The Basel problem with trigonometric Fourier series [math-0016]

2025-11-07 · Lîm Tsú-thuàn

An idea is define f(x)=x2f(x) = x^2 on [−π,π][-\pi, \pi]. Its trigonometric Fourier series was:

a02+∑n=1∞(ancos⁡(nx)+bnsin⁡(nx))\frac{a_0}{2} + \sum_{n=1}^{\infty}(a_n \cos(nx)+b_n \sin(nx))

which is periodic and converges to f(x)f(x) in [−π,π][-\pi,\pi].

Observing that f(x)f(x) is even, hence

bn=1π∫−ππf(x)sin⁡(nx) dx=0b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)\sin(nx)\,dx = 0

for all n=1,2,3,…n = 1,2,3,\dots. Now compute a0a_0

a0=1π∫−ππx2 dx=2π∫0πx2 dx=2π[x33]0π=2ππ33=2π23\begin{aligned} a_0 &= \frac{1}{\pi} \int_{-\pi}^{\pi} x^2\,dx = \frac{2}{\pi}\int_{0}^{\pi} x^2\,dx \\ &= \frac{2}{\pi} \left[\frac{x^3}{3}\right]_0^\pi \\ &= \frac{2}{\pi} \frac{\pi^3}{3} = \frac{2\pi^2}{3} \end{aligned}

and each ana_n is

an=1π∫−ππx2cos⁡(nx) dx=2π∫0πx2cos⁡(nx) dx\begin{aligned} a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi} x^2\cos(nx)\,dx \\ &= \frac{2}{\pi} \textcolor{red}{\int_{0}^{\pi} x^2\cos(nx)\,dx} \end{aligned}

Now let's focus on integral by part

∫x2cos⁡(nx) dx=x2(∫cos⁡(nx) dx)−∫2x(∫cos⁡(nx) dx) dx=x2sin⁡(nx)n−2∫xsin⁡(nx)n dx\begin{aligned} \textcolor{red}{\int x^2\cos(nx)\,dx} &= x^2(\int \cos(nx)\,dx) - \int 2x (\int \cos(nx)\,dx)\,dx \\ &= x^2 \frac{\sin(nx)}{n} - 2 \textcolor{blue}{\int x \frac{\sin(nx)}{n}\,dx} \end{aligned}

and again

∫xsin⁡(nx)n dx=x∫sin⁡(nx)n dx−∫1(∫sin⁡(nx)n dx) dx=x−cos⁡(nx)n2−∫−cos⁡(nx)n2 dx=x−cos⁡(nx)n2−−sin⁡(nx)n3\begin{aligned} \textcolor{blue}{\int x \frac{\sin(nx)}{n}\,dx} &= x \int\frac{\sin(nx)}{n}\,dx - \int 1 (\int\frac{\sin(nx)}{n}\,dx)\,dx \\ &= x \frac{-\cos(nx)}{n^2} - \int \frac{-\cos(nx)}{n^2}\,dx \\ &= x \frac{-\cos(nx)}{n^2} - \frac{-\sin(nx)}{n^3} \end{aligned}

It seems complicated, but in fact we have sin⁡(nπ)=0\sin(n\pi) = 0, hence we can ignore them in this definite integral

an=2π[2xcos⁡(nx)n2]0π=2π2πcos⁡(nπ)n2=4cos⁡(nπ)n2=4(−1)nn2=(−1)n4n2\begin{aligned} a_n &= \frac{2}{\pi} \left[ \frac{2 x \cos(nx)}{n^2} \right]_0^\pi \\ &= \frac{2}{\pi} \frac{2 \pi \cos(n \pi)}{n^2} \\ &= \frac{4 \cos(n \pi)}{n^2} \\ &= \frac{4 (-1)^n}{n^2} = (-1)^n \frac{4}{n^2} \end{aligned}

Therefore, if we compute f(π)f(\pi) can get

f(π)=π2=π23+∑n=1∞((−1)n4n2cos⁡(nπ))=π23+∑n=1∞((−1)n(−1)n4n2)=π23+∑n=1∞((−1)2n4n2)=π23+∑n=1∞(4n2)=π23+4∑n=1∞(1n2)\begin{aligned} f(\pi) &= \pi^2 \\ &= \frac{\pi^2}{3} + \sum_{n=1}^\infty ((-1)^n \frac{4}{n^2} \cos(n\pi)) \\ &= \frac{\pi^2}{3} + \sum_{n=1}^\infty ((-1)^n (-1)^n \frac{4}{n^2}) \\ &= \frac{\pi^2}{3} + \sum_{n=1}^\infty ((-1)^{2n} \frac{4}{n^2}) \\ &= \frac{\pi^2}{3} + \sum_{n=1}^\infty (\frac{4}{n^2}) \\ &= \frac{\pi^2}{3} + 4 \sum_{n=1}^\infty (\frac{1}{n^2}) \end{aligned}

Hence

2π23=4∑n=1∞(1n2)π26=∑n=1∞(1n2)\frac{2\pi^2}{3} = 4 \sum_{n=1}^\infty (\frac{1}{n^2}) \\ \frac{\pi^2}{6} = \sum_{n=1}^\infty (\frac{1}{n^2})