Fourier Series on the Unit Interval

[Euge]

Evaluate the Fourier series

$$\frac{1}{\pi^2}\sum_{k = 1}^\infty \frac{\cos 2\pi kx}{k^2}$$ for $0 \le x \le 1$.

 

[anuttarasammyak]

Spoiler

The formula equals

$$-4 Re \int^x ds \int^s dt \sum_{k=1}^\infty e^{2\pi kti}$$

$$=-4 Re \int^x ds \int^s dt \frac{e^{2\pi ti}}{1-e^{2\pi ti}}$$

$$=2Re \int^x ds \int^s dt (1-i \cot \pi t )$$

$$=x^2+Bx+C$$

$$=x(x-1)+\frac{1}{6}$$

$$=x^2-x+\frac{1}{6}$$

because for x=0,1 it is $\zeta(2)/\pi^2=1/6$. Minimum for x=1/2 is $-\frac{1}{12}$ which coincides with -$\eta(2)/\pi^2$ using eta function or alternate zeta function. Ref. https://mathworld.wolfram.com/DirichletEtaFunction.html

 

[julian]

Spoiler

We will evaluate the sum:

\begin{align*}
\frac{1}{\pi^2} \sum_{k=1}^\infty (-1)^k \frac{\cos 2 \pi k x}{k^2}
\end{align*}

for $-\frac{1}{2} \leq x \leq \frac{1}{2}$ first and then make the shift $x \mapsto x - \frac{1}{2}$.

I will express the sum via a complex contour integration. It employs the fact that the function

\begin{align*}
\dfrac{\cos 2 \pi z x}{\sin \pi z}
\end{align*}

has simple poles at all integer values except when $\cos 2 \pi z x = 0$, as we now verify. First consider the case where $\cos 2 \pi z x \not= 0$ for $z=n$,

\begin{align*}
\dfrac{\cos 2 \pi z x}{\sin \pi z} & = \dfrac{\cos 2 \pi n x}{\sin \pi [n + (z-n)]}
\nonumber \\
& = \dfrac{\cos 2 \pi n x}{(-1)^n [\pi (z-n) - \frac{1}{3!} \pi^3 (z-n)^3 + \cdots]}
\nonumber \\
& = (-1)^n \dfrac{\cos 2 \pi n x}{(z-n) \pi [1 - \frac{1}{3!} \pi^2 (z-n)^2 + \cdots]}
\nonumber \\
& = (-1)^n \dfrac{\cos 2 \pi n x}{(z-n) \pi} + \cdots
\end{align*}

Now consider the case where $\cos 2 \pi z x = 0$ for some $z=n$,

\begin{align*}
\left| \lim_{z \rightarrow n} \dfrac{\cos 2 \pi z x}{\sin \pi z} \right| & = \left| \lim_{z \rightarrow n} \dfrac{\frac{d}{dz} \cos 2 \pi z x}{\frac{d}{dz} \sin \pi z} \right|
\nonumber \\
& = \left| \lim_{z \rightarrow n} \dfrac{- 2 \pi x \sin 2 \pi z x}{\pi \cos \pi z} \right| < \infty
\end{align*}

So when $\cos 2 \pi z x$ vanishes for $z = n$, there is no pole at $n$. This allows us to write

\begin{align*}
\frac{1}{\pi^2} \sum_{k=1}^\infty (-1)^k \frac{\cos 2 \pi k x}{k^2} = \frac{1}{2 \pi^2 i} \oint_C \dfrac{\cos 2 \pi z x}{z^2 \sin \pi z} dz
\end{align*}

where the contour $C$ is defined in fig (a).

We have

\begin{align*}
2 \frac{1}{\pi^2} \sum_{k=1}^\infty (-1)^k \frac{\cos 2 \pi k x}{k^2} & = \frac{1}{\pi^2} \sum_{k=1}^\infty (-1)^k \frac{\cos 2 \pi k x}{k^2} + \frac{1}{\pi^2} \sum_{k=-1}^{-\infty} (-1)^k\frac{\cos 2 \pi k x}{k^2}
\nonumber \\
& = \frac{1}{2 \pi^2 i} \oint_{C+C'} \dfrac{\cos 2 \pi z x}{z^2 \sin \pi z} dz
\end{align*}

where the contour $C'$ is defined in fig (a). We complete the path of integration along semicircles at infinity (see fig (b)) since the integrand vanishes there. Since the resulting enclosed area contains no singularities except at $z=0$, we can shrink this contour down to an infinitesimal circle $C_0$ around the origin (see fig (c)). So that

\begin{align*}
\frac{1}{\pi^2} \sum_{k=1}^\infty (-1)^k \frac{\cos 2 \pi k x}{k^2} = \frac{1}{4 \pi^2 i} \oint_{C_0} \dfrac{\cos 2 \pi z x}{z^2 \sin \pi z} dz
\end{align*}

We expand the integrand in powers of $z$ about $z=0$ and isolate the $z^{-1}$ term. We get

\begin{align*}
\dfrac{\cos 2 \pi z x}{z^2 \sin \pi z} & = \dfrac{1 - \frac{1}{2!} 2^2 \pi^2 x^2 z^2 + \cdots}{z^2 [\pi z - \frac{1}{3!} \pi^3 z^3 + \cdots]}
\nonumber \\
& = \dfrac{1 - 2 \pi^2 x^2 z^2 + \cdots}{z^3 \pi [1 - \frac{1}{6} \pi^2 z^2 + \cdots]}
\nonumber \\
& = \dfrac{1 - 2 \pi^2 x^2 z^2 + \cdots}{z^3 \pi} (1 + \frac{1}{6} \pi^2 z^2 + \cdots)
\nonumber \\
& = \cdots + \left( \dfrac{1}{6} - 2 x^2 \right) \pi \frac{1}{z} + \cdots
\end{align*}

So that for $- \frac{1}{2} \leq x \leq \frac{1}{2}$,

\begin{align*}
\frac{1}{\pi^2} \sum_{k=1}^\infty (-1)^k \frac{\cos 2 \pi k x}{k^2} & = \frac{1}{4 \pi^2 i} \oint_{C_0} \dfrac{\cos 2 \pi z x}{z^2 \sin \pi z} dz
\nonumber \\
& = \frac{1}{4 \pi^2 i} (-2 \pi i) \left( \dfrac{1}{6} - 2 x^2 \right) \pi
\nonumber \\
& = x^2 - \frac{1}{12}
\end{align*}

Finally, we do the shift $x \mapsto x - \frac{1}{2}$ and obtain:

\begin{align*}
\frac{1}{\pi^2} \sum_{k=1}^\infty \frac{\cos 2 \pi k x}{k^2} = \left( x - \frac{1}{2} \right)^2 - \frac{1}{12} = x^2 - x + \frac{1}{6} .
\end{align*}

for $0 \leq x \leq 1$.

 

[I like Serena]

Spoiler

The Fourier cosine series for an even periodic function $f(x)$ with period $2L$ is
\begin{array}{cl}f(x)&=\frac{c_0}2 + \sum_{k=1}^\infty c_k \cos\frac{k\pi x}{L}\\
c_k &= \frac 2L\int_0^L f(x) \cos\frac{k\pi x}{L}\,dx\end{array}
Pick $L=\frac 12$ and let $f(x)$ be the desired function. It follows that for $k>0$: $$\pi^2 c_k=4\pi^2\int_0^{1/2} f(x) \cos(2k\pi x)\,dx=\frac 1{k^2}$$
We have for $k>0$:
$$\begin{cases}4\pi^2\int_0^{1/2} x \cos(2k\pi x)\,dx = \frac{(-1)^k-1}{k^2}\\
4\pi^2\int_0^{1/2} x^2 \cos(2k\pi x)\,dx = \frac{(-1)^k}{k^2}\end{cases}\implies
4\pi^2\int_0^{1/2} (x^2-x) \cos(2k\pi x)\,dx = \frac{1}{k^2}
$$
Furthermore, $f(0)=\frac 1{\pi^2}\sum_{k=1}^\infty \frac 1{k^2}=\frac 16$.
So the function is $f(x)=x^2-x+\frac 16$ on the interval [0,1], which is indeed an even periodic function if we extend it periodically.