Misadventure #1

By Jean Pierre Mutanguha

Now seems like a good time to write about my failed math research. There are several questions I've never been able to answer and today I present the most frustrating one - an identity.

\[\left(1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots\right)^2 = 1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49} + \cdots \]

I discovered this identity while attempting to calculate \( \int_0^\infty \frac{1}{1+x^4}\,dx \). This evaluated to give me:1

\[\sum_{n=0}^\infty (-1)^n\left(\frac{1}{4n+1} +\frac{1}{4n+3}\right) = \frac{\pi}{\sqrt8}\]

I instantly recognized that

\[\sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}\]

from other proofs of Basel's problem. Thus squaring the former series would produce the latter. On the other hand, if I could prove the identity with a different and preferably simpler approach, then I would have a new and original proof of \( \zeta(2) = \frac{\pi^2}{6} \).

I eventually seeked help from Math Stack Exchange. The only answer received two weeks later was not very satisfactory as it reverted back to integrals and looked very similar to previous solutions to Basel's problem.2

The identity looked so simple that going back to integrals seemed like a step backwards. Surely there should be a proof that relies symbol manipulation? Or a proof that showed the difference between the two series tended to 0 as the number of terms increased?

A comment in on my question linked to a different proof that did something similar. The proof uses a heuristic argument to show that

\[ \sum_{k=-\infty}^{\infty} \dfrac1{(2k+1)^2} = \left(\sum_{k=-\infty}^{\infty} \dfrac{(-1)^k}{(2k+1)} \right)^2 \]

I attempted to prove the identity along the same lines but I never could show that the difference between the two series vanished.

So eight months later, I still have no nice proof for this beautiful identity: my first mathematical misadventure.

  1. \[ \begin{aligned} \int_0^\infty \frac{1}{1+x^4}\,dx ~ &= \int_0^1 \frac{1+x^2}{1+x^4}\,dx \ &= \int_0^1 1+x^2-x^4-x^6+x^8+x^{10} - \cdots \,dx \\ &= 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots\end{aligned}\] And, \[ \begin{aligned} \int_0^\infty \frac{1}{1+x^4}\,dx ~ &= \int_0^1 \frac{1+x^2}{1+x^4}\,dx \ &= \frac{1}{2}\int_0^1 \frac{1}{x^2-\sqrt{2}x+1} + \frac{1}{x^2+\sqrt{2}x+1} \,dx \\ &= \frac{1}{\sqrt{2}} \left[ \arctan(\sqrt{2}-1) + \arctan(\sqrt{2}+1) \right] \ &= \frac{\pi}{\sqrt{8}}\end{aligned}\] ↩

  2. Using similar techniques as previous solutions to Basel's problem meant that the new proof contained nothing original as noted by the author of the post. Secondly, I was originally trying to use a single simple integral \( \displaystyle \int \frac{1}{1+x^4}\,dx.\) That solution eventually introduced the double integral I was trying to avoid \( \displaystyle \int \int \frac{x}{(1+x^2)(1+x^2y^2)} \,dx\,dy.\) ↩