An Analytic Proof of the Extraordinary Euler Sum |

In Deriving the Extraordinary Euler Sum , we derived one of Euler’s most celebrated results:

$\displaystyle 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + ... = \frac{\pi^2}{6}\quad\quad\quad(1)$

Now, we aim to provide a rigorous proof of this statement.

First, we expressed the partial sum of the left-hand side as follows:

$\displaystyle \sum\limits_{i=1}\frac{1}{i^2} = \sum\limits_{i=1}\frac{1}{(2i)^2} + \sum\limits_{i=0}\frac{1}{(2i+1)^2}$

This splits the partial sum into two parts: one involving the squares of even numbers and the other involving the squares of odd numbers. Simplifying the even part gives us:

$\displaystyle \sum\limits_{i=1}\frac{1}{i^2} = \frac{1}{4}\sum\limits_{i=1}\frac{1}{i^2} + \sum\limits_{i=0}\frac{1}{(2i+1)^2}$

Rearranging terms $\displaystyle \sum\limits_{i=1}\frac{1}{i^2}$ on one side, we obtain:

$\displaystyle \frac{3}{4}\sum\limits_{i=1}\frac{1}{i^2}=\sum\limits_{i=0}\frac{1}{(2i+1)^2}$

Since $\displaystyle \sum_{i=1}\frac{1}{i^2}$ converges to $\displaystyle \sum_{i = 1}^{\infty}\frac{1}{i^2}$ (see My Shot at Harmonic Series)

$\displaystyle \implies \sum\limits_{i=0} \frac{1}{(2i+1)^2}$ converges to $\displaystyle \sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2} = \frac{3}{4}\sum\limits_{i=1}^{\infty}\frac{1}{i^2},$

to prove (1), it suffices to demonstrate that

$\displaystyle \frac{4}{3}\sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2} = \frac{\pi^2}{6}$

or equivalently,

$\displaystyle \sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2} = \frac{\pi^2}{8}\quad\quad\quad(2)$

Let $\displaystyle I = \int\limits_{0}^{\infty}\frac{\arctan(x)}{1+x^2}\;dx$ and $J(\beta) = \displaystyle \int\limits_{0}^{\infty}\frac{\arctan(\beta x)}{1+x^2}\;dx,$ we have

$\displaystyle J(1) = I = \frac{1}{2}\arctan^2(x)\bigg|_{0}^{\infty} = \frac{1}{2}\left(\frac{\pi}{2}\right)^2=\frac{\pi^2}{8}\quad\quad\quad(3)$

and

$J(0) = 0\quad\quad\quad(4)$

Differentiating $\displaystyle J(\beta)$ with respect to $\beta:$

$\frac{d}{d\beta}J(\beta) = \displaystyle\frac{d}{d\beta}\int\limits_{0}^{\infty}\frac{\arctan(\beta x)}{1+x^2}\;dx = \int\limits_{0}^{\infty}\frac{\partial}{\partial \beta}\left(\frac{\arctan(\beta x)}{1+x^2}\right)\;dx$

$= \displaystyle \int\limits_{0}^{\infty}\frac{1}{1+x^2}\cdot\frac{x}{\beta^2x^2+1}\;dx$

That is,

$\frac{d}{d\beta}J(\beta) = \displaystyle \int\limits_{0}^{\infty}\frac{x}{(1+x^2)(\beta^2x^2+1)}\;dx$

Integrating with respect to $\beta$ from $0$ to $1:$

$\displaystyle\int\limits_{0}^{1}\frac{d}{d\beta}J(\beta)\;d\beta = \int\limits_{0}^{1}\int\limits_{0}^{\infty}\frac{x}{(1+x^2)(\beta^2 x^2+1)}\;dx\;d\beta$

and expressing the integrand in partial fractions:

yields

$J(1) - J(0) = \displaystyle\int\limits_{0}^{1}\int\limits_{0}^{\infty}\frac{1}{\beta^2-1}\left(\frac{\beta^2x}{\beta^2x^2+1} - \frac{x}{x^2+1}\right)\;dx\;d\beta$

$= \displaystyle\int\limits_{0}^{1}\int\limits_{0}^{\infty}\frac{1}{\beta^2-1}\cdot\frac{1}{2}\cdot\left(\frac{2\beta^2x}{\beta^2x^2+1} - \frac{2x}{x^2+1}\right)\;dx\;d\beta$

$= \displaystyle\int\limits_{0}^{1}\frac{1}{2(\beta^2-1)}\int\limits_{0}^{\infty}\left(\frac{2\beta^2x}{\beta^2x^2+1} - \frac{2x}{x^2+1}\right)\;dx\;d\beta$

$= \displaystyle\int\limits_{0}^{1}\frac{1}{2(\beta^2-1)}\log\left(\frac{\beta^2 x^2+1}{x^2+1}\right)\bigg|_{0}^{\infty}\;d\beta$

$= \displaystyle\int\limits_{0}^{1}\frac{1}{2(\beta^2-1)}\lim\limits_{x \rightarrow \infty}\log\left(\frac{\beta^2 x^2+1}{x^2+1}\right)\;d\beta = \displaystyle\int\limits_{0}^{1}\frac{1}{2(\beta^2-1)}\lim\limits_{x \rightarrow \infty}\log\left(\frac{\beta^2 + \frac{1}{x^2}}{1+\frac{1}{x^2}}\right)\;d\beta$

$= \displaystyle\int\limits_{0}^{1}\frac{1}{2(\beta^2-1)}\log(\beta^2)\;d\beta=\boxed{\int\limits_{0}^{1}\frac{\log(\beta)}{\beta^2-1}\;d\beta=\displaystyle\int\limits_{0}^{1}\sum\limits_{i=0}^{\infty}\frac{\beta^{2i}}{2i+1}\;d\beta \quad\quad(\star)}$

$= \displaystyle\sum\limits_{i=0}^{\infty}\int\limits_{0}^{1}\frac{\beta^{2i}}{2i+1}\;d\beta= \displaystyle\sum\limits_{i=0}^{\infty}\left(\frac{1}{2i+1}\underline{\int\limits_{0}^{1}\beta^{2i}\;d\beta}\right)=\sum\limits_{i=0}^{\infty}\frac{1}{2i+1}\cdot\underline{\frac{\beta^{2i+1}}{2i+1}\bigg|_{0}^{1}}$

$= \displaystyle\sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2}$

i.e.,

$J(1)-J(0) = \displaystyle\sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2}.$

By (3) and (4),

$\displaystyle\frac{\pi^2}{8}-0 =\sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2}$

$\displaystyle\sum\limits_{i=0}^{\infty}\frac{1}{(2i+1)^2} = \frac{\pi^2}{8}$

which is (2)

Prove $\boxed{\int\limits_{0}^{1}\frac{\log(\beta)}{\beta^2-1}\;d\beta=\displaystyle\int\limits_{0}^{1}\sum\limits_{i=0}^{\infty}\frac{\beta^{2i}}{2i+1}\;d\beta \quad\quad(\star)}$

Since

$\displaystyle\frac{d}{d\beta}v = \frac{1}{\beta^2-1} = \frac{1}{\beta-1}-\frac{1}{\beta+1}$

means

$v = \displaystyle \frac{1}{2}\int\frac{1}{\beta-1}-\frac{1}{\beta+1}\;d\beta = \frac{1}{2}\log\left(\frac{|\beta-1|}{|\beta+1|}\right) \overset{\beta \le 1}{=}\frac{1}{2}\log\left(\frac{1-\beta}{\beta+1}\right),$

we have

$\displaystyle\int\limits_{0}^{1}\frac{\log(\beta)}{\beta^2-1}\;d\beta = \int\limits_{0}^{1}\log(\beta)\cdot\frac{d}{d\beta}\left(\frac{1}{2}\log\left(\frac{1-\beta}{1+\beta}\right)\right)\;d\beta$

$= \displaystyle\log(\beta)\cdot\frac{1}{2}\log\left(\frac{1-\beta}{1+\beta}\right)\bigg|_{0}^{1} - \int\limits_{0}^{1}\frac{1}{\beta}\cdot\frac{1}{2}\log\left(\frac{1-\beta}{1+\beta}\right)\;d\beta$

$= \displaystyle\frac{1}{2}\cdot\underline{\log\left(\beta\right)\log\left(\frac{1-\beta}{1+\beta}\right)\bigg|_{0}^{1}} - \int\limits_{0}^{1}\frac{1}{\beta}\cdot\frac{1}{2}\log\left(\frac{1-\beta}{1+\beta}\right)\;d\beta$

$\boxed{\log\left(\beta\right)\log\left(\frac{1-\beta}{1+\beta}\right)\bigg|_{0}^{1}=0\quad\quad\quad(\star\star)}$

$= \displaystyle\int\limits_{0}^{1}\frac{1}{\beta}\cdot\frac{1}{2}\cdot\log\left(\frac{1+\beta}{1-\beta}\right)\;d\beta$

Expand $\log(\frac{1+\beta}{1-\beta})$ into its Maclaurin series:

$= \displaystyle \int\limits_{0}^{1}\frac{1}{\beta}\cdot\frac{1}{2}\cdot2\sum\limits_{i=0}^{\infty}\frac{\beta^{2i+1}}{2i+1}\;d\beta = \int\limits_{0}^{1}\sum\limits_{i=0}^{\infty}\frac{\beta^{2i}}{2i+1}\;d\beta$

Prove $\boxed{\log\left(\beta\right)\log\left(\frac{1-\beta}{1+\beta}\right)\bigg|_{0}^{1}=0\quad\quad\quad(\star\star)}$

For $\displaystyle k \in \mathbb{R}, \log(x)\log(1+kx) = \log(x)\cdot x \cdot \frac{1}{x}\cdot\log(1+kx) = \log(x)x \cdot \frac{\log(1+kx)}{x}$

We have

$\displaystyle\lim\limits_{x \rightarrow 0}\log(x)x = \lim\limits_{x \rightarrow 0} \frac{\log(x)}{\frac{1}{x}}=\lim\limits_{x \rightarrow 0} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim\limits_{x \rightarrow 0}-x = 0$

$\displaystyle \lim\limits_{x \rightarrow 0}\frac{\log(1+kx)}{x} = \lim\limits_{x \rightarrow 0}\frac{\frac{k}{1+kx}}{1} = k.$

As a result,

$\displaystyle \lim\limits_{x \rightarrow 0} \log(x)\log(1+kx) = \lim\limits_{x \rightarrow 0} \log(x)x \cdot \lim\limits_{x \rightarrow 0} \frac{\log(1+kx)}{x}=0\cdot k = 0.$ i.e.,

$\displaystyle \lim\limits_{x \rightarrow 0} \log(x)\log(1+kx) = 0, \quad k \in \mathbb{R}\quad\quad\quad(\star\star-1)$

Moreover,

$\displaystyle \lim\limits_{\beta \rightarrow 1}(\log(\beta)\cdot\log(1-\beta)-\log(\beta)\log(1+\beta))$

$\displaystyle= \lim\limits_{\beta \rightarrow 1}\log(\beta)\cdot\log(1-\beta)-\underbrace{\lim\limits_{\beta \rightarrow 1}\log(\beta)\cdot\log(1+\beta)}_{0}$

Let $\displaystyle x = 1-\beta \implies \beta=1-x$ and $\beta \rightarrow 1 \implies x \rightarrow 0$

$\displaystyle = \lim\limits_{x \rightarrow 0}\log(x)\log(1-x)$

$\displaystyle = \lim\limits_{x \rightarrow 0}\log(x)\log(1+(-1) x) \overset{(\star\star-1)}{=} 0$

gives

$\displaystyle \lim\limits_{\beta \rightarrow 1}(\log(\beta)\log(1-\beta)-\log(\beta)\log(1+\beta))=0.\quad\quad\quad(\star\star-2)$

It follows that

$\displaystyle \log(\beta)\log(\frac{1-\beta}{1+\beta})\bigg|_{0}^{1}$

$\displaystyle = \lim\limits_{\beta \rightarrow 1}\log(\beta)(\log(1-\beta) - \log(1+\beta)) - \lim\limits_{\beta \rightarrow 0}\log(\beta)(\log(1-\beta) - \log(1+\beta))$

$\displaystyle \overset{(\star\star-2)}{=} 0 -\lim\limits_{\beta \rightarrow 0}\log(\beta)\log(1+(-1)\cdot\beta) + \lim\limits_{\beta \rightarrow 0} \log(\beta)\log(1+1\cdot\beta)\overset{(\star\star-1)}{=} 0$

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