Pointwise convergence of Fourier series of $x \mapsto x \sin(\pi x), x\in[-1,1]$

Question

Consider $x \mapsto x \sin(\pi x), x\in[-1,1]$.

The task is to comput the Fourier series of this function and determine the pointwise limit of the Fourier series

$$c_0+\sum_{k=1}^{\infty}\Big(a_k\cos(k\pi x)+b_k \sin(k \pi x)\Big)$$

My solution is $c_0=\frac{1}{\pi}, a_1=\frac{-1}{2\pi}$ and $a_k=\frac{2(-1)^{k+1}}{(k^2-1)\pi}$ for $k>1$ so the Fourier series is:

$$\frac{1}{\pi}-\frac{\cos(\pi x)}{2\pi}+\sum_{k=2}^{\infty}\frac{2\cos(k\pi x)}{\pi(k^2-1)}(-1)^{k+1}$$

How can I now determine the pointwise limit of this series?

Answer 1

We need to check the $3$ Dirichlet conditions for this function.

1. $$\int\limits_{-1}^1 \left| f(x)\right | dx < \infty $$ since $f$ is continuous. So $f$ is absolutely integrable.

2. $$\int_{-1}^1 |f'(x)|dx < \infty $$ since $f'$ is continuous, hence $f$ is of bounded variation.

3. Of course, $f$ is continuous everywhere on $[-1, 1]$.

Hence the Dirichlet conditions hold, and we can say that the Fourier series of $f$ converge to $f$ at every point of continuity of $f$, hence at every point where it's defined, since $f$ is continuous.

Answer 2

If i) $f$ is continuous on $[-1,1]$ and ii) the Fourier series of $f$ converges uniformly on that interval, then the Fourier series of $f$ converges uniformly (hence pointwise) on $[-1,1]$ to $f.$ This is an elementary result.

In our problem we have $f(x) = x\sin (\pi x),$ so i) is satisfied, and ii) the Fourier series of $f$ converges uniformly by the Weierstrass M test. Thus the given series converges uniformly to $x\sin (\pi x)$ on $[-1,1].$