MediaWiki:Sitenotice

Problem

Compute $$\sum_{j=1}^\infty \frac{1}{j^4}$$ using Parseval identity.

Hint: consider \(f(x) = \frac{1}{2} x^2\)

Solution Sketch

1. Calculate the Fourier coefficients \(\hat f(x)\)
2. Apply Parseval identity
3.  ???
4. Profit

Solution

Recall the definition of the Fourier coefficients for \(2\pi\)-periodic functions in \(C^\infty(\mathbb{R})\): $$c_k = \frac{1}{2\pi} \int_0^{2\pi} f(x) \: e^{-ikx} \: dx$$

We consider the continuous continuation of \(f(x)\) on \([-\pi, \pi]\) and calculate the Fourier coefficients:

For \(k=0\): $$\hat f(0) = \frac{1}{2\pi}\int_{-\pi}^\pi \frac{x^2}{2} dx = \frac{\pi^2}{6}$$

and for \(k \neq 0\) (twice we integrate by parts): $$\begin{align} \hat f(k) &= \frac{1}{2\pi} \int_{-\pi}^\pi \frac{x^2}{2} e^{-ikx}\: dx \\ &= \frac{1}{2\pi} \int_0^\pi x^2 \cos(kx)\: dx \\ &= - \frac{1}{k\pi} \int_0^\pi x \sin(kx)\: dx \\ &= \left[ \frac{1}{k^2\pi}x \cos(kx) \right]_0^\pi - \frac{1}{k^2\pi} \int_0^\pi \cos(kx)\: dx \\ &= \frac{\cos(k\pi)}{k^2} \\ &= \frac{(-1)^k}{k^2} \end{align}$$

Now, remember Parseval's identity: $$\frac{1}{2\pi} \int_{-\pi}^\pi \left| f(x) \right|^2 \: dx = \sum_{k=-\infty}^\infty \left| \hat f(k) \right|^2 \tag{*}$$

Applied to our \(f(x)\): $$\frac{\pi^4}{20} = \frac{1}{2\pi} \int_{-\pi}^\pi \left| \frac{1}{2} x^2 \right|^2 \: dx \overset{(*)}{=} \sum_{k=-\infty}^\infty \left| \hat f(k) \right|^2 = \frac{\pi^4}{36} + 2 \sum_{k=1}^\infty \frac{1}{k^4}$$ And thus, it follows that $$\sum_{k=1}^\infty \frac{1}{k^4} = \frac{\pi^4 (\frac{1}{20} - \frac{1}{36})}{2} = \frac{\pi^4}{90}$$