Fourier serie property for $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$

Question

If $$f(x) \sim \frac{1}{2} a_0 + \sum_{n=1}^\infty (a_n\cos nx+b_n \sin nx)$$ For $x\in[0,2\pi)$, how can I show that: $$ \sum_{n=1}^\infty \frac{b_{n}}{n} = \frac{1}{2\pi}\int_{0}^{2\pi}(\pi -x) f(x)\,dx$$

Find the Fourier coefficients of $f(x)=\pi -x$, $x\in[0,2\pi)$ and deduce that (using the first part of this problem): $$\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$$

I do not know how to prove this, I tried computing the coefficients of the series but I am really mess with this proof. Please if you can, help me.

I tried this but I do not know if it is correct...

And for the other part I tried this:

Answer 1

As Mattos said, You need to calculate:

$\frac{1}{2\pi}\int_{0}^{2\pi}(\pi -x) f(x)\,dx$, by substituting into this integral,

$f(x) =\frac{1}{2} a_0 + \sum_{n=1}^\infty (a_n\cos nx+b_n \sin nx)$

Use integration by parts for the xcosnx and xsinnx terms.

You should get $\sum_{n=1}^\infty \frac{b_{n}}{n}$ as a result (everything else will cancel).

Then for the second part, after you get the fourier terms of $f(x)=\pi-x$, substitute $f(x)=\pi-x$ and the $b_n$ formula into

$\sum_{n=1}^\infty \frac{b_{n}}{n} = \frac{1}{2\pi}\int_{0}^{2\pi}(\pi -x) f(x)\,dx$

Evaluate the integral on the right hand side, and do a little simplification and you should get:

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

EDIT, here's the integral for the first part:

I'll calculate the integral in pieces. First the $a_0$ term:

$\frac{1}{2\pi}\int_{0}^{2\pi}(\pi -x)(\frac{1}{2}a_0)\,dx$

$=\frac{a_0}{4\pi}\int_{0}^{2\pi}(\pi -x)dx$

$=\frac{a_0}{4\pi}(\pi x -0.5x^2)$ (from 0 to 2$\pi$)

$=0$

Now the $a_n$ part:

$\frac{1}{2\pi}\int_{0}^{2\pi}(\pi -x)a_ncos(nx)\,dx$

$=\frac{a_n}{2\pi}\int_{0}^{2\pi}(\pi -x)cos(nx)\,dx$

$=\frac{a_n}{2\pi}\int_{0}^{2\pi}\pi cos(nx)\,dx-\frac{a_n}{2\pi}\int_{0}^{2\pi}x cos(nx)\,dx$

The first integral goes to 0. So we have:

$=-\frac{a_n}{2\pi}\int_{0}^{2\pi}x cos(nx)\,dx$

Use integration by parts

$u = x$

$dv = cos(nx)dx$

this gives $v = \frac{1}{n}sin(nx)$

So using $\int udv = uv - \int vdu$

We have:

$\int_{0}^{2\pi}x cos(nx)\,dx = \frac{x}{n}sin(nx)$ (from 0 to $2\pi$) - $\int_{0}^{2\pi}\frac{1}{n}sin(nx)\,dx$

this comes out to 0.

Now the $b_n$ part:

$\frac{1}{2\pi}\int_{0}^{2\pi}(\pi -x)b_nsin(nx)\,dx$

$=\frac{b_n}{2\pi}\int_{0}^{2\pi}(\pi -x)sin(nx)\,dx$

$=\frac{b_n}{2\pi}\int_{0}^{2\pi}\pi sin(nx)\,dx-\frac{b_n}{2\pi}\int_{0}^{2\pi}x sin(nx)\,dx$

The first integral goes to 0. So we have:

$=-\frac{b_n}{2\pi}\int_{0}^{2\pi}x sin(nx)\,dx$

Use integration by parts

$u = x$

$dv = sin(nx)dx$

this gives $v = -\frac{1}{n}cos(nx)$

So using $\int udv = uv - \int vdu$

We have:

$\int_{0}^{2\pi}x sin(nx)\,dx = -\frac{x}{n}cos(nx)$ (from 0 to $2\pi$) + $\int_{0}^{2\pi}\frac{1}{n}cos(nx)\,dx$

this comes out to $-\frac{2\pi}{n}$.

So:

$-\frac{b_n}{2\pi}\int_{0}^{2\pi}x sin(nx)\,dx = (-\frac{b_n}{2\pi})(-\frac{2\pi}{n}) = \frac{b_n}{n}$

Now just add the summation sign.

EDIT For the second part, calculate

$\frac{1}{2\pi}\int_{0}^{2\pi}(\pi -x)^2\,dx$

$\frac{1}{2\pi}(\frac{x^3}{3}-\pi x^2+\pi^2x)$ evaluated between $2\pi$ and 0.

This comes out to:

$\frac{1}{2\pi} (\frac{8\pi^3}{3}-4\pi^3+2\pi^3)$

$=\frac{\pi^2}{3}$

$f(x) = \pi-x$, calculate fourier components

$a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}(\pi-x)dx$

$a_0 = 2\pi$

$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}(\pi-x)cos(nx)dx$

$a_n = 0$

$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}(\pi-x)sin(nx)dx$

$b_n = -\frac{sin(nx)+n(\pi-x)\cos(nx)}{\pi n^2}$ (evaluation at $\pi$ minus evaluation at -$\pi$)

$b_n = \frac{2}{n}$

So finally, substituting into:

$\sum_{n=1}^\infty \frac{b_{n}}{n} = \frac{1}{2\pi}\int_{0}^{2\pi}(\pi -x) f(x)\,dx$

we have

$\sum_{n=1}^\infty \frac{2}{n^2} = \frac{\pi^2}{3}$

divide both sides by 2

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