The Basel problem then follows easily.

• I guess this doesn't strictly answer your question, I thought I'd share it anyway. – user45861 Jan 13 '13 at 21:40
• Sorry, but how do you go from $\sum_{k=1}^{\infty}\frac{1-(-1)^k}{k^2}$ to $\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}$? I got $\sum_{k=1}^{\infty}\frac{2}{(2k-1)^2}$ – Ian Mateus Jan 13 '13 at 21:50
• You're quite right. I managed to introduce a few errors. I'll edit them. – user45861 Jan 13 '13 at 21:55
• I'd be happy if you clarify a little more what you did next, I couldn't understand it. – Ian Mateus Jan 13 '13 at 21:57
• That should be a bit better. – user45861 Jan 13 '13 at 22:14