# Evaluating a summation of inverse squares over odd indices

$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$$

I want to evaluate this sum when $n$ takes only odd values.

Note that $$\sum_{n \text{ is even}} \dfrac1{n^2} = \sum_{k=1}^{\infty} \dfrac1{(2k)^2} = \dfrac14 \sum_{k=1}^{\infty} \dfrac1{k^2} = \dfrac{\zeta(2)}4$$ Also, $$\sum_{k=1}^{\infty} \dfrac1{k^2} = \sum_{k \text{ is odd}} \dfrac1{k^2} + \sum_{k \text{ is even}} \dfrac1{k^2}$$ Hence, $$\sum_{k \text{ is odd}} \dfrac1{k^2} = \dfrac34 \zeta(2)$$

This can be shown in a similar way to Euler's proof of $$\zeta(2) = \frac{\pi^2}{6}$$, which starts with the function $$\frac{sin(x)}{x}$$ (i.e. the sinc function). Here we start with the cosine function which can be expressed as the infinite product

$$cos(x) = \prod_{n=1}^\infty \left(1-\frac{4x^2}{\pi^2(2n-1)^2}\right)$$ $$= \left(1- \frac{4x^2}{\pi^2}\right)\left(1- \frac{4x^2}{9\pi^2}\right)\left(1- \frac{4x^2}{25\pi^2}\right) ...$$ $$=1-x^2\left(\frac{4}{\pi^2}+\frac{4}{9\pi^2}+\frac{4}{25\pi^2}+...\right)+...$$

$$cos(x)$$ can also be expressed by the following Maclaurin series expansion: $$cos(x) = \sum_{n=1}^\infty \frac{(-1)^n}{(2n)!}x^{2n} = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!} +...$$

Comparing the $$x^2$$ coefficients gives:

$$-\frac{1}{2!} = -\frac{4}{\pi^2}\left(1+\frac{1}{9} + \frac{1}{25} + ...\right)$$

Thus,

$$\sum_{n=1}^\infty \frac{1}{(2n-1)^2} = \frac{\pi^2}{8}$$

Here, we present a way forward that does not require prior knowledge of the value of the series $\sum_{n=1}\frac{1}{n^2}=\frac{\pi^2}{6}$, the Riemann-Zeta Function, or dilogarithm function. Rather, we apply straightforward analysis that includes application of the residue theorem.

To that end, note that we can write the series of interest as

\begin{align} \sum_{n=0}^\infty \frac{1}{(2n+1)^2}&=\sum_{n=1}^N \int_0^1 x^{2n}\,dx\int_0^1y^{2n}\,dy\\\\ &=\int_0^1\int_0^1 \sum_{n=0}^\infty(x^2y^2)^n\,dx\,dy\\\\ &=\int_0^1\int_0^1 \frac{1}{1-x^2y^2}\,dx\,dy\\\\ &=\frac12\int_0^1\frac{\log(1+x)-\log(1-x)}{x}\,dx\tag 1 \end{align}

Then, we enforce the substitution $x\to \frac{x-1}{x+1}$ in $(1)$ to obtain

$$\frac12\int_0^1\frac{\log(1+x)-\log(1-x)}{x}\,dx=\int_1^\infty \frac{\log(x)}{x^2-1}\,dx\tag 2$$

Next, enforcing the substitution $x\to 1/x$ in $(2)$ reveals

$$\int_1^\infty \frac{\log(x)}{x^2-1}\,dx=\int_0^1 \frac{\log(x)}{x^2-1}\,dx \tag 3$$

Adding $(2)$ and $(3)$ and dividing by $(2)$ yields

$$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx$$

Moving to the complex plane, we evaluate the integral $J$ defined by

$$J=\oint_C \frac{\log^2(z)}{z^2-1}\,dz$$

where $C$ is the classical keyhole contour with (i) the branch cut along the non-negative real axis and (ii) with deformations around $z=1$. Applying the residue theorem, it is easy to see that $J=i\pi^3$. Therefore, we find that \begin{align} J&=i\pi^3\\\\ &=\int_{0}^{\infty}\frac{\log^2(x)}{x^2-1}\,dx-\text{PV}\int_0^\infty \frac{\left(\log(x)+i2\pi\right)^2}{x^2-1}\,dx\\\\ &=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx\\\\ &+\color{blue}{(4\pi^2)\text{PV}\left(\int_0^\infty \frac{1}{x^2-1}\,dx\right)}\\\\ &+\color{red}{(4\pi^2)\lim_{\epsilon \to 0^+}\int_{\pi}^{2\pi} \frac{1}{(1+\epsilon e^{i\phi})^2-1}\,(i\epsilon e^{i\phi})\,d\phi}\\\\ &=-i4\pi\int_0^\infty \frac{\log(x)}{x^2-1}\,dx+\color{blue}{0}+\color{red}{i2\pi^3}\tag 4 \end{align}

Finally, solving $(4)$ for the integral of interest yields

$$\frac12\int_0^\infty \frac{\log(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$

and hence we find that the series of interest is

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\frac{\pi^2}{8}}$$

This is found using Fourier series in Brown and Churchill's Fourier Series and Boundary Value Problems by expanding $$f(x)=\begin{cases}x & 0<x<\pi\\ 0 & -\pi<x\leq 0 \end{cases}$$

into $$\frac{a_0}{2}+\sum_{n=1}^{\infty}\Big[a_n\cos(nx)+b_n\sin(nx)\Big],$$ a Fourier series that converges to $f$ when $-\pi<x<\pi$, its periodic extension over values of $x$ at which the periodic extension is continuous, and the mean value of one-sided limits of the periodic extension at discontinuities.

The coefficient $a_n$ is \begin{align} a_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx\\ &=\frac{1}{\pi}\int_{0}^{\pi}x\cos(nx)dx+\frac{1}{\pi}\int_{-\pi}^{0}0\cos(nx)dx. \end{align} When $n\neq 0$, integration by parts gives us \begin{align}a_n&=\frac{1}{\pi}\Big[\frac{1}{n}x\sin(nx)\Big|_0^{\pi}-\int_0^\pi\frac{1}{n}\sin(nx)dx\Big]\\ &=\frac{1}{\pi}\Big[0+\frac{1}{n^2}\cos(nx)\Big|_0^{\pi}\Big]\\ &=\frac{1}{\pi n^2}(\cos(n\pi)-\cos(0))\\ &=\frac{(-1)^n-1}{\pi n^2}. \end{align} When $n=0$, \begin{align} a_n&=\frac{1}{\pi}\int_{0}^{\pi}x\cos(0x)dx\\ &=\frac{1}{\pi}\int_{0}^{\pi}xdx\\ &=\frac{1}{\pi}\big(\frac{1}{2}x^2\Big|_{0}^{\pi}\big)\\ &=\frac{1}{2\pi}\pi^2\\ &=\frac{\pi}{2}. \end{align}

The coefficient $b_n$ is \begin{align} b_n&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\\ &=\frac{1}{\pi}\int_{0}^{\pi}x\sin(nx)dx+\frac{1}{\pi}\int_{-\pi}^{0}0\sin(nx)dx. \end{align} Again using integration by parts, this becomes \begin{align} b_n&=\frac{1}{\pi}\Big[\frac{-1}{n}x\cos(nx)\Big|_{0}^{\pi}-\int_0^{\pi}\frac{-1}{n}\cos(nx) \Big]\\ &=\frac{1}{\pi}\Big[\frac{-\pi}{n}\cos(\pi n)+\frac{1}{n^2}\sin(nx)\Big|_0^{\pi} \Big]\\ &=\frac{-1}{n}\cos(\pi n)\\ &=\frac{(-1)^{n+1}}{n}. \end{align}

So we have $$\frac{\pi}{4}+\sum_{n=1}^{\infty}\Big[ \frac{(-1)^n-1}{\pi n^2}\cos(nx)+\frac{(-1)^{n+1}}{n}\sin(nx)\Big]$$

as our Fourier series.

This converges to $f$ when $x=0$, so \begin{align}f(0)=0&=\frac{\pi}{4}+\sum_{n=1}^{\infty}\Big[ \frac{(-1)^n-1}{\pi n^2}\cos(n0)+\frac{(-1)^{n+1}}{n}\sin(n0)\Big]\\ \frac{-\pi}{4}&=\sum_{n=1}^{\infty}\Big[ \frac{(-1)^n-1}{\pi n^2}\cos(n0)+\frac{(-1)^{n+1}}{n}\sin(n0)\Big]\\ &=\sum_{n=1}^{\infty}\frac{(-1)^n-1}{\pi n^2}. \end{align}

When $n$ is even, $(-1)^n-1=1-1=0$, so the summand will be $0$. Discarding these $0$-summands gives us \begin{align} \frac{-\pi}{4}&=\sum_{\substack{n\in \Bbb{N};\\n\text{ odd}}}\frac{(-1)^n-1}{\pi n^2}\\ &=\sum_{\substack{n\in \Bbb{N};\\n\text{ odd}}}\frac{(-1)-1}{\pi n^2}\\ &=\sum_{\substack{n\in \Bbb{N};\\n\text{ odd}}}\frac{-2}{\pi n^2}\\ &=\frac{-2}{\pi}\sum_{\substack{n\in \Bbb{N};\\n\text{ odd}}}\frac{1}{n^2}\\ \frac{\pi^2}{8}&=\sum_{\substack{n\in \Bbb{N};\\n\text{ odd}}}\frac{1}{n^2}. \end{align}