$g_n(y)=\sum_{k=1}^{n} \frac{\sin ky}{k},x \in [0,\pi]$ Evaluate $\lim_{n\to \infty} \int_0^{\pi} g_n (x)dy$

Question

$g_n(y)=\sum_{k=1}^{n} \frac{\sin ky}{k},x \in [0,\pi]$. I want to evaluate $\lim_{n\to \infty} \int_0^{\pi} g_n (x)dy$. So i try to find the pointwise limit function first, following this approach in this post Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$. I find it to be $\lim\limits_{n\to\infty}\ \sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2}=g(x) ,\, x\in(0,2\pi).$ Then I guess that (g_n) converge uniformly on $[0,\pi]$. I tried to prove it by Dirichlet test:
take $h_n(x)=\frac{1}{n}\quad \text{and} \quad u_n(x)=\sin(nx) , \text{then} |\sum_{k=1}^n\sin(kx)| \leq \frac{1}{sin(\frac{x}{2})} \leq \frac{1}{\sin(\frac{\delta}{2})}$.
Then I do whether it is right and do not know how to continue to prove. Suppose it is indeed uniformly convergent.And I want to show $\lim_{n\to \infty}\int_0^{\pi}g_n(x)dx=\int_0^{\pi} f(x)dx$.

Answer 1

You have taken a good first step in showing that $g_n(x) \to g(x)$ uniformly on the interval $[\delta,\pi]$ for any $\delta$ where $0 < \delta < \pi$.

However, as proved here, convergence is not uniform on $[0,\pi]$ and we can't use the standard theorem to conclude immediately that

$$\tag{*}\lim_{n \to \infty}\int_0^\pi g_n(x) dx = \int_0^\pi g(x) dx$$

Nevertheless, we can prove (*) holds using the additional fact that the sequence of partial sums is uniformly bounded -- that is, there exists $M > 0$ such that

$$|g_n(x)| = \left|\sum_{k=1}^n \frac{\sin kx}{k} \right|\leqslant M$$

for all $n \in \mathbb{N}$ and all $x \in [0,\pi]$. For a proof of this fact, which is not trivial to show, see here. Furthermore, because $g_n(x) \to g(x)$ pointwise, it follows that $|g(x)| \leqslant M$ for all $x \in [0,\pi]$.

We can write

$$\left|\int_0^\pi (g_n(x) - g(x)) \, dx \right| =\left|\int_0^\delta (g_n(x) - g(x)) \, dx + \int_\delta^\pi (g_n(x) - g(x)) \, dx \right| \\ \leqslant \int_0^\delta |g_n(x)| \, dx + \int_0^\delta |g(x)| \, dx + \left|\int_\delta^\pi (g_n(x) - g(x)) \, dx \right| $$

Using the bounds $|g_n(x)|, |g(x)| \leqslant M$ we find that the first and second integrals on the RHS are each bounded by $M \delta$. Choosing $\delta \leqslant \epsilon/(4M)$ we get

$$\left|\int_0^\pi (g_n(x) - g(x)) \, dx \right| \leqslant 2M\delta + \left|\int_\delta^\pi (g_n(x) - g(x)) \, dx \right| \leqslant \frac{\epsilon}{2} + \left|\int_\delta^\pi (g_n(x) - g(x)) \, dx \right| $$

Since $g_n \to g$ uniformly on $[\delta,\pi]$, it follows that $\int_\delta^\pi g_n(x) \to \int_\delta^\pi g(x) \, dx$ and given $\epsilon > 0$ there exists $N$ such that for all $n > N$ we have $\left|\int_\delta^\pi (g_n(x) - g(x)) \, dx \right| < \epsilon/2$.

Therefore, for all $n > N$ we have

$$\left|\int_0^\pi g_n(x)\, dx -\int_0^\pi g(x)\, dx\right|= \left|\int_0^\pi (g_n(x) - g(x)) \, dx \right| \leqslant \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon,$$

and (*) is true.

Answer 2

In p. 133 of 'Fourier series' by Edwards it is shown that if $a_n$ decreases to $0$ then $\sum_k a_k \sin kx$ converges boundedly iff $na_n$ is bounded and it converges uniformly iff $na_n \to 0$. In our case $a_n=\frac 1 n$ so the series is not uniformly convergent. However it converges boundedly in the sense the partial sums $g_n(x)$ are uniformly bounded by one fixed constant. By Bounded convergence Theorem we get $\lim \int g_n(x)dx=\int \lim g_n(x)dx$.