Fall 2001 Problem 2

Exercise 12.3 (Basel problem with Fourier analysis)  

1. Find the Fourier coefficients $\widehat{f}(k)$ for the function $f(x)=x$ with respect to the exponential system $e^{2\pi i k x}$ ( $k\in\mathbb{Z}$) on $[-\frac{1}{2}, \frac{1}{2}]$.
2. Use the result of part (a) to compute

   $$\sum_{k=1}^\infty \frac{1}{k^2}.$$ (12.16)

Proof. Introduce the inner product on $L^1\cap L^2$

$$\langle f, g \rangle = \int_{-\frac{1}{2}}{\frac{1}{2}} f(x)\overline{g(x)}dx.$$ (12.17)

Then the functions $\{e^{2\pi i k x}\}$ form an orthonormal system and

$$f(x) = \sum_{k=-\infty}^\infty \langle f, e^{2\pi i k x} \rangle e^{2\pi i k x} L^2$$ (12.18)

Let us compute these inner products for the given $f(x)=x$.

$$\langle x, e^{2\pi i k x} \rangle = \int_{-\frac{1}{2}}^{\frac{1}{2}} x e^{-2\pi i k x}dx$$ (12.19)

$$= \int x\left(e^{-2\pi i k x}/(-2\pi i k)\right)'dx$$ (12.20)

$$= \left. \frac{xe^{-2\pi i k x}}{-2\pi i k} \right\vert _{-\frac{1}{2}}^{\frac{1}{2}} - \int (e^{-2\pi i k x}/(-2\pi i k))dx$$ (12.21)

$$= \frac{1}{-2\pi i k}\left[\frac{e^{-\pi i k}}{2} + \frac{e^{\pi i k}}{2}\right] - \int (e^{-2\pi i k x}/(-2\pi i k))dx$$ (12.22)

For the part:

$$\int (e^{-2\pi i k x}/(-2\pi i k))dx = \int (e^{-2\pi i k x}/(-2\pi i k))'/(-2\pi i k)dx$$ (12.23)

$$= -\left.\frac{e^{-2\pi i k x}}{4\pi^2k^2} \right\vert _{-\frac{1}{2}}^{\frac{1}{2}}$$ (12.24)

$$= -\frac{1}{4\pi^2k^2}\left[e^{-\pi i k}-e^{\pi i k}\right]$$ (12.25)

Combining these shows

$$\langle x, e^{2\pi i k x} \rangle = \frac{1}{-2\pi i k}\left[\frac{e^{-\pi i k}}{2} + \frac{e^{\pi i k}}{2}\right] + \frac{1}{4\pi^2k^2}\left[e^{-\pi i k}-e^{\pi i k}\right]$$ (12.26)

$$= \frac{i\cos(\pi k)}{2\pi k} - \frac{i\sin(\pi k)}{2\pi^2k^2}$$ (12.27)

If $k=0$,

$$\langle x, 1\rangle = \int_{-\frac{1}{2}}^{\frac{1}{2}} x dx =0$$ (12.28)

so

$$x = \sum_{k=-\infty}^\infty \widehat{f}(k)e^{2\pi i kx}$$ (12.29)

$$= \sum_{k\neq 0} \left[\frac{i\cos(\pi k)}{2\pi k} - \frac{i\sin(\pi k)}{2\pi^2k^2}\right] e^{2\pi i kx}$$ (12.30)

$$= \sum_{k\neq 0} \frac{i(-1)^k}{2\pi k}e^{2\pi i kx}$$ (12.31)

Apply the Parseval identity for our $f(x)=x$

$$\Vert f\Vert _2 = \sum_{k=-\infty}^\infty \vert\widehat{f}(k)\vert^2$$ (12.32)

$$= \sum_{k\neq 0} \frac{1}{4\pi^2k^2}$$ (12.33)

$$= 2\sum_{k=1}^\infty \frac{1}{4\pi^2 k^2}$$ (12.34)

Then

$$\Vert f\Vert _2 = \int x^2 dx = \left. \frac{x^3}{3} \right\vert _{-1/2}^{1/2} = \frac{1/8}{3} + \frac{1/8}{3} = \frac{1}{12}$$ (12.35)

so that

$$\frac{1}{12} = \frac{1}{2\pi^2}\sum_{k=1}^\infty \frac{1}{k^2}$$ (12.36)

and finally

$$\frac{\pi^2}{6} = \sum_{k=1}^\infty \frac{1}{k^2}.$$ (12.37)

$\qedsymbol$