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

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?

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.

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 $$[-1,1].$$